
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


1) Ta có: \(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\cdot\sqrt{6}-\left(\frac{5}{2}\sqrt{2}+12\right)\)
\(=\left(2\sqrt{3}-6\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{25}{4}\cdot2}+12\right)\)
\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{50}{4}}+12\right)\)
\(=-12\sqrt{2}+12-\frac{5\sqrt{2}}{2}-12\)
\(=\frac{-24\sqrt{2}-5\sqrt{2}}{2}\)
\(=\frac{-29\sqrt{2}}{2}\)
2) Ta có: \(\frac{26}{2\sqrt{3}+5}-\frac{4}{\sqrt{3}-2}\)
\(=\frac{26\left(5-2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}+\frac{4}{2-\sqrt{3}}\)
\(=\frac{26\left(5-2\sqrt{3}\right)}{25-12}+\frac{4\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=2\left(5-2\sqrt{3}\right)+4\left(2+\sqrt{3}\right)\)
\(=10-4\sqrt{3}+8+4\sqrt{3}\)
\(=18\)
3) ĐK để phương trình có nghiệm là: x≥0
Ta có: \(\sqrt{x^2-6x+9}=2x\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x\)
\(\Leftrightarrow\left|x-3\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x\\x-3=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3-2x=0\\x-3+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x-3=0\\3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=3\\3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={1}
4) ĐK để phương trình có nghiệm là: \(x\ge\frac{1}{2}\)
Ta có: \(\sqrt{4x^2+1}=2x-1\)
\(\Leftrightarrow\left(\sqrt{4x^2+1}\right)^2=\left(2x-1\right)^2\)
\(\Leftrightarrow4x^2+1=4x^2-4x+1\)
\(\Leftrightarrow4x^2+1-4x^2+4x-1=0\)
\(\Leftrightarrow4x=0\)
hay x=0(loại)
Vậy: S=∅

2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)

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

1) \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
= \(\frac{ \left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}+\frac{\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)
= \(\frac{\left(\sqrt{7}+\sqrt{5}\right)^2+\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\) = \(\frac{\left(\sqrt{7}\right)^2+2\sqrt{7}.\sqrt{5}+\left(\sqrt{5}\right)^2+\left(\sqrt{7}\right)^2-2\sqrt{7}.\sqrt{5}+\left(\sqrt{5}\right)^2}{\left(\sqrt{7}\right)^2-\left(\sqrt{5}\right)^2}\)
= \(\frac{7+2\sqrt{35}+5+7-2\sqrt{35}+5}{7-5}\) = \(\frac{24}{2}=12\)
2) \(x+2y-\sqrt{\left(x^2-4xy+4y^2\right)^2}\left(x\ge2y\right)\)
= \(x+2y-\sqrt{\left(x-2y\right)^4}\) = \(x+2y-|x-2y|\)
= \(x+2y-\left(x-2y\right)\) = \(x+2y-x+2y=4y\)
3)\(4x+\sqrt{\left(x-12\right)^2}\left(x\ge2\right)\)
= \(4x+x-12=5x-12\)

ghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghgh
by
duc

làm tạm câu này vậy
a/\(\left(x^2-x+1\right)^4+4x^2\left(x^2-x+1\right)^2=5x^4\)
\(\Leftrightarrow\left(x^2-x+1\right)^4+4x^2\left(x^2-x+1\right)+4x^4=9x^4\)
\(\Leftrightarrow\left\{\left(x^2-x+1\right)^2+2x^2\right\}=\left(3x^2\right)^2\)
\(\Leftrightarrow\left(x^2-x+1\right)^2+2x^2=3x^2\)(vì 2 vế đều không âm)
\(\Leftrightarrow\left(x^2-x+1\right)=x^2\)
\(\Leftrightarrow\left|x\right|=x^2-x+1\)\(\left(x^2-x+1=\left(x-\frac{1}{4}\right)^2+\frac{3}{4}>0\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x=x^2-x+1\\-x=x^2-x+1\end{cases}\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\x^2+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x^2+1=0\left(vo.nghiem\right)\end{cases}}}\)
Vậy...