\(x=-1\)Giao lưu thôi nhé

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

\(\Leftrightarrow\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(\sqrt{\left(x+5\right)\left(x+2\right)}+1\right)=3\)

Đặt \(\hept{\begin{cases}\sqrt{x+5}=a\left(a\ge0\right)\\\sqrt{x+2}=b\left(b\ge0\right)\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a^2-b^2=3\\\left(a-b\right)\left(ab+1\right)=3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a^2-b^2=3\\\left(a-b\right)\left(ab+1-a-b\right)=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a^2-b^2=3\\\left(a-b\right)\left(a-1\right)\left(b-1\right)=0\end{cases}}\)

Với a = b thì

\(\sqrt{x+5}=\sqrt{x+2}\Leftrightarrow0x=3\left(l\right)\)

Với a = 1 thì

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

Với b = 1 thì

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

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: \(S=\left\{0;-\sqrt{5}\right\}\)

\(ĐK:x,y\in R\)

Từ 2 PT \(\Leftrightarrow\sqrt{\left(x+1\right)^2+\left(y-1\right)^2}=\sqrt{\left(x-5\right)^2+\left(y+1\right)^2}\)

\(\Leftrightarrow x^2+2x+y^2-2y+2=x^2-10x+y^2+2y+26\\ \Leftrightarrow12x-4y-24=0\\ \Leftrightarrow3x-y-6=0\\ \Leftrightarrow y=3x-6\)

Thay vào \(PT\left(1\right)\Leftrightarrow\sqrt{\left(x-1\right)^2+\left(3x-8\right)^2}=\sqrt{\left(x+1\right)^2+\left(3x-7\right)^2}\)

\(\Leftrightarrow10x^2-50x+65=10x^2-40x+50\\ \Leftrightarrow10x=15\Leftrightarrow x=\dfrac{3}{2}\Leftrightarrow y=-\dfrac{3}{2}\)

Vậy hệ có nghiệm \(\left(x;y\right)=\left(\dfrac{3}{2};-\dfrac{3}{2}\right)\)

ĐKXĐ: \(0\le x\le5\).

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{5-x}=b\end{matrix}\right.\left(a,b\ge0\right)\).

PT đã cho tương đương với: \(\left(8-ab\right)\left(a-b\right)=2\left(a-b\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\ab=6\end{matrix}\right.\).

+) \(a=b\Leftrightarrow\sqrt{x}=\sqrt{5-x}\Leftrightarrow x=2,5\left(TMĐK\right)\).

+) \(ab=6\Leftrightarrow\sqrt{x\left(5-x\right)}=6\Leftrightarrow x^2-5x+6=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(TMĐK\right)\\x=3\left(TMĐK\right)\end{matrix}\right.\).

Vậy...

ĐK: \(0\le x\le5\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{5-x}=b\end{matrix}\right.\left(a,b\ge0\right)\)

\(pt\Leftrightarrow\left(8-ab\right)\left(a-b\right)=2\left(a^2-b^2\right)\)

\(\Leftrightarrow\left(a-b\right)\left(8-ab-2a-2b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\ab+2a+2b=8\end{matrix}\right.\)

TH1: \(a=b\Leftrightarrow\sqrt{x}=\sqrt{5-x}\Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

TH2: \(ab+2a+2b=8\)

\(\Leftrightarrow\sqrt{5x-x^2}+2\sqrt{5-x}+2\sqrt{x}=8\)

\(\Leftrightarrow\left(\sqrt{5-x}+\sqrt{x}-3\right)\left(\sqrt{5-x}+\sqrt{x}+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{5-x}+\sqrt{x}=-7\left(l\right)\\\sqrt{5-x}+\sqrt{x}=3\end{matrix}\right.\)

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

\(\Leftrightarrow5+2\sqrt{5x-x^2}=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

Vậy ...