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DP
16 tháng 8 2017
\(4\cdot\left(x-12\right)=2x+164\)
\(\Leftrightarrow4x-48=2x+164\)
\(\Leftrightarrow4x-2x=164+48\)
\(\Leftrightarrow2x=212\Rightarrow x=106\)
Sửa đề : \(\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{401.405}+x=2\frac{1}{405}\)
\(\Leftrightarrow\frac{1}{4}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{401}-\frac{1}{405}\right)+x=\frac{811}{405}\)
\(\Leftrightarrow\frac{1}{4}\cdot\left(1-\frac{1}{405}\right)+x=\frac{811}{405}\)
\(\Leftrightarrow\frac{1}{4}\cdot\frac{404}{405}+x=\frac{811}{405}\)
\(\Leftrightarrow\frac{101}{405}+x=\frac{811}{405}\Rightarrow x=\frac{142}{81}\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
12 tháng 3 2020
a) Ta có: \(\frac{x+1}{3}=\frac{2}{6}\)
⇔\(x=\frac{2\cdot3}{6}-1=\frac{6}{6}-1=1-1=0\)
Vậy: x=0
b) Ta có: \(\frac{x-1}{4}=\frac{1}{-2}\)
⇔\(x=\frac{1\cdot4}{-2}+1=\frac{4}{-2}+1=-1\)
Vậy: x=-1
c) Ta có: \(\frac{-1}{6}=\frac{3}{2x}\)
⇔\(2x=\frac{3\cdot6}{-1}=-18\)
hay x=-9
Vậy: x=-9
d) Ta có: \(\frac{x+1}{3}=\frac{3}{x+1}\)
⇔\(\left(x+1\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy: x∈{2;-4}
e) Ta có: \(\frac{4}{5}=\frac{-12}{9-x}\)
⇔\(9-x=\frac{-12\cdot5}{4}=-15\)
hay x=24
Vậy: x=24
f) Ta có: \(\frac{x-1}{-4}=\frac{-4}{x-1}\)
⇔\(\left(x-1\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
Vậy: x∈{5;-3}
g) Ta có: \(\frac{5-x}{2}=\frac{2}{5-x}\)
⇔\(\left(5-x\right)^2=4\)
⇔\(\left[{}\begin{matrix}5-x=2\\5-x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)
Vậy: x∈{3;7}
h) Ta có: \(\frac{4-x}{-5}=\frac{-5}{4-x}\)
⇔\(\left(4-x\right)^2=25\)
⇔\(\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\)
Vậy: x∈{-1;9}
24 tháng 7 2019
a) \(3x-\frac{1}{5}=\frac{4+x}{2}\)
=> \(\frac{15x-1}{5}=\frac{4+x}{2}\)
=> \(\left(15x-1\right).2=\left(4+x\right).5\)
=> \(30x-2=20+5x\)
=> \(30x-5x=20+2\)
=> \(25x=22\)
=> \(x=\frac{22}{25}\)
b) \(\frac{4}{3}x-1=\frac{4\left(x+1\right)}{3}-\frac{1}{3}\)
=> \(\frac{1}{3}x=\frac{4x+4-1}{3}\)
=> \(\frac{1}{3}x=\frac{4x+3}{3}\)
=> \(3x=3\left(4x+3\right)\)
=> \(3x=12x+9\)
=> \(3x-12x=9\)
=> \(-9x=9\)
=> \(x=9:\left(-9\right)=-1\)
24 tháng 7 2019
a) \(\frac{4}{3}x-1=\frac{x}{5}\)
=> \(\frac{4}{3}x-\frac{1}{5}x=1\)
=> \(\frac{17}{15}x=1\)
=> \(x=1:\frac{17}{15}=\frac{15}{17}\)
b) \(x+50\%=\frac{4\left(x+1\right)}{3}-\frac{1}{3}\)
=> \(x+\frac{1}{2}=\frac{4x+4-1}{3}\)
=> \(\frac{2x+1}{2}=\frac{4x+3}{3}\)
=> \(\left(2x+1\right).3=2.\left(4x+3\right)\)
=> \(6x+3=8x+6\)
=> \(6x-8x=6-3\)
=> \(-2x=3\)
=> \(x=3:\left(-2\right)=-\frac{3}{2}\)
24 tháng 7 2019
\(a,\frac{4}{3}x-1=\frac{x}{5}\)
\(\Rightarrow\frac{4}{3}x=\frac{x}{5}+1\)
\(\Rightarrow\frac{4x}{3}=\frac{x+5}{5}\)
\(\Rightarrow20x=3x+15\)
\(\Rightarrow17x=15\)
\(\Rightarrow x=\frac{15}{17}\)
\(b,x+50\%=\frac{4\left(x+1\right)}{3}-\frac{1}{3}\)
\(\Rightarrow x+\frac{1}{2}=\frac{4x+4-1}{3}\)
\(\Rightarrow\frac{2x+1}{2}=\frac{4x+3}{3}\)
\(\Rightarrow3\left(2x+1\right)=\left(4x+3\right).2\)
\(\Rightarrow6x+3=8x+6\)
\(\Rightarrow2x=-3\)
\(\Rightarrow x=-\frac{3}{2}\)
\(c,-200\%.x+\frac{4}{3}=\frac{7}{4}\left(x+1\right)\)
\(\Rightarrow-2x+\frac{4}{3}=\frac{7}{4}x+\frac{7}{4}\)
\(\Rightarrow\frac{-6x+4}{3}=\frac{7x+7}{4}\)
\(\Rightarrow4\left(-6x+4\right)=\left(7x+7\right)3\)
\(\Rightarrow-24x+16=21x+21\)
\(\Rightarrow45x=-5\)
\(\Rightarrow x=-\frac{1}{7}\)
MV
23 tháng 4 2017
9) \(\dfrac{x}{4}=\dfrac{9}{x}\)
Theo định nghĩa về hai phân số bằng nhau, ta có:
\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
8)
\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)
7)
\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)
6)
\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)
5)
\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)
4)
\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
3)
\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)
2)
\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)
1)
\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)
TN
19 tháng 7 2016
a)\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{2013}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2013}\)
đề sai
b)\(\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(x+2004=0\).Do \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
\(x=-2004\)
c)\(\frac{x+5}{205}-1+\frac{x+4}{204}-1+\frac{x+3}{203}-1=\frac{x+166}{366}-1+\frac{x+167}{367}-1+\frac{x+168}{368}-1\)
\(\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}=\frac{x-200}{366}+\frac{x-200}{367}+\frac{x-200}{368}\)
\(\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}-\frac{x-200}{366}-\frac{x-200}{367}-\frac{x-200}{368}=0\)
\(\left(x-200\right)\left(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\right)=0\)
\(x-200=0\).Do\(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\ne0\)
\(x=200\)
d)chịu
9 tháng 4 2018
\(b)\) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(1-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{100}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(100=2x+4\)
\(\Leftrightarrow\)\(2x=96\)
\(\Leftrightarrow\)\(48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
9 tháng 4 2018
\(a)\) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{47.49}=\frac{24}{x+1}\)
\(\Leftrightarrow\)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(1-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{48}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(49=x+1\)
\(\Leftrightarrow\)\(x=48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
PA
28 tháng 11 2016
\(8-12x+6x^2-x^3\)
\(=\left(2-x\right)^3\)
\(125x^3-75x^2+15x-1\)
\(=\left(5x-1\right)^3\)
\(x^2-xz-9y^2+3yz\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
\(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
x = 3
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