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a)\(\sqrt{x^2+48}=4x-3+\sqrt{x^2+35}\)
\(\Leftrightarrow\sqrt{x^2+48}-7=4x-4+\sqrt{x^2+35}-6\)
\(\Leftrightarrow\dfrac{x^2+48-49}{\sqrt{x^2+48}+7}=4\left(x-1\right)+\dfrac{x^2+35-36}{\sqrt{x^2+35}+6}\)
\(\Leftrightarrow\dfrac{x^2-1}{\sqrt{x^2+48}+7}-4\left(x-1\right)-\dfrac{x^2-1}{\sqrt{x^2+35}+6}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x+1}{\sqrt{x^2+48}+7}-4-\dfrac{x+1}{\sqrt{x^2+35}+6}\right)=0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
b)\(\left(\sqrt{x-1}+1\right)^3+2\sqrt{x-1}=2-x\)
\(pt\Leftrightarrow\left(\sqrt{x-1}+1\right)^3-1+2\sqrt{x-1}=1-x\)
\(\Leftrightarrow\left(\sqrt{x-1}+1-1\right)\left(\left(\sqrt{x-1}+1\right)^6+\left(\sqrt{x-1}+1\right)^3+1\right)+2\sqrt{x-1}-\left(1-x\right)=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\left(\sqrt{x-1}+1\right)^6+\left(\sqrt{x-1}+1\right)^3+1\right)+2\sqrt{x-1}+x-1=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\left(\sqrt{x-1}+1\right)^6+\left(\sqrt{x-1}+1\right)^3+3+\sqrt{x-1}\right)=0\)
Dễ thấy: \(\left(\sqrt{x-1}+1\right)^6+\left(\sqrt{x-1}+1\right)^3+3+\sqrt{x-1}>0\)
\(\Rightarrow\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
\(pt\Leftrightarrow\sqrt{x^2+48}-7=4x-4+\sqrt{x^2+35}-6\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x^2+48}-7\right)\left(\sqrt{x^2+48}+7\right)}{\sqrt{x^2+48}+7}=4\left(x-1\right)+\dfrac{\left(\sqrt{x^2+35}-6\right)\left(\sqrt{x^2+35}+6\right)}{\sqrt{x^2+35}+6}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+48}+7}-4\left(x-1\right)-\dfrac{\left(x+1\right)\left(x-1\right)}{\sqrt{x^2+35}+6}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x+1}{\sqrt{x^2+48}+7}-4-\dfrac{x+1}{\sqrt{x^2+35}+6}\right)=0\)
Do : \(\dfrac{x+1}{\sqrt{x^2+48}+7}-4-\dfrac{x+1}{\sqrt{x^2+35}+6}\ne0\)
\(\Rightarrow x=1\)
\(\sqrt{x^2+48}=4x-3+\sqrt{x^2+35}\Leftrightarrow\sqrt{x^2+48}-7=4x-4+\sqrt{x^2+35}-6\)
\(\Leftrightarrow\frac{x^2+48-49}{\sqrt{x^2+48}+7}=4x-4+\frac{x^2+35-36}{\sqrt{x^2+35}+6}\Leftrightarrow\frac{x^2-1}{\sqrt{x^2+48}+7}=4\left(x-1\right)+\frac{x^2-1}{\sqrt{x^2+35}+6}\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{x+1}{\sqrt{x^2+48}+7}-4-\frac{x+1}{\sqrt{x^2+35}+6}\right)=0\)\(\Leftrightarrow x-1=0\Leftrightarrow x=1\).
VÌ \(\sqrt{x^2+48}-\sqrt{x^2+35}>0\)
=> \(x>\frac{3}{4}\)
Phương trình tương đương
\((x+6-\sqrt{x^2+48})+3\left(x-1\right)+\left(\sqrt{x^2+35}-6\right)=0\)
=> \(\frac{12\left(x-1\right)}{x+6+\sqrt{x^2+48}}+3\left(x-1\right)+\frac{x^2-1}{\sqrt{x^2+35}+6}=0\)
\(\hept{\begin{cases}x=1\\\frac{12}{x+6+\sqrt{x^2+48}}+3+\frac{x+1}{\sqrt{x^2+35}+6}=0\left(2\right)\end{cases}}\)
Phương trình (2) vô nghiệm do x>3/4=> VT>0
\(\sqrt{x^2+48}=4x-3+\sqrt{x^2+35}\)
\(\Leftrightarrow\sqrt{x^2+48}-7=4x-4+\sqrt{x^2+35}-6\)
\(\Leftrightarrow\frac{x^2+48-49}{\sqrt{x^2+48}+7}=4\left(x-1\right)+\frac{x^2+35-36}{\sqrt{x^2+35}+6}\)
\(\Leftrightarrow\frac{x^2-1}{\sqrt{x^2+48}+7}-4\left(x-1\right)-\frac{x^2-1}{\sqrt{x^2+35}+6}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{x+1}{\sqrt{x^2+48}+7}-4-\frac{x+1}{\sqrt{x^2+35}+6}\right)=0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
bài 2 rút gọn :
a) \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}-3\right)^2}\)
= \(\left|1-\sqrt{2}\right|+\left|\sqrt{2}-3\right|\)
=\(\sqrt{2}-1+3-\sqrt{2}\)
=2
b) \(\sqrt{4-2\sqrt{3}}+\sqrt{7}-\sqrt{48}\)
= \(\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{7}-4\sqrt{3}\)
= \(\sqrt{3}-1+\sqrt{7}-4\sqrt{3}\)
= \(\sqrt{7}-3\sqrt{3}+1\)
c)