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\(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a/ \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt[]{x-3}\right)}\right]:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt[]{x-3}}\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}+\sqrt{x}-3}{\sqrt{x}-3}\right].\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
=> \(R=\frac{3\sqrt{x}-3}{\sqrt{x}-3}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
b/ Để R<-1 => \(\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< -1\)
<=> \(3\sqrt{x}-3< -\sqrt{x}-1\)
<=> \(4\sqrt{x}< 2\)=> \(\sqrt{x}< \frac{1}{2}\) => \(-\frac{1}{4}< x< \frac{1}{4}\)
Chỗ => R = \(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\) là sao vậy ạ?
a, \(\sqrt{\left(\sqrt{2}\right)^2+2\times2\times\sqrt{2}+2^2}\)+ \(\sqrt{2^2+2\times2\times\sqrt{2}+\left(\sqrt{2}\right)^2}\)
= \(\sqrt{\left(\sqrt{2}+2\right)^2}\)+ \(\sqrt{\left(2-\sqrt{2}\right)^2}\)
= \(\sqrt{2}+2+2-\sqrt{2}\)
= 4
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) \(ĐKXĐ:x\ge0;x\ne3\)
b) \(A=\left(\frac{x-2\sqrt{3x}+3}{x-3}\right)\left(\sqrt{4x}+\sqrt{12}\right)\)
\(\Leftrightarrow A=\left(\frac{\left(\sqrt{x}-\sqrt{3}\right)^2}{\left(\sqrt{x}-\sqrt{3}\right)\left(\sqrt{x}+\sqrt{3}\right)}\right)\left(2\sqrt{x}+2\sqrt{3}\right)\)
\(\Leftrightarrow A=\left(\frac{\sqrt{x}-\sqrt{3}}{\sqrt{x}+\sqrt{3}}\right).2\left(\sqrt{x}+\sqrt{3}\right)\)
\(\Leftrightarrow A=2\left(\sqrt{x}-\sqrt{3}\right)\)
\(\Leftrightarrow A=2\sqrt{x}-2\sqrt{3}\)
c) Thay \(x=4-2\sqrt{3}\)vào A, ta có :
\(A=2\sqrt{4-2\sqrt{3}}-2\sqrt{3}\)
\(\Leftrightarrow A=2\sqrt{\left(1-\sqrt{3}\right)^2}-2\sqrt{3}\)
\(\Leftrightarrow A=2\left(\sqrt{3}-1\right)-2\sqrt{3}\)
\(\Leftrightarrow A=2\sqrt{3}-2-2\sqrt{3}\)
\(\Leftrightarrow A=-2\)
a: \(\sqrt{9\left(1+4x+4x^2\right)}=3\cdot\left|2x+1\right|\)
\(=3\left|-2\sqrt{3}+1\right|\)
\(=3\left(2\sqrt{3}-1\right)=6\sqrt{3}-3\)
b: \(\sqrt{4a^2\left(b^2+9-6b\right)}\)
\(=2\cdot\left|a\right|\cdot\left|b-3\right|\)
\(=2\cdot3\cdot\left|-\sqrt{2}-3\right|\)
\(=6\left(3+\sqrt{2}\right)\)