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mình giúp bài 3 cho
\(\sqrt{25x-125}-3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=6\left(ĐKXĐ:x\ge5\right)\)
\(< =>\sqrt{25\left(x-5\right)}-3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=6\)
\(< =>\sqrt{25}.\sqrt{x-5}-3\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=6\)
\(< =>5.\sqrt{x-5}-3.\frac{\sqrt{x-5}}{3}-\frac{1}{3}.3.\sqrt{x-5}=6\)
\(< =>5.\sqrt{x-5}-\sqrt{x-5}-\sqrt{x-5}=6\)
\(< =>3\sqrt{x-5}=6< =>\sqrt{x-5}=2\)
\(< =>x-5=4< =>x=4+5=9\left(tmđk\right)\)
c) \(C=\frac{\left(2\sqrt{x}+x\right)\left(\sqrt{x}+1\right)-\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+\sqrt{x}+1-\left(\sqrt{x}+2\right)}{x+\sqrt{x}+1}=\)
\(C=\frac{x\sqrt{x}+2x+x+2\sqrt{x}-x\sqrt{x}+1}{\left(\left(\sqrt{x}\right)^3-1\right)\left(\sqrt{x}+1\right)}\times\frac{x+\sqrt{x}+1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\times\frac{x+\sqrt{x}+1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\times\frac{1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{x-1}\times\frac{1}{x-1}=\frac{3x+2\sqrt{x}+1}{\left(x-1\right)^2}.\)
a: \(P=\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}:\left(\dfrac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\cdot\dfrac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
b: Thay \(x=\dfrac{3-2\sqrt{2}}{4}\) vào P, ta được:
\(P=\left(3\cdot\dfrac{\sqrt{2}-1}{2}-5\right):\left(2\cdot\dfrac{\sqrt{2}-1}{2}+1\right)\)
\(=\dfrac{\left(\dfrac{3}{2}\sqrt{2}-\dfrac{3}{2}-5\right)}{\sqrt{2}}\)
\(=\dfrac{\dfrac{3}{2}\sqrt{2}-\dfrac{13}{2}}{\sqrt{2}}=\dfrac{6-13\sqrt{2}}{4}\)
B3: \(\sqrt{x^4-4x^3+2x^2+4x+1}=3x-1\)
\(pt\Leftrightarrow x^4-4x^3+2x^2+4x+1=\left(3x-1\right)^2\)
\(\Leftrightarrow x^4-4x^3+2x^2+4x+1=9x^2-6x+1\)
\(\Leftrightarrow x^4-4x^3-7x^2+10x=0\)
\(\Leftrightarrow x\left(x^3-4x^2-7x+10\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}\) (thỏa mãn (mấy cái kia loại hết))
a)\(\frac{3-\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}}=\sqrt{3}-1\)
b)\(\frac{2\sqrt{2}+\sqrt{6}}{4+\sqrt{12}}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2\left(2+\sqrt{3}\right)}=\frac{\sqrt{2}}{2}\)
c)\(\frac{1-\sqrt{a^3}}{a-1}=\frac{1-\sqrt{a}^3}{-\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}=\frac{-\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}=\frac{-1-\sqrt{a}-a}{1+\sqrt{a}}\)
d)\(\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}=\frac{\sqrt{5+2\sqrt{5}+1}}{\sqrt{5}+1}=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{5}+1}=\frac{\left|\sqrt{5}+1\right|}{\sqrt{5}+1}=\frac{\sqrt{5}+1}{\sqrt{5}+1}=1\)
e)\(\frac{\sqrt{5+2\sqrt{6}}}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3+2\sqrt{6}+2}}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}{\sqrt{3}+\sqrt{2}}=\frac{\left|\sqrt{3}+\sqrt{2}\right|}{\sqrt{3}+\sqrt{2}}=1\)
Mau la \(\sqrt{X - 3} \) that sao