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ĐKXĐ : \(-4\le x\le4\)
TA CÓ : \(\left(\sqrt{x+4}-2\right)\left(\sqrt{4-x}+2\right)=2x\)
\(\Leftrightarrow\left[\left(\sqrt{x+4}-2\right)\left(\sqrt{x+4}+2\right)\right]\left(\sqrt{4-x}+2\right)=2x\left(\sqrt{x+4}+2\right)\)
\(\Leftrightarrow\left[x+4-4\right]\left(\sqrt{4-x}+2\right)-2x\left(\sqrt{x+4}+2\right)=0\)
\(\Leftrightarrow x\left(\sqrt{4-x}+2\right)-2x\left(\sqrt{x+4}+2\right)=0\)
\(\Leftrightarrow x\left[\sqrt{4-x}+2-2\sqrt{x+4}-4\right]=0\)
\(\Leftrightarrow x=0\)HOẶC \(\sqrt{4-x}-2\sqrt{x+4}-2=0\)
VỚI \(\sqrt{4-x}-2\sqrt{x+4}-2=0\)
\(\Leftrightarrow\sqrt{4-x}-2=2\sqrt{x+4}\)
\(\Leftrightarrow4-x+4-4\sqrt{4-x}=4x+16\)
\(\Leftrightarrow8-x-4x-16=4\sqrt{4-x}\)
\(\Leftrightarrow-5x-8=4\sqrt{4-x}\)ĐK : \(-4\le x\le\frac{-8}{5}\)
\(\Leftrightarrow\left[-\left(5x+8\right)\right]^2=16\left(4-x\right)\)
\(\Leftrightarrow25x^2+64+80x=64-16x\)
\(\Leftrightarrow25x^2+96x=0\Leftrightarrow x\left(25x+96\right)=0\)
\(\Leftrightarrow x=0\)HOẶC \(x=\frac{-96}{25}\)(THỎA MÃN ĐK )
VẬY PT CÓ 2 NGHIỆM \(x\in\left[0;\frac{-96}{25}\right]\)
P/S : CÁCH CỦA MÌNH KHÁ DÀI VÀ CHI TIẾT QUÁ . BẠN CÓ THỂ THAM KHẢO CÁCH KHÁC NHANH HƠN :>
\(ĐK:x\ge2\)
\(x^2-5x+4=2\sqrt{2x-4}\)
<=>\(x^2-5x+4=2\sqrt{2\left(x-2\right)}\)
<=>\(x^2-5x+4+x-2+2=\left(x-2\right)+2\sqrt{2\left(x-2\right)}+2\)
<=>\(x^2-4x+4=\left(\sqrt{x-2}+2\right)^2\)
<=>\(\left(x-2\right)^2=\left(\sqrt{x-2}+2\right)^2\)
<=> \(\left(x-2-\sqrt{x-2}-2\right)\left(x-2+\sqrt{x-2}+2\right)=0\)
<=>\(\left(x-\sqrt{x-2}-4\right)\left(x+\sqrt{x-2}\right)=0\)
Xét \(x-\sqrt{x-2}-4=0\)
<=>\(x^2-8x+16=x-2\)
<=>\(x^2-9x+18=0\)
=> x=6;3(nhận)
Xet1\(x+\sqrt{x-2}=0\)
Do x\(\ge2\)=> pt vô nghiệm
Vậy ...
Em xin phép làm bài EZ nhất :)
4,ĐK :\(\forall x\in R\)
Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))
\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)
\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)
\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy ....
\(\sqrt{1+2005^2+\dfrac{2005^2}{2006^2}}=\dfrac{1}{2006}\sqrt{2006^2+2005^2+\left(2005.2006\right)^2}\)
\(=\dfrac{1}{2006}\sqrt{\left(2006-2005\right)^2+2.2005.2006+\left(2005.2006\right)^2}\)
\(=\dfrac{1}{2006}\sqrt{1+2.2005.2006+\left(2005.2006\right)^2}\)
\(=\dfrac{1}{2006}\sqrt{\left(2005.2006+1\right)^2}=\dfrac{2005.2006+1}{2006}=2005+\dfrac{1}{2006}\)
Phương trình tương đương:
\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=2005+\dfrac{1}{2006}+\dfrac{2005}{2006}\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2006\)
TH1: \(x\ge2\): \(x-1+x-2=2006\Rightarrow2x=2009\Rightarrow x=\dfrac{2009}{2}\)
TH2: \(x\le1\) : \(1-x+2-x=2006\Rightarrow-2x=2003\Rightarrow x=\dfrac{-2003}{2}\)
TH3: \(1< x< 2:\) \(x-1+2-x=2006\Rightarrow3=2006\) (vô nghiệm)
Vậy \(\left[{}\begin{matrix}x=\dfrac{2009}{2}\\x=\dfrac{-2003}{2}\end{matrix}\right.\)
\(\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)
\(\Leftrightarrow\sqrt{2x-5-6\sqrt{2x-5}+9}+\sqrt{2x-5+2\sqrt{2x-5}+1}=4\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}+1\right)^2}=4\)
\(\Leftrightarrow\left|\sqrt{2x-5}-3\right|+\left|\sqrt{2x-5}+1\right|=4\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{2x-5}-3+\sqrt{2x-5}+1=4\\\sqrt{2x-5}-3+\sqrt{2x-5}+1=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2\sqrt{2x-5}-2=4\\2\sqrt{2x-5}-2=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2\sqrt{2x-5}=6\\2\sqrt{2x-5}=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{2x-5}=3\\\sqrt{2x-5}=-1\left(L\right)\end{cases}}\)
\(\Leftrightarrow2x-5=9\)
\(\Leftrightarrow x=7\)