\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\left(đk:x\ge-2\right)\)

Đặt \(a=\sqrt{x+5},b=\sqrt{x+2}\left(đk:a,b\ge0,a\ne b\right)\)

\(\Rightarrow\left\{{}\begin{matrix}ab=\sqrt{\left(x+5\right)\left(x+2\right)}=\sqrt{x^2+7x+10}\\a^2-b^2=x+5-x-2=3\end{matrix}\right.\)

PT trở thành: \(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)

\(\Leftrightarrow\left(a-b\right)\left(ab+1\right)=\left(a-b\right)\left(a+b\right)\)

\(\Leftrightarrow\left(a-b\right)\left(ab+1-a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(b-1\right)\left(a-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\left(loại\right)\\a=1\\b=1\end{matrix}\right.\)

+ Với a=1

\(\Rightarrow\sqrt{x+5}=1\Leftrightarrow x+5=1\Leftrightarrow x=-4\left(ktm\right)\)

+ Với b=1

\(\Rightarrow\sqrt{x+2}=1\Leftrightarrow x+2=1\Leftrightarrow x=-1\left(tm\right)\)

Vậy \(S=\left\{-1\right\}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+5}=a\\\sqrt{x+2=b}\end{matrix}\right.\)

Thì được:

\(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)

\(\Leftrightarrow\left(a-1\right)\left(b-1\right)\left(a-b\right)=0\)

Làm tiếp

\(ĐK:x\ge-2\)

\(PT\Leftrightarrow\dfrac{x+5-x-2}{\sqrt{x+5}+\sqrt{x+2}}\left(1+\sqrt{x^2+7x+10}\right)=3\\ \Leftrightarrow\dfrac{3\left(1+\sqrt{\left(x+5\right)\left(x+2\right)}\right)}{\sqrt{x+5}+\sqrt{x+2}}=3\\ \Leftrightarrow1+\sqrt{\left(x+5\right)\left(x+2\right)}=\sqrt{x+5}+\sqrt{x+2}\\ \Leftrightarrow\left(\sqrt{x+5}-1\right)\left(1-\sqrt{x+2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=1\\\sqrt{x+2}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+5=1\\x+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\ \Leftrightarrow x=-1\)

<=><=>(X+1)(Y+1)=6 và (x+1)^3+(y+1)^3=35đặt X+1;Y+1 biến đổi vế 2 giải ra đc(1;2);(2;1)

b,<=>\(\left[\sqrt{2}+1\right]^x+\left[\sqrt{2}-1\right]^x=6\)

<=>\(2\sqrt{2}^x+2=6\)

<=>x=2

mong mọi người giải giúp em vs