\(3,\\a,đkx\le1\\ \sqrt{25}.\sqrt{\left(1-x\right)^2}=10\\ 5\left(1-x\right)=10\\ 1-x=2\\ x=-1\left(thoanman\right)\\ b,đkx\ge-\dfrac{1}{2}\\ \left(\sqrt{2x+1}\right)^2=\left(\sqrt{x-9}\right)^2\\ 2x+1=x-9\\ 2x-x=-9-1\\ x=-10\left(kothoanman\right)=>\varnothing\\\)
\(a)5(1-x)-10=0\)\(=>x=1\)
\(b)2x+1=x-9=>x=-10\)(bình phuong 2 vế )
c)\(DK.x>=3\)
\(3\sqrt{\dfrac{25\left(x-3\right)}{9}}+2\sqrt{x-3}-\dfrac{1}{2}2\sqrt{x-3}=12\)
\(3\sqrt{x-3}+\sqrt{x-3}=12=>4\sqrt{x-3}=12\)
=>\(\sqrt{x-3}=3\)=>\(x-3=9=>x=12\)
D)
Mình làm trên Word nên hoi khó nhìn
\(d;pt\Leftrightarrow\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{\left(x+2\right)\left(x+5\right)}\right)=3\)\(\left(1\right)\)\(\left(đkxđ:\left[{}\begin{matrix}x\ge-2\\x\le-5\end{matrix}\right.\right)\)
\(đặt:\)\(\sqrt{x+5}=a;\sqrt{x+2}=b\left(a;b\ge0\right)\Rightarrow a^2-b^2=3\)
\(\Rightarrow\left(1\right)\Leftrightarrow\left(a-b\right)\left(1+ab\right)=a^2-b^2\)
\(\Leftrightarrow\left(a-b\right)\left(a-1\right)\left(b-1\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\\a=1\\b=1\end{matrix}\right.\)
\(với:a=b\Rightarrow\sqrt{x+5}=\sqrt{x+2}\Leftrightarrow x+5=x+2\Leftrightarrow3=0\left(loại\right)\) \(với:a=1\Rightarrow\sqrt{x+5}=1\Leftrightarrow x=-4\left(loại\right)\)
\(với:b=1\Rightarrow\sqrt{x+2}=1\Leftrightarrow x=-1\left(tm\right)\)
Bài 3:
d) \(ĐKXĐ:x\ge-2\).
\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\)
\(\Leftrightarrow\left(\sqrt{x+5}-\sqrt{x+2}\right)\left[1+\sqrt{\left(x+5\right)\left(x+2\right)}\right]=3\)
\(\Rightarrow\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(\sqrt{x+5}+\sqrt{x+2}\right)\left[1+\sqrt{\left(x+5\right)\left(x+2\right)}\right]=3\left(\sqrt{x+5}+\sqrt{x+2}\right)\)
\(\Leftrightarrow\left[\left(x+5\right)-\left(x+2\right)\right]\left[1+\sqrt{\left(x+5\right)\left(x+2\right)}\right]=3\left(\sqrt{x+5}+\sqrt{x+2}\right)\)
\(\Leftrightarrow1+\sqrt{\left(x+5\right)\left(x+2\right)}=\sqrt{x+5}+\sqrt{x+2}\)
\(\Rightarrow\left[1+\sqrt{\left(x+5\right)\left(x+2\right)}\right]^2=\left(\sqrt{x+5}+\sqrt{x+2}\right)^2\)
\(\Leftrightarrow1+2\sqrt{\left(x+5\right)\left(x+2\right)}+\left(x+5\right)\left(x+2\right)=x+5+2\sqrt{\left(x+5\right)\left(x+2\right)}+x+2\)
\(\Leftrightarrow1+x^2+7x+10=2x+7\)
\(\Leftrightarrow x^2+5x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(nhận\right)\\x=-4\left(loại\right)\end{matrix}\right.\)
- Thử lại, ta có \(x=-1\) là nghiệm của phương trình.
- Vậy tập nghiệm \(S=\left\{-1\right\}\)