Giải các phương trình sau:
a) x − 2 x + x x + 2 = 2 ;
b) 2 x + 1 − 1 x − 2 = 3 x − 11 x + 1 x − 2 ;
c) 5 + 96 x 2 − 16 = 2 x − 1 x + 4 + 3 x − 1 x − 4 ;
d) 2 x + 2 − 2 x 2 + 16 x 3 + 8 = 5 x 2 − 2 x + 4 .
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a) Ta có: \(\left|x^2-x+2\right|-3x-7=0\)
\(\Leftrightarrow\left|x^2-x+2\right|=3x+7\)
\(\Leftrightarrow x^2-x+2=3x+7\)(Vì \(x^2-x+2>0\forall x\))
\(\Leftrightarrow x^2-x+2-3x-7=0\)
\(\Leftrightarrow x^2-4x-5=0\)
\(\Leftrightarrow x^2-5x+x-5=0\)
\(\Leftrightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
Vậy: S={5;-1}
bạn giải giúp mình câu b nữa với
mai mình phải nộp bài rồi!!!
a.\(x^2-25=8\left(5-x\right)\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)-8\left(5-x\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+8\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-13\end{matrix}\right.\)
b.\(\dfrac{x-2}{x+2}-\dfrac{2\left(x-11\right)}{x^2-4}=\dfrac{3}{x-2}\) ; \(ĐK:x\ne\pm2\)
\(\Leftrightarrow\dfrac{\left(x-2\right)\left(x-2\right)-2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\left(x-2\right)^2-2\left(x-11\right)=3\left(x+2\right)\)
\(\Leftrightarrow x^2-4x+4-2x+22=3x+6\)
\(\Leftrightarrow x^2-9x+20=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
\(a,\\ \Leftrightarrow3x-x=10-6\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\\ b,\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
\(a,\Leftrightarrow3x-x=10-6\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=\dfrac{4}{2}=2\)
Vậy phương trình có tập nghiệm S = \(\left\{2\right\}\)
\(b,\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow x+1=0\) hoặc \(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=-1\) \(\Leftrightarrow x=2\)
Vậy phương trình có tập nghiệm S = \(\left\{-1;2\right\}\)
b: =>(x-5)2+6(x-5)=0
=>(x-5)(x+1)=0
=>x=5 hoặc x=-1
c: \(\Leftrightarrow x^2-x+2=2\)
=>x(x-1)=0
=>x=0(loại) hoặc x=1(nhận)
\(a,\Leftrightarrow x^2+2x+1+2x+3-2\sqrt{2x+3}+1=0\\ \Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1\\2x+3=1\end{matrix}\right.\Leftrightarrow x=-1\left(N\right)\)
\(b,\Leftrightarrow3x^2+3x-2\sqrt{x^2+x}=0\left(x\le-1;x\ge0\right)\\ \Leftrightarrow3x\left(x-1\right)-2\sqrt{x\left(x+1\right)}=0\\ \Leftrightarrow\sqrt{x\left(x+1\right)}\left(3\sqrt{x\left(x-1\right)}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x\left(x-1\right)=0\\\sqrt{x\left(x-1\right)}=\dfrac{2}{3}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x^2-x-\dfrac{4}{9}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\9x^2-9x-4=0\left(1\right)\end{matrix}\right.\)
\(\Delta\left(1\right)=81-4\left(-4\right)\cdot9=225\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{9-15}{18}\\x=\dfrac{9+15}{18}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(N\right)\\x=1\left(N\right)\\x=-\dfrac{1}{3}\left(L\right)\\x=\dfrac{4}{3}\left(N\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{4}{3}\end{matrix}\right.\)
\(a,=>x^3-2x^2+4x+2x^2-4x+8-x^3+2x-15=0\)
\(< =>2x-7=0< =>x=\dfrac{7}{2}\)
b,\(=>x\left(x^2-25\right)-\left(x+2\right)\left(x^2-2x+4\right)-3=0\)
\(< =>x^3-25x-x^3+2x^2-4x-2x^2+4x-8-3=0\)
\(< =>-25x-11=0\)
\(< =>x=-0,44\)