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Bài rút gọn
\(\sqrt{\left(x-1\right)^2}-x=\left|x-1\right|-x\)
\(=\left(x-1\right)-x=x-1-x=-1\left(x>1\right)\)
Bài gpt:
\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)
Đk:\(-1\le x\le3\)
\(pt\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}+\sqrt{x-3}\right)=0\)
Dễ thấy:\(\sqrt{x-2}+\sqrt{x-3}=0\) vô nghiệm
Nên \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
1) Sửa đề: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
Ta có: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
Ta có: \(x=3-2\sqrt{2}\)
\(=2-2\cdot\sqrt{2}\cdot1+1\)
\(=\left(\sqrt{2}-1\right)^2\)
Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\), ta được:
\(A=\frac{-5\cdot\sqrt{\left(\sqrt{2}-1\right)^2}+2}{\sqrt{\left(\sqrt{2}-1\right)^2}+3}\)
\(=\frac{-5\cdot\left(\sqrt{2}-1\right)+2}{\sqrt{2}-1+3}\)
\(=\frac{-5\sqrt{2}+5+2}{\sqrt{2}+2}\)
\(=\frac{-5\sqrt{2}+7}{\sqrt{2}+2}\)
Vậy: Khi \(x=3-2\sqrt{2}\) thì \(A=\frac{-5\sqrt{2}+7}{\sqrt{2}+2}\)
2) Ta có: \(B=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\)
\(=\frac{\left(x+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}+x+2\sqrt{x}+2+x+x\sqrt{x}-\sqrt{x}-1-\left(2x+2\sqrt{x}+x\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{2x+2x\sqrt{x}+\sqrt{x}+1-2x-2\sqrt{x}-x\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(x-1\right)}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
Ta có: \(x=7-2\sqrt{6}\)
\(=6-2\sqrt{6}\cdot1+1\)
\(=\left(\sqrt{6}-1\right)^2\)
Thay \(x=\left(\sqrt{6}-1\right)^2\) vào biểu thức \(B=\frac{\sqrt{x}}{x+\sqrt{x}+1}\), ta được:
\(B=\frac{\sqrt{\left(\sqrt{6}-1\right)^2}}{\left(\sqrt{6}-1\right)^2+\sqrt{\left(\sqrt{6}-1\right)^2}+1}\)
\(=\frac{\sqrt{6}-1}{7-2\sqrt{6}+\sqrt{6}-1+1}\)
\(=\frac{\sqrt{6}-1}{7-\sqrt{6}}\)
Vậy: Khi \(x=7-2\sqrt{6}\) thì \(B=\frac{\sqrt{6}-1}{7-\sqrt{6}}\)
3) Ta có: \(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\frac{\sqrt{x}\left(x-3\sqrt{x}-x-9\right)}{\left(\sqrt{x}+3\right)\left(2\sqrt{x}+4\right)}\)
\(=\frac{\sqrt{x}\left(-3\sqrt{x}-9\right)}{\left(\sqrt{x}+3\right)\cdot2\cdot\left(\sqrt{x}+2\right)}\)
\(=\frac{-3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(2\sqrt{x}+4\right)}\)
\(=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
Ta có: \(x=7-4\sqrt{3}\)
\(=4-2\cdot2\cdot\sqrt{3}+3\)
\(=\left(2-\sqrt{3}\right)^2\)
Thay \(x=\left(2-\sqrt{3}\right)^2\) vào biểu thức \(C=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\), ta được:
\(C=\frac{-3\cdot\sqrt{\left(2-\sqrt{3}\right)^2}}{2\cdot\sqrt{\left(2-\sqrt{3}\right)^2}+4}\)
\(=\frac{-3\cdot\left(2-\sqrt{3}\right)}{2\cdot\left(2-\sqrt{3}\right)+4}\)
\(=\frac{-6+3\sqrt{3}}{4-2\sqrt{3}+4}\)
\(=\frac{-6+3\sqrt{3}}{8-2\sqrt{3}}\)
Vậy: Khi \(x=7-4\sqrt{3}\) thì \(C=\frac{-6+3\sqrt{3}}{8-2\sqrt{3}}\)
a) ĐKXĐ : \(7\le x\le9\)
đặt \(A=\sqrt{x-7}+\sqrt{9-x}\)
\(\Rightarrow A^2=2+2\sqrt{\left(x-7\right)\left(9-x\right)}\le2+\left(x-7\right)+\left(9-x\right)=4\)
\(\Rightarrow A\le2\)
Mà \(x^2-16x+66=\left(x-8\right)^2+2\ge2\)
\(\Rightarrow VT=VP=2\)
do đó : \(x-7=9-x\Leftrightarrow x=8\)( t/m )
b) ĐKXĐ : \(x\le1\)
Ta có : \(\sqrt{1-x}+\sqrt{\left(x-1\right)\left(x-2\right)}-\left|x-2\right|\sqrt{\frac{x-1}{x-2}}=3\)
\(\Leftrightarrow\sqrt{1-x}+\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{\left(x-1\right)\left(x-2\right)}=3\)
\(\Leftrightarrow\sqrt{1-x}=3\Leftrightarrow x=-8\left(tm\right)\)