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a)\(\sqrt{\frac{\left(x-2\right)^4}{\left(3-x\right)^2}}+\frac{x^2-1}{x-3}=\frac{\sqrt{\left(x-2\right)^4}}{\sqrt{\left(3-x\right)^2}}+\frac{x^2-1}{x-3}=\frac{\left(x-2\right)^2}{x-3}+\frac{x^2-1}{x-3}=\frac{x^2-4x+4+x^2-1}{x-3}=\frac{2x^2-4x+3}{x-3}\)
Tại x=0,5 thay vào ta có:
\(A=\frac{2\cdot\left(0,5\right)^2-4\cdot0,5+3}{0,5-3}=-\frac{3}{5}\)
b)\(4x-\sqrt{8}+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\frac{\sqrt{x^2\left(x+2\right)}}{\sqrt{x+2}}=4x-\sqrt{8}+\frac{\sqrt{x^2}\cdot\sqrt{x+2}}{\sqrt{x+2}}\)\(=4x-\sqrt{8}+x^2\)
Tại \(x=-\sqrt{2}\) thay vào ta có:
\(B=4\cdot\left(-\sqrt{2}\right)+\left(-\sqrt{2}\right)^2=2-4\sqrt{2}\)
a) \(x+3+\sqrt{x^2-6x+9}\left(x\le3\right)\)
\(=x+3+\sqrt{\left(x-3\right)^2}\)
\(=x+3+\left|x-3\right|\)
\(=x+3-\left(x-3\right)\)
\(=x+3-x+3\)
\(=6\)
b) \(\sqrt{x^2+4x+4}-\sqrt{x^2}\left(-2\le x\le0\right)\)
\(=\sqrt{\left(x+2\right)^2}-\sqrt{x^2}\)
\(=\left|x+2\right|-\left|x\right|\)
\(=x+2-\left(-x\right)\)
\(=x+2+x\)
\(=2x+2=2\left(x+1\right)\)
c) \(\frac{\sqrt{x^2-2x+1}}{x-1}\left(x>1\right)\)
\(=\frac{\sqrt{\left(x-1\right)^2}}{x-1}\)
\(=\frac{\left|x-1\right|}{x-1}\)
\(=\frac{x-1}{x-1}=1\)
d) \(\left|x-2\right|+\frac{\sqrt{x^2-4x+4}}{x-2}\)
\(=\left|x-2\right|+\frac{\sqrt{\left(x-2\right)^2}}{x-2}\)
\(=\left|x-2\right|+\frac{\left|x-2\right|}{x-2}\)
\(=\left|x-2\right|+\frac{-\left(x-2\right)}{x-2}\)
\(=\left|x-2\right|-1\)
\(=-\left(x-2\right)-1\)
\(=-x+2-1\)
\(=-x+1=-\left(x-1\right)\)
\(1,ĐKXĐ:x\ge0;x\ne4\)
\(A=\left(1+\frac{2}{\sqrt{x}}\right)\left(\frac{\sqrt{x}-2+\sqrt{x}+2-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(A=\left(1+\frac{2}{\sqrt{x}}\right)\left(\frac{2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(A=\left(1+\frac{2}{\sqrt{x}}\right)\left(\frac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(A=\left(\frac{\sqrt{x}+2}{\sqrt{x}}\right)\left(\frac{2}{\sqrt{x}+2}\right)\)
\(A=\frac{2}{\sqrt{x}}\)
\(2,A>\frac{1}{2}\)
\(\Leftrightarrow\frac{2}{\sqrt{x}}>\frac{1}{2}\)
\(\Leftrightarrow\frac{2}{\sqrt{x}}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{4}{2\sqrt{x}}-\frac{\sqrt{x}}{2\sqrt{x}}>0\)
\(\Leftrightarrow\frac{4-\sqrt{x}}{2\sqrt{x}}>0\)
Do \(\sqrt{x}>0\Rightarrow2\sqrt{x}>0\)
\(\Rightarrow4-\sqrt{x}>0\)
\(\Leftrightarrow-\sqrt{x}>-4\)
\(\Leftrightarrow\sqrt{x}< 4\)
\(\Leftrightarrow x< 16\)
Kết hợp với ĐKXĐ thì \(0\le x< 16\)và \(x\ne4\)
\(3,A=-2\sqrt{x}+5\)
\(\Leftrightarrow\frac{2}{\sqrt{x}}=-2\sqrt{x}+5\)
\(\Leftrightarrow\sqrt{x}\left(-2\sqrt{x}+5\right)=2\)
\(\Leftrightarrow-2x+5\sqrt{x}-2=0\)
\(\Leftrightarrow-2x+2.5\sqrt{x}+2.5\sqrt{x}-2=0\)
\(\Leftrightarrow\left(-2x+2.5\sqrt{x}\right)+\left(2.5\sqrt{x}-2\right)=0\)
Đến đây thì mình chịu
Bạn tự giải nốt nhé
HỌC TỐT
a) P = \(\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}}{\sqrt{x}+2}\right)\)\(.\)\(\frac{x-4}{\sqrt{4x}}\)
= \(\frac{\sqrt{x}.\left(\sqrt{x}+2\right)+\sqrt{x}.\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)\(.\)\(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{4x}}\)
= \(\frac{x+2\sqrt{x}+x-2\sqrt{x}}{\sqrt{4x}}\)
= \(\frac{2x}{2\sqrt{x}}\)= \(\sqrt{x}\)
b) x = \(3-2\sqrt{2}\)=\(2-2\sqrt{2}+1\)= \(\left(\sqrt{2}-1\right)^2\)
Thay x = \(\left(\sqrt{2}-1\right)^2\) vào P ta được
P = \(\sqrt{\left(\sqrt{2}-1\right)^2}\)= \(\sqrt{2}-1\)
Rút gọn:
\(A=\sqrt{\frac{1}{x^2-4x+4}}+\frac{-4}{x^2-2^2}\)
\(=\sqrt{\frac{1}{\left(x-2\right)^2}}-\frac{4}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{1}{x-2}-\frac{4}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{4}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x+2-4}{\left(x-2\right)\left(x+2\right)}=\frac{x-2}{\left(x-2\right)\left(x+2\right)}=\frac{1}{x+2}\)
Thay x=3 vào A ta được \(\frac{1}{3+2}=\frac{1}{5}\)