tìm x:
a,2(x+5)-x^2-5x=0
b,x^2-2x=8
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C
3
31 tháng 7 2021
a) \(\text{5x(x-2)+(2-x)=0}\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\text{x(2x-5)-10x+25=0}\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)
31 tháng 7 2021
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)
\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)
\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)
31 tháng 10 2021
\(a,\Leftrightarrow\left(x-2\right)\left(5x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{2}{5}\end{matrix}\right.\\ b,\Leftrightarrow2x^2+2x-x^2+4x-4-6=0\\ \Leftrightarrow x^2+6x-10=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{19}\\x=-3-\sqrt{19}\end{matrix}\right.\\ c,\Leftrightarrow2x^2-2x+9x-9=0\\ \Leftrightarrow\left(2x+9\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{2}\\x=1\end{matrix}\right.\)
17 tháng 8 2021
a: Ta có: \(40x^4+5x=0\)
\(\Leftrightarrow5x\left(8x^3+1\right)=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b: Ta có: \(8x^2-2x-1=0\)
\(\Leftrightarrow8x^2-4x+2x-1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 12 2021
a.
\(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
b.
\(6x^2-7x+2=0\)
\(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow3x\left(2x-1\right)-2\left(2x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
TL
3
1 tháng 7 2021
\(a,\) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+6\)
\(< =>20x-5x^2+5x^2-12-x-6=0\)
\(< =>19x-18=0\)
\(< =>x=\dfrac{18}{19}\)
\(b,\left(2x-7\right)\left(5+4x\right)-8\left(x^2-4x+5\right)=-30\)
\(< =>10x+8x^2-35-28x-8x^2+24x-40+30=0\)
\(< =>6x-45=0< =>x=\dfrac{45}{6}=7,5\)
PT
1 tháng 7 2021
a) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+\Rightarrow6\\ \Leftrightarrow20x-5x^2+5x^2-12=x+6\\ \Leftrightarrow20x-12=x+6\\\Rightarrow20x-x=6+12\\ \Rightarrow19x=18\\ \Rightarrow x=\dfrac{18}{19}\)
b) \(\left(2x-7\right)\left(5+4x\right)-8\left(x^2-3x+5\right)=-30\\ \Rightarrow10x+8x^2-35-28x-8x^2+24x-40=-30\\ \Rightarrow6x-75=-30\\ \Rightarrow6x=45\\ \Rightarrow x=\dfrac{15}{2}\)
S
1
25 tháng 12 2021
a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)
C
1
7 tháng 9 2021
a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)
\(\Leftrightarrow2x-1=0\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x^3-\left(x+1\right)\left(x^2-x+1\right)=x\)
\(\Leftrightarrow x^3-x^3-1=x\)
hay x=-1
c: Ta có: \(56x^4+7x=0\)
\(\Leftrightarrow7x\left(8x^3+1\right)=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d: Ta có: \(x^2-5x-24=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
HU
1
3 tháng 5 2021
a,(2x+7)+135=0 b, 1/2x-2/5=1/5
2x+7=0-135 1/2x=1/5+2/5
2x+7=-135 1/2x=3/5
2x=-135-7 x=3/5:1/2
2x=-142 x=6/5
x=-142:2 Vậy x=6/5
x=-71
Vậy x=-71
c, 10-|x+1|=5 d, 1/2x+150%x=2014
|x+1|=10-5 2x=2014
|x+1|=5 x=2014:2
*TH1:x+1=5 *TH2:x+1=-5 x=1007
x=5-1 x=-5-1 Vậy x=1007
x=4 x=-6
Vậy x=4 hoặc x=-6
C
2
NT
24 tháng 8 2021
a. \(x^2-2x+2\left|x-1\right|-7=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-2x+2\left(x-1\right)-7=0\\x^2-2x-2\left(x-1\right)-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-9=0\\x^2-4x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=9\\\left(x-5\right)\left(x+1\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm3\\x=5\\x=-1\end{matrix}\right.\)
24 tháng 8 2021
b: Ta có: \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)
\(\Leftrightarrow\left(x^2+5x\right)^2+10\cdot\left(x^2+5x\right)=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
13 tháng 11 2021
\(a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
a)\(2\left(x+5\right)-x^2-5x=0\)
=>\(2x+10-x^2-5x=0\)
=>\(10-x^2-3x=0\)
=>x=2
ý b bạn giải PT theo cách tương tự nhegg
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