Giải các phương trình sau:
a) x 2 –l0x = -25; b) 4 x 2 - 4x = -1;
c) ( 1 - 2 x ) 2 = ( 3 x - 2 ) 2 ; d) ( x - 2 ) 3 + ( 5 - 2 x ) 3 =0.
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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.\(\dfrac{x-2}{x+2}-\dfrac{2\left(x-11\right)}{x^2-4}=\dfrac{3}{x-2}\) ; \(ĐK:x\ne\pm2\)
\(\Leftrightarrow\dfrac{\left(x-2\right)\left(x-2\right)-2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\left(x-2\right)^2-2\left(x-11\right)=3\left(x+2\right)\)
\(\Leftrightarrow x^2-4x+4-2x+22=3x+6\)
\(\Leftrightarrow x^2-9x+20=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
Lời giải:
a) $|4x^2-25|=0$
$\Leftrightarrow 4x^2-25=0$
$\Leftrightarrow (2x-5)(2x+5)=0$
$\Rightarrow x=\pm \frac{5}{2}$
b)
$|x-2|=3$
\(\Rightarrow \left[\begin{matrix} x-2=-3\\ x-2=3\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-1\\ x=5\end{matrix}\right.\)
c)
\(|x-3|=2x-1\Rightarrow \left\{\begin{matrix} 2x-1\geq 0\\ \left[\begin{matrix} x-3=2x-1\\ x-3=1-2x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\end{matrix}\right.\Rightarrow x=\frac{4}{3}\)
d)
$|x-5|=|3x-2|$
\(\Rightarrow \left[\begin{matrix} x-5=3x-2\\ x-5=2-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-3}{2}\\ x=\frac{7}{4}\end{matrix}\right.\)
a) ĐKXĐ: \(x\notin\left\{5;-5\right\}\)
Ta có: \(\dfrac{-\left(x^2+5\right)}{x^2-25}=\dfrac{3}{x+5}+\dfrac{x}{x-5}\)
\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\dfrac{x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-x^2-5}{\left(x-5\right)\left(x+5\right)}\)
Suy ra: \(3x-15+x^2+5x+x^2+5=0\)
\(\Leftrightarrow2x^2+8x-10=0\)
\(\Leftrightarrow2x^2+10x-2x-10=0\)
\(\Leftrightarrow2x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\2x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={1}
`a,(-(x^2+5))/(x^2-25)=3/(x+5)+x/(x-5)`
`ĐK:x ne +-5`
`pt<=>-x^2+5=3(x-5)+x(x+5)`
`<=>-x^2+5=3x-15+x^2+5x`
`<=>-x^2+5=x^2+8x-15`
`<=>2x^2+8x-20=0`
`<=>x^2+4x-5=0`
`<=>x^2-x+5x-5=0`
`<=>x(x-1)+5(x-1)=0`
`<=>` $\left[ \begin{array}{l}x=1\\x=-5\end{array} \right.$
Vậy `S={1,-5}`
a) \(PT\Leftrightarrow3x-2x=2-3\Leftrightarrow x=-1\)
Vậy: \(S=\left\{-1\right\}\)
b) \(PT\Leftrightarrow-2x+3x=-7+22\Leftrightarrow x=15\)
Vậy: \(S=\left\{15\right\}\)
c) \(PT\Leftrightarrow8x-5x=3+12\Leftrightarrow3x=15\Leftrightarrow x=5\)
Vậy: \(S=\left\{5\right\}\)
d) \(PT\Leftrightarrow x+4x-2x=12+25-1\Leftrightarrow3x=36\Leftrightarrow x=12\)
Vậy: \(S=\left\{12\right\}\)
e) \(PT\Leftrightarrow x+2x+3x-3x=19+5\Leftrightarrow3x=24\Leftrightarrow x=8\)
Vậy: \(S=\left\{8\right\}\)
a)3x-2=2x-3
=>x=-1
b)7-2x=22-3x
=>x=15
c)8x-3=5x+12
=>3x=15
=>x=5
d)x-12+4x=25+2x-1
=>3x=12
=>x=4
e)x+2x+3x-19=3x+5
=>3x=24
=>x=8
a: =>3x^2-3x-2x+2=0
=>(x-1)(3x-2)=0
=>x=2/3 hoặc x=1
b: =>2x^2=11
=>x^2=11/2
=>\(x=\pm\dfrac{\sqrt{22}}{2}\)
c: Δ=5^2-4*1*7=25-28=-3<0
=>PTVN
f: =>6x^4-6x^2-x^2+1=0
=>(x^2-1)(6x^2-1)=0
=>x^2=1 hoặc x^2=1/6
=>\(\left[{}\begin{matrix}x=\pm1\\x=\pm\dfrac{\sqrt{6}}{6}\end{matrix}\right.\)
d: =>(5-2x)(5+2x)=0
=>x=5/2 hoặc x=-5/2
e: =>4x^2+4x+1=x^2-x+9 và x>=-1/2
=>3x^2+5x-8=0 và x>=-1/2
=>3x^2+8x-3x-8=0 và x>=-1/2
=>(3x+8)(x-1)=0 và x>=-1/2
=>x=1
a) x = 5. b) x = 1 2 .
c) x = 3 5 hoặc x = 1. d) x = 3.
\(a,x^2-10x=-25\)
\(< =>x^2-10x+25=0\)
\(< =>\left(x-5\right)^2=0< =>x=5\)
b, \(4x^2-4x=-1\)
\(< =>4x^2-4x+1=0\)
\(< =>\left(2x-1\right)^2=0< =>x=\frac{1}{2}\)