giải phương trình ( e cần gấp giúp e vs ạ )
1) \(\sqrt{x}\)+\(\sqrt{3x-2}\)=x2+1
2) x2-2x-2\(\sqrt{2x+1}\)-2=0
3)x2-x+2\(\sqrt{x-1}\)=0
4) 2x2+2x-1=\(\sqrt{4x+1}\)
5) 8x2-3x+6=4x\(\sqrt{3x^2+x+2}\)
6)\(\sqrt{x-2000}\)+\(\sqrt{y-2001}\)+\(\sqrt{z-2002}\)=\(\frac{1}{2}\)(x+y+z)- 3000
7) x+4\(\sqrt{x+3}\)+2\(\sqrt{3-2x}\)=1
1/
ĐKXĐ: \(x\ge\frac{2}{3}\)
\(\Leftrightarrow2x^2+2-2\sqrt{x}-2\sqrt{3x-2}=0\)
\(\Leftrightarrow2x^2-4x+2+\left(x+1-2\sqrt{x}\right)+\left(3x-1-2\sqrt{3x-2}\right)=0\)
\(\Leftrightarrow2\left(x-1\right)^2+\frac{\left(x-1\right)^2}{x+1+2\sqrt{x}}+\frac{9\left(x-1\right)^2}{3x-1+2\sqrt{3x-2}}=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(2+\frac{1}{x+1+2\sqrt{x}}+\frac{9}{3x-1+2\sqrt{3x-2}}\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\) (ngoặc to luôn dương)
\(\Leftrightarrow x=1\)
2/
ĐKXĐ: \(x\ge-\frac{1}{2}\)
\(\Leftrightarrow x^2-3x-4+\left(x+2-2\sqrt{2x+1}\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+1\right)+\frac{x^2-4x}{x+2+2\sqrt{2x+1}}=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+1\right)+\frac{x\left(x-4\right)}{x+2+2\sqrt{2x+1}}=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+1+\frac{x}{x+2+2\sqrt{2x+1}}\right)=0\)
\(\Leftrightarrow x-4=0\) (ngoặc to luôn dương)
\(\Leftrightarrow x=4\)
3/
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow x\left(x-1\right)+2\sqrt{x-1}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(x\sqrt{x-1}+2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=0\\x\sqrt{x-1}=-2\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow x-1=0\Rightarrow x=1\)
4/
Bạn coi lại đề, nghiệm pt này rất xấu (là nghiệm của pt bậc 3 không phân tích được về hằng đẳng thức, chương trình Việt Nam ko học)
5/
\(\Leftrightarrow8x^2-3x+6-4x\sqrt{3x^2+x+2}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{3x^2+x+2}+3x^2+x+2\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{3x^2+x+2}\right)^2+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-\sqrt{3x^2+x+2}=0\\x-2=0\end{matrix}\right.\) \(\Rightarrow x=2\)
6/
ĐKXĐ: ....
\(\Leftrightarrow\left(x-2000-2\sqrt{x-2000}+1\right)+\left(y-2001-2\sqrt{y-2001}+1\right)+\left(z-2002-2\sqrt{z-2002}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2000}-1=0\\\sqrt{y-2001}-1=0\\\sqrt{z-2002}-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2001\\y=2002\\z=2003\end{matrix}\right.\)
7/
ĐKXĐ: \(-3\le x\le\frac{2}{3}\)
\(\Leftrightarrow2x+8\sqrt{x+3}+4\sqrt{3-2x}=2\)
\(\Leftrightarrow8\sqrt{x+3}+4\sqrt{3-2x}-\left(3-2x\right)+1=0\)
\(\Leftrightarrow8\sqrt{x+3}+\sqrt{3-2x}\left(4-\sqrt{3-2x}\right)+1=0\)
Do \(x\ge-3\Rightarrow3-2x\le9\Rightarrow\sqrt{3-2x}\le3\)
\(\Rightarrow4-\sqrt{3-2x}>0\)
\(\Rightarrow VT>0\)
Phương trình vô nghiệm (bạn coi lại đề)
