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HN
16 tháng 12 2018
\(a,\frac{2x+4}{10}+\frac{2-x}{15}=\frac{\left(2x+4\right).3}{10.3}+\frac{\left(2-x\right).2}{15.2}\)
\(=\frac{6x+12}{30}+\frac{4-2x}{30}=\frac{6x+12+4-2x}{30}=\frac{4x+16}{30}\)
\(=\frac{4.\left(x+4\right)}{30}=\frac{2\left(x+4\right)}{15}\)
\(b,\frac{3x}{10}+\frac{2x-1}{15}+\frac{2-x}{20}=\frac{3x.6}{10.6}+\frac{\left(2x-1\right).4}{15.4}+\frac{\left(2-x\right).3}{20.3}\)
\(=\frac{18x}{60}+\frac{8x-4}{60}+\frac{6-3x}{60}=\frac{18x+8x-4+6-3x}{60}=\frac{23x+2}{60}\)
\(c,\frac{x+1}{2x-2}+\frac{x^2+3}{2-2x^2}=\frac{x+1}{2\left(x-1\right)}+\frac{x^2+3}{2\left(1-x^2\right)}=\frac{x+1}{2\left(x-1\right)}+\frac{-x^2-3}{2\left(x^2-1\right)}\)
\(=\frac{x+1}{2\left(x-1\right)}+\frac{-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)\(=\frac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\frac{-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+2x+1-x^2-3}{2\left(x-1\right)\left(x+1\right)}=\frac{2x-2}{2\left(x-1\right)\left(x+1\right)}=\frac{2\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\)\(=\frac{1}{x+1}\)
DN
21 tháng 8 2018
\(A=\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3-2\right)\)
\(\Rightarrow A=\left(x^3+8\right)-\left(x^3-2\right)\)
\(\Rightarrow A=x^3+8-x^3+2\)
\(\Rightarrow A=\left(x^3-x^3\right)+\left(8+2\right)\)
\(\Rightarrow A=10\)
21 tháng 8 2018
\(A=\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3-2\right)\)
\(=x^3+8-x^3+2\)
\(=10\)
\(B=\left(x+2\right)\left(x-2\right)\left(x^2+2x+4\right)\left(x^2-2x+4\right)\)
\(=\left(x+2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x^2+2x+4\right)\)
\(=\left(x^3+8\right)\left(x^3-8\right)\)
\(=x^6-64\)
\(C=\left(x^2+3x+1\right)^2+\left(3x-1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)\)
\(=\left(x^2+3x+1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x+1-3x+1\right)^2\)
\(=\left(x^2+2\right)^2\)
\(D=\left(3x^3+3x+1\right)\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1+3x\right)\left(3x^3+1-3x\right)-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1\right)^2-9x^2-\left(3x^3+1\right)^2\)
\(=-9x^2\)
\(E=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)-\left(2x^2+1\right)^2\)
\(=\left(2x^2+1+2x\right)\left(2x^2+1-2x\right)-\left(2x^2+1\right)^2\)
\(=\left(2x^2+1\right)^2-4x^2-\left(2x^2+1\right)^2\)
\(=-4x^2\)
QD
20 tháng 8 2017
a)\(\dfrac{2x^2-10xy}{2xy}+\dfrac{5y-x}{y}+\dfrac{x+2y}{x}\)
\(=\dfrac{2x\left(x-5y\right)}{2xy}+\dfrac{5y-x}{y}+\dfrac{x+2y}{x}\)
\(=\dfrac{x-5y}{y}+\dfrac{5y-x}{y}+\dfrac{x+2y}{x}\)
\(=\dfrac{x\left(x-5y\right)+x\left(5y-x\right)+y\left(x+2y\right)}{xy}\)
\(=\dfrac{x^2-5xy+5xy-x^2+xy+2y^2}{xy}\)
\(=\dfrac{y\left(x+2y\right)}{xy}\)
NN
24 tháng 11 2017
b) \(\dfrac{x+1}{2x-2}+\dfrac{x^2+3}{2-2x^2}\)
