![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-2\left(\sqrt{1+x}+\sqrt{1-x}\right)+7=\sqrt{\left(5-2x\right)\left(5+2x\right)}-2\sqrt{1-x^2}\)
ĐKCĐ: \(-1\le x\le1\)
\(\Leftrightarrow2\left(\sqrt{\left(1-x\right)}-1\right)\left(\sqrt{1+x}-1\right)+5-\sqrt{\left(5-2x\right)\left(5+2x\right)}=0\)
\(\Leftrightarrow2x^2\left[\frac{2}{5+\sqrt{\left(5-2x\right)\left(5+2x\right)}}-\frac{1}{\left(\sqrt{1-x}+1\right)\left(\sqrt{1+x}+1\right)}\right]\)
Đặt: \(A=\frac{2}{5+\sqrt{\left(5-2x\right)\left(5+2x\right)}}-\frac{1}{\left(\sqrt{1-x}+1\right)\left(\sqrt{1+x}+1\right)}\)
Có: \(A\le\frac{2}{5+\sqrt{\left(5-2\right)\left(5-2\right)}}-\frac{1}{\sqrt{1-x^2}+1+\sqrt{1-x}+\sqrt{1+x}}< \frac{2}{5+3}-\frac{1}{1+1+2}=0\)
\(\Rightarrow x=0\) là nghiệm của pt
![](https://rs.olm.vn/images/avt/0.png?1311)
f) \(\left(\sqrt{6x+1}-\sqrt{6x-1}\right)^2=\left(\sqrt{6x+1}\right)^2-2\sqrt{\left(6x+1\right)\left(6x-1\right)}+\left(\sqrt{6x-1}\right)^2\)
\(=6x+1+6x-1-2\sqrt{36x^2-1}=12x-2\sqrt{36x^2-1}\)
tương tự các câu khác mình làm tắt chút nha:
c) \(\left(\sqrt{2x+3}+\sqrt{2x-3}\right)^2=2x+3+2x-3-2\sqrt{\left(2x+3\right)\left(2x-3\right)}=4x+2\sqrt{4x^2-9}\)
d) \(\left(\sqrt{2x+y}+\sqrt{2x-y}\right)^2=2x+y+2x-y-2\sqrt{\left(2x+y\right)\left(2x-y\right)}=4x-2\sqrt{4x^2-y^2}\)
\(\left(\sqrt{5x-2}-\sqrt{5x+2}\right)^2=5x-2+5x+2-2\sqrt{\left(5x-2\right)\left(5x+2\right)}=10x-2\sqrt{25x^2-4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\left(x-\sqrt{2}\right)+3\left(x^2-2\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)+3\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(1+3x+3\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{2}=0\\3x+3\sqrt{2}+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\3x=-3\sqrt{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=\dfrac{-3\sqrt{2}-1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{2};\dfrac{-3\sqrt{2}-1}{3}\right\}\)
b) Ta có: \(x^2-5=\left(2x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
\(\Leftrightarrow\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)-\left(2x-\sqrt{5}\right)\left(x+\sqrt{5}\right)=0\)
\(\Leftrightarrow\left(x+\sqrt{5}\right)\left(x-\sqrt{5}-2x+\sqrt{5}\right)=0\)
\(\Leftrightarrow-x\left(x+\sqrt{5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\x+\sqrt{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\sqrt{5}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-\sqrt{5}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\sqrt{2x+3}+\sqrt{2x-3}\right)^2=\sqrt{2x-3}^2+2\sqrt{\left(2x+3\right)\left(2x-3\right)}+\sqrt{2x-3}^2=2x-3+2x+3+2\sqrt{4x^2-9}=4x+2\sqrt{4x^2-9}\)
\(\left(\sqrt{2x+y}+\sqrt{2x-y}\right)^2=2x+y+2x-y+2\sqrt{\left(2x+y\right)\left(2x-y\right)}=4x+2\sqrt{4x^2-y^2}\)
\(\left(\sqrt{2x}\right)^2=2x\)
\(\left(\sqrt{2x}\right)^2=\left[\left(2x\right)^{\frac{1}{2}}\right]^2=\left(2x\right)^{\frac{1}{2}.2}=\left(2x\right)^1=2x\)