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6 tháng 7 2021
a) \(\left(\sqrt{125}+\sqrt{45}-2\sqrt{80}\right).\sqrt{5}=\left(5\sqrt{5}+3\sqrt{5}-8\sqrt{5}\right).\sqrt{5}\)
\(=0.\sqrt{5}=0\)
b) \(\frac{5-2\sqrt{6}}{\sqrt{2}-\sqrt{3}}=\frac{\left(5-2\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}=\frac{\left(5\sqrt{2}+5\sqrt{3}-4\sqrt{3}-6\sqrt{2}\right)}{-1}\)
\(=-\left(-\sqrt{2}+\sqrt{3}\right)=\sqrt{2}-\sqrt{3}\)
AN
6 tháng 7 2021
a,\(\left(\sqrt{125}+\sqrt{45}-2\sqrt{80}\right).\sqrt{5}\)
\(=\left(5\sqrt{5}+3\sqrt{5}-8\sqrt{5}\right).\sqrt{5}\)
\(=0.\sqrt{5}\)
\(=0\)
b,\(\frac{5-2\sqrt{6}}{\sqrt{2}-\sqrt{3}}\)
\(=\frac{\left(5-2\sqrt{6}\right).\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right).\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{\sqrt{3}-\sqrt{2}}{-1}\)
\(=\sqrt{2}-\sqrt{3}\)
23 tháng 7 2019
ĐK: \(x-9\ne0\Rightarrow x\ne9\)
\(\sqrt{x}\ge0\Rightarrow x\ge0\)
\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)
\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)
ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)
\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\left(\frac{1+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-3\right)}\)
2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)
\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
20 tháng 10
a: Ta có: \((x - \sqrt{x^2+5})(x + \sqrt{x^2+5}) = x^2 - (x^2+5) = -5\)
=>\(x + \sqrt{x^2+5} = \frac{-5}{x - \sqrt{x^2+5}}\)
Ta có: \((x - \sqrt{x^2+5})(y - \sqrt{y^2+5}) = 5\)
=>\((x-\sqrt{x^2+5})(y-\sqrt{y^2+5})\left(x+\sqrt{x^2+5}\right)=5\left(x+\sqrt{x^2+5}\right)\)
=>\(-5 (y - \sqrt{y^2+5}) = 5 (x + \sqrt{x^2+5})\)
=>\(-(y-\sqrt{y^2+5})=x+\sqrt{x^2+5}\)
=>\(\sqrt{y^2+5}-y=x+\sqrt{x^2+5}\quad(1)\)
Ta có: \((x - \sqrt{x^2+5})(y - \sqrt{y^2+5}) = 5\)
=>\((x-\sqrt{x^2+5})(y-\sqrt{y^2+5})\left(y+\sqrt{y^2+5}\right)=5\left(y+\sqrt{y^2+5}\right)\)
=>\(-5 (x - \sqrt{x^2+5}) = 5 (y + \sqrt{y^2+5})\)
=>\(\sqrt{x^2+5}-x=y+\sqrt{y^2+5}\quad(2)\)
Từ (1),(2) suy ra -x-y=x+y
=>x+y=0
=>x=-y
\(M=x^{2015}+y^{2015}=\left(-y\right)^{2015}+y^{2015}=0\)
b: Đặt \(u=\sqrt[3]{5 + 2\sqrt{13}};v=\sqrt[3]{5-2\sqrt{13}}\)
=>\(uv = \sqrt[3]{(5 + 2\sqrt{13})(5 - 2\sqrt{13})} = \sqrt[3]{5^2 - (2\sqrt{13})^2} = \sqrt[3]{25 - 52} = \sqrt[3]{-27} = -3\)
x=u+v
=>\(x^3 = (u + v)^3 = u^3 + v^3 + 3uv(u + v)\)
=>\(x^3=(5+2\sqrt{13})+(5-2\sqrt{13})+3(-3)x\)
=>\(x^3=10-9x\)
=>\(x^3+9x-10=0\)
=>(x-1)(x^2+x+10)=0
=>x-1=0
=>x=1
\(A=x^{2015}-x^{2016}+2017\)
\(=1^{2015}-1^{2016}+2017\)
=2017
YK
11 tháng 7 2018
A=\(\frac{2\sqrt{2}-\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}}{3}\)
ĐÚNG KO ♥
18 tháng 5 2019
\(a)\)\(A=\sqrt{3\left(\sqrt{2}\right)^2-2\left(\sqrt{2}\right)-\sqrt{2}.\sqrt{2}-1}=\sqrt{6-2\sqrt{2}-2-1}\)
\(A=\sqrt{3-2\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-\sqrt{1}\right|=\sqrt{2}-1\)
\(b)\)\(C=\left(1-\sqrt{2}\right)^2+\sqrt{8}-2=1-2\sqrt{2}+2+2\sqrt{2}-2=1\)
\(c_1)\)\(D=\sqrt{\left(1-\sqrt{x}\right)^2}.\sqrt{x+1+2\sqrt{x}}=\sqrt{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}\) ( đề sai r mem :3 )
\(D=\left|\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\right|=\left|x-1\right|\)
\(c_2)\)\(D=\left|x-1\right|=\left|2020-1\right|=2019\)
A = 2.\(\sqrt{80}\) - 2.\(\sqrt{245}\) + 2.\(\sqrt{180}\)
A = 2.\(\sqrt{16.5}\) - 2.\(\sqrt{49.5}\) + 2.\(\sqrt{36.5}\)
A = 2.\(\sqrt{16}\).\(\sqrt{5}\) - .2.\(\sqrt{49}\).\(\sqrt{5}\) + 2.\(\sqrt{36}\).\(\sqrt{5}\)
A = 2.4.\(\sqrt{5}\) - 2.7.\(\sqrt{5}\) + 2.6.\(\sqrt{5}\)
A = 8.\(\sqrt{5}\) - 14.\(\sqrt{5}\) + 12.\(\sqrt{5}\)
A = -6\(\sqrt{5}\) + 12\(\sqrt{5}\)
A = 6\(\sqrt{5}\)
A=2(4\(\sqrt{5}\)-7\(\sqrt{5}\)+6\(\sqrt{5}\))
A= 6\(\sqrt{5}\)
6 căn 5
=8√5 - 14√5 + 12√5
=6√5
Giá trị của biểu thức A là \(\mathbf{6}\sqrt{\mathbf{5}}\).
A=6_/5
\(A=6\sqrt5\)
6√5
A=6 căn 5
K bt lm
6√5
A=12 căn bậc 2 của 5
6√5
6 căn bậc 2 của 5
6√5