\(=\dfrac{x+1}{2x-2}-\dfrac{x^2+3}{2x^2-2}\)
\(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{x^2+3}{2\left(x^2-1\right)}\)
\(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{x^2+3}{2\left(x-1\right)\left(x+1\right)}\) MTC: \(2\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x^2+3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)-\left(x^2+3\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)
e) \(\dfrac{2x^2-xy}{x-y}+\dfrac{xy+y^2}{y-x}+\dfrac{2y^2-x^2}{x-y}\)
\(=\dfrac{2x^2-xy}{x-y}-\dfrac{xy+y^2}{x-y}+\dfrac{2y^2-x^2}{x-y}\)
\(=\dfrac{\left(2x^2-xy\right)-\left(xy+y^2\right)+\left(2y^2-x^2\right)}{x-y}\)
\(=\dfrac{2x^2-xy-xy-y^2+2y^2-x^2}{x-y}\)
\(=\dfrac{x^2-2xy+y^2}{x-y}\)
\(=\dfrac{\left(x-y\right)^2}{x-y}\)
\(=x-y\)
25 tháng 2 2020
1) \(\frac{x+1}{2x-2}+\frac{x^2+3}{2-2x^2}\)
\(=\frac{-4x^2+8x-4}{-4x^3+4x^2+4x-4}\)
\(=\frac{-x^2+2x-1}{-x^3+x^2+x-1}\)
\(=\frac{\left(-x+1\right)\left(x-1\right)}{\left(-x-1\right)\left(x-1\right)\left(x-1\right)}\)
\(=\frac{1}{x+1}\)
2) \(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)
\(=\frac{-16x^3+16x^2-4x}{-16x^4+16x^3-4x^2}\)
\(=\frac{-16x^2+16x-4}{-16x^3+16x^2-4x}\)
\(=\frac{-4x^2+4x-1}{-4x^3+4x^2-x}\)
\(=\frac{\left(-2x+1\right)\left(2x-1\right)}{x\left(-2x+1\right)\left(2x-1\right)}\)
\(=\frac{1}{x}\)
13 tháng 11 2017
Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
AH
Akai Haruma
Giáo viên
4 tháng 7 2018
1)
ĐK: \(x,y\neq 0\); \(x+y\neq 0\)
\(\frac{x^2-y^2}{6x^2y^2}: \frac{x+y}{12xy}\)
\(=\frac{x^2-y^2}{6x^2y^2}. \frac{12xy}{x+y}=\frac{(x-y)(x+y).12xy}{6x^2y^2(x+y)}=\frac{2(x-y)}{xy}\)
2) ĐK: \(x\neq \frac{\pm 1}{2}; 0; 1\)
\(\frac{5x}{2x+1}: \frac{3x(x-1)}{4x^2-1}=\frac{5x}{2x+1}.\frac{4x^2-1}{3x(x-1)}\)
\(=\frac{5x(2x-1)(2x+1)}{(2x+1).3x(x-1)}=\frac{5(2x-1)}{3(x-1)}\)
AH
Akai Haruma
Giáo viên
4 tháng 7 2018
3) ĐK: \(x\neq \frac{\pm 1}{2}; 0\)
\(\left(\frac{2x-1}{2x+1}-\frac{2x-1}{2x+1}\right): \frac{4x}{10x-5}=0: \frac{4x}{10x-5}=0\)
4) ĐK: \(x\neq \frac{\pm 1}{3}\)
\(\frac{2}{9x^2+6x+1}-\frac{3x}{9x^2-1}=\frac{2}{(3x+1)^2}-\frac{3x}{(3x-1)(3x+1)}\)
\(=\frac{2(3x-1)}{(3x+1)^2(3x-1)}-\frac{3x(3x+1)}{(3x-1)(3x+1)^2}\)
\(=\frac{6x-2-9x^2-3x}{(3x+1)^2(3x-1)}=\frac{-9x^2+3x-2}{(3x-1)(3x+1)^2}\)
5) ĐK: \(x\neq \pm 1; \frac{-7\pm \sqrt{89}}{4}\)
\(\left(\frac{5}{x^2+2x+1}+\frac{2x}{x^2-1}\right): \frac{2x^2+7x-5}{3x-3}\)
\(=\left(\frac{5}{(x+1)^2}+\frac{2x}{(x-1)(x+1)}\right). \frac{3(x-1)}{2x^2+7x-5}\)
\(=\frac{5(x-1)+2x(x+1)}{(x-1)(x+1)^2}. \frac{3(x-1)}{2x^2+7x-5}=\frac{2x^2+7x-5}{(x+1)^2(x-1)}.\frac{3(x-1)}{2x^2+7x-5}\)
\(=\frac{3}{(x+1)^2}\)
8 tháng 8 2018
\(x^3+2x^2+2x+1=\left(x^3+1\right)+\left(2x^2+2x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+2x\right)=\left(x+1\right)\left(x^2+x+1\right)\)
