ĐK:\(-1\le x\le1\)

\(pt\Leftrightarrow\left(\sqrt{1+x}+\sqrt{1-x}\right)\left(2+2\sqrt{\left(1+x\right)\left(1-x\right)}\right)=8\)

Đặt \(\hept{\begin{cases}\sqrt{1+x}=a\\\sqrt{1-x}=b\end{cases}\left(a,b\ge0\right)}\) thì có:

\(\Leftrightarrow\hept{\begin{cases}a^2+b^2=2\\\left(a+b\right)\left(2+2ab\right)=8\end{cases}}\). Xét \(pt\left(2\right)\) có:

\(\left(a+b\right)\left(a^2+b^2+2ab\right)=8\)

\(\Leftrightarrow\left(a+b\right)^3=8=2^3\Leftrightarrow a+b=2\)

Hay \(\sqrt{1+x}+\sqrt{1-x}=2\)

Bình phương 2 vế rồi thu gọn được x=0

ĐK:\(-1\le x\le1\)

\(pt\Leftrightarrow\left(\sqrt{1+x}+\sqrt{1-x}\right)\left(2+2\sqrt{\left(1+x\right)\left(1-x\right)}\right)=8\)

Đặt \(\hept{\begin{cases}\sqrt{1+x}=a\\\sqrt{1-x}=b\end{cases}\left(a,b\ge0\right)}\) thì có:

\(\Leftrightarrow\hept{\begin{cases}a^2+b^2=2\\\left(a+b\right)\left(2+2ab\right)=8\end{cases}}\). Xét \(pt\left(2\right)\) có:

\(\left(a+b\right)\left(a^2+b^2+2ab\right)=8\)

\(\Leftrightarrow\left(a+b\right)^3=8=2^3\Leftrightarrow a+b=2\)

Hay \(\sqrt{1+x}+\sqrt{1-x}=2\)

Bình phương 2 vế rồi thu gọn được x=0

.

a: ĐKXĐ: x>=1

Đặt \(a=\sqrt{5x-1};b=\sqrt{x-1}\)

\(a^2-b^2=\left(5x-1\right)-\left(x-1\right)=4x\)

\(\frac{a^2+b^2}{2}=\frac{5x-1+x-1}{2}=\frac{6x-2}{2}=3x-1\)

\(ab=\sqrt{\left(5x-1\right)\left(x-1\right)}=\sqrt{5x^2-6x+1}\)

Thay vào phương trình ban đầu, ta được: \(\left(a+b\right)\left(\frac{a^2+b^2}{2}-ab\right)=\left(a-b\right)\left(a+b\right)\)

=>\(\frac{\left(a+b\right)\left(a-b\right)^2}{2}=\left(a-b\right)\left(a+b\right)\)

=>\(\frac{\left(a-b\right)^2}{2}=\left(a-b\right)\)

=>\(\left(a-b\right)^2=2\left(a-b\right)\)

=>(a-b)(a-b-2)=0

TH1: a-b=0

=>a=b

=>5x-1=x-1

=>x=0(loại)

TH2: a-b-2=0

=>\(\sqrt{5x-1}-\sqrt{x-1}-2=0\)

=>\(\sqrt{5x-1}=\sqrt{x-1}+2\)

=>\(5x-1=x-1+4+2\cdot2\cdot\sqrt{x-1}\)

=>\(4\sqrt{x-1}+x+3=5x-1\)

=>\(4\sqrt{x-1}=5x-1-x-3=4x-4\)

=>\(\sqrt{x-1}=x-1\)

=>\(\sqrt{x-1}\left(\sqrt{x-1}-1\right)=0\)

=>\(\left[\begin{array}{l}x-1=0\\ x-1=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\left(nhận\right)\\ x=2\left(nhận\right)\end{array}\right.\)

\(\left(\sqrt{x^2+1}+x\right)\left(\sqrt{y^2+1}-y\right)=1\)

\(\Leftrightarrow\sqrt{x^2+1}+x=\sqrt{y^2+1}+y\) (1)

Tương tự ta có: \(\sqrt{y^2+1}-y=\sqrt{x^2+1}-x\) (2)

Cộng vế (1) và (2) \(\Rightarrow x-y=y-x\Rightarrow x=y\)

Thế xuống dưới:

\(3\sqrt{3x-2}+x\sqrt{6-x}=10\)

Đặt \(\sqrt{6-x}=a\Rightarrow\left\{{}\begin{matrix}0\le a\le\frac{4\sqrt{3}}{3}\\x=6-a^2\end{matrix}\right.\)

\(\Rightarrow a^3-6a+10-3\sqrt{16-3a^2}=0\)

\(\Leftrightarrow\left(a^3-3a-2\right)+3\left(4-a-\sqrt{16-3a^2}\right)=0\)

\(\Leftrightarrow\left(a-2\right)\left(a+1\right)^2+\frac{12a\left(a-2\right)}{4-a+\sqrt{16-a^2}}=0\)

\(\Leftrightarrow\left(a-2\right)\left[\left(a+1\right)^2+\frac{12a}{4-a+\sqrt{16-a^2}}\right]=0\)

\(\Leftrightarrow a=2\Leftrightarrow...\)