\(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
\(x^4+2x^3+2x^2+2x+1=x^4+x^2+2x^3+x^2+2x+1\)
\(=x^2\left(x^2+1\right)+2x\left(x^2+1\right)+\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+2x+1\right)\)
\(=\left(x^2+1\right)\left(x+1\right)^2\)
\(x^4-2x^3+2x-1=\left(x^4-1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+1-2x\right)=\left(x^2-1\right)\left(x-1\right)^2\)
8 tháng 8 2018
\(x^3+2x^2+2x+1=\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\)
\(=x^2.\left(x+1\right)+x.\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right).\left(x^2+x+1\right)\)
\(x^3-4x^2+12x-27\)
\(=\left(x^3-x^2\right)-\left(3x^2-3x\right)+\left(9x-27\right)\)
\(=x^2.\left(x-1\right)-3x.\left(x-1\right)+9.\left(x-3\right)\)
\(=\left(x-1\right).\left(x^2-3x\right)+9.\left(x-3\right)\)
\(=x.\left(x-1\right).\left(x-3\right)+9.\left(x-3\right)\)
\(=\left(x-3\right)\left[x.\left(x-1\right)+9\right]\)
CN
27 tháng 12 2017
a) (x - 2)2 - (x - 3)(x + 3) = 17
⇔ (x2 - 4x + 4) - (x2 - 9) = 17
⇔ x2 - 4x + 4 - x2 + 9 = 17
⇔ 13 - 4x = 17
⇔ - 4x = -4
⇔ x = 1
b) 4(x - 3)2 - (2x - 1)(2x + 1) = 10
⇔ [2(x - 3)]2 - (4x2 - 1) = 10
⇔ (2x - 6)2 - 4x2 + 1 = 10
⇔ 4x2 - 24x + 36 - 4x2 + 1 = 10
⇔ - 24x = -27
⇔ x = \(\dfrac{9}{8}\)
c) (x - 4)2 - (x - 2)(x + 2) = 36
⇔ x2 - 8x + 16 - x2 + 4 = 36
⇔ -8x = 16
⇔ x = -2
d) (2x + 3)2 - (2x - 1)(2x + 1) = 10
⇔ 4x2 + 12x + 9 - 4x2 + 1 = 10
⇔ 12x = 0
⇔ x = 0
27 tháng 12 2017
Tìm x ,biết :
a, \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=17\)
\(\Rightarrow x^2-4x+4-x^2+9=17\)
\(\Rightarrow-4x+13=17\)
\(\Rightarrow-4x=4\)
\(\Rightarrow x=-1\)
b,\(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Rightarrow4\left(x^2-6x+9\right)-4x^2+1=10\)
\(\Rightarrow4x^2-24x+36-4x^2+1=10\)
⇒ \(-24x+37=10\)
\(\Rightarrow-24x=-27\)
\(\Rightarrow x=\dfrac{-27}{-24}=\dfrac{9}{8}\)
c,\(\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=36\)
⇒ \(x^2-8x+16-x^2+4=36\)
⇒ \(-8x+20=36\)
⇒ \(-8x=16\Rightarrow x=-2\)
d,\(\left(2x+3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Rightarrow4x^2+12x+9-4x^2+1=10\)
\(\Rightarrow12x+10=10\)
\(\Rightarrow12x=0\Rightarrow x=0\)
UK
9 tháng 10 2017
Bài a,b,c,e,g,i thì đặt điều kiện rồi bình phương 2 vế rồi giải, bài j chuyển vế rồi bình phương
Chỉ trình bày lời giải, tự tìm điều kiện nha :v
d) \(\sqrt{x+2\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-1}+1=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Rightarrow x-1=1\Leftrightarrow x=2\)
f) \(\sqrt{x+4\sqrt{x-4}}=2\)
\(\Leftrightarrow\sqrt{x-4+2.2\sqrt{x-4}+4}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-4}+2=2\)
\(\Leftrightarrow\sqrt{x-4}=0\)
\(\Rightarrow x-4=0\Leftrightarrow x=4\)
Thần kinh dung dinh giẫm phải đinh
Đề thiếu chi tiết nhé bạn,phải có Kết quả sau dấu "=" thì mới tìm X cho bạn được
x^2
ai hoi