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CM
2
10 tháng 9 2021
\(a,\Leftrightarrow6x^2-6x^2-11x+10=-12\\ \Leftrightarrow-11x=-22\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^3+27-x^3-2x=12-5x\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\\ c,\Leftrightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x-8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
10 tháng 9 2021
a: ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)
\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)
\(\Leftrightarrow-11x=-22\)
hay x=2
b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=12-5x\)
\(\Leftrightarrow x^3+27-x^3-2x+5x=12\)
\(\Leftrightarrow x=-5\)
L
1 tháng 8 2021
a) \(\left(x-4\right)^2-\left(x-4\right)=0\)
\(\left(x-4\right)\left(x-4-1\right)=0\)
\(\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
b) \(5x^2\left(x-7\right)+7\left(x-7\right)=0\)
\(\left(x-7\right)\left(5x^2+7\right)=0\)
\(\left[{}\begin{matrix}x-7=0\\5x^2+7=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=7\\x^2=\dfrac{-7}{5}\end{matrix}\right.\)
\(x=7\)
c) \(x^2\left(x-3\right)-\left(x-3\right)=0\)
\(\left(x-3\right)\left(x^2-1\right)=0\)
\(\left[{}\begin{matrix}x=3\\x=\pm1\end{matrix}\right.\)
1 tháng 8 2021
a) (x - 4)^2=(x - 4)
(x - 4) (x -4)=(x -4 )
(x - 4) (x - 4)-(x - 4)=0
(x-4) (x-4-1)=0
(x-4) (x-5)=0
TH1:x-4=0 TH2:x-5=0
x=4 x=5
27 tháng 8 2021
`a)5x(x-1)-(x+2)(5x-7)=6`
`<=>5x^2-5x-(5x^2-7x+10x-14)=6`
`<=>5x^2-5x-(5x^2+3x-14)=6`
`<=>-8x+14=6`
`<=>8x=8<=>x=1`
Vậy `x=1`
`b)(x+2)^2-(x^2-4)=0`
`<=>x^2+4x+4-x^2+4=0`
`<=>4x+8=0`
`<=>4x=-8`
`<=>x=-2`
Vậy `x=-2`
1 tháng 12 2021
\(a,\Leftrightarrow x^2+14x+49-x^2+3x=12\\ \Leftrightarrow17x=-37\Leftrightarrow x=-\dfrac{37}{17}\\ b,\Leftrightarrow x^2-x-2x+2=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
1 tháng 12 2021
a) \(x^2+2x7+49-x^2+3x=12\Leftrightarrow17x=-37\Leftrightarrow x=\dfrac{-37}{17}\)
b) \(x^2-2x-x+2=0\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)=\left(0\right)\Leftrightarrow x=1,x=2\)
18 tháng 8 2023
(2-x)^3+(2+x)^3-12x(x+1)=0
=>\(8-12x+6x^2-x^3+8+12x+6x^2+x^3-12x\left(x+1\right)=0\)
=>\(12x^2+16-12x^2-12x=0\)
=>16-12x=0
=>4-3x=0
=>x=4/3
RH
31 tháng 8 2021
a) (x - 3)2 - 5.(x - 2) + 5 = 0.
<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0
<=> x^2 - 11x + 24 = 0
<=> (x-3)(x-8)=0
<=> x = 3 hoặc x = 8
Bài 1.
26 tháng 10 2021
a: \(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
c: \(\Leftrightarrow\left(x-1\right)\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
26 tháng 10 2021
a) \(x^2-6x=0\\ \Leftrightarrow x\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\Leftrightarrow\left(3x-1-x-5\right)\left(3x-1+x+5\right)=0\\ \Leftrightarrow\left(2x-6\right)\left(4x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c) \(9x^2\left(x-1\right)=x-1\\ \Leftrightarrow\left(9x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(3x+1\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
d) \(x^2-4=\left(x-2\right)^2\\ \Leftrightarrow\left(x-2\right)\left(x+2-x+2\right)=0\\ \Leftrightarrow4\left(x-2\right)=0\\ \Leftrightarrow x=2\)
e) \(\Leftrightarrow\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
f) \(x^3-0,36=0\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)
g) \(\Leftrightarrow\left(5x-1\right)\left(x-2018\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2018\end{matrix}\right.\)
h) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
18 tháng 5 2021
a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)
\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)
\(\Leftrightarrow-9x=18\)
hay x=-2
Vậy: S={-2}
b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)
\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
\(\Leftrightarrow14x=7\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)
\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)
\(\Leftrightarrow5.2x=-6.5\)
hay \(x=-\dfrac{5}{4}\)
Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)
18 tháng 5 2021
d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x+16=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
Vậy: S={-5}
e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
Vậy: S={0}
13 tháng 12 2023
a: \(3\left(x-3\right)-6x=0\)
=>\(3x-9-6x=0\)
=>-3x-9=0
=>3x+9=0
=>3x=-9
=>\(x=-\dfrac{9}{3}=-3\)
b: Đề thiếu vế phải rồi bạn
c: \(2\left(x-3\right)+3x=9\)
=>2x-6+3x=9
=>5x-6=9
=>5x=6+9=15
=>x=15/5=3
d: \(x\left(x-11\right)+2\left(x-11\right)=0\)
=>\(\left(x-11\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-11=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-2\end{matrix}\right.\)
e: \(x\left(x+2\right)+8=x^2\)
=>\(x^2+2x+8=x^2\)
=>2x+8=0
=>2x=-8
=>x=-8/2=-4
f: \(8\left(x+1\right)+2x=-2\)
=>\(8x+8+2x=-2\)
=>10x=-2-8=-10
=>\(x=-\dfrac{10}{10}=-1\)
g: 12-3(x+2)=0
=>3(x+2)=12
=>x+2=12/3=4
=>x=4-2=2
a) x^2 + 14x + 49 - x^2 + 3x = 12
<=> 17x = -37
<=> x = -37/17
b) x^2 + 2x +1 - x^2 + 4 = 0
<=> 2x = -5
<=> x = -5/2
a) \(\left(x+7\right)^2-x\left(x-3\right)=12\)
\(\Leftrightarrow x^2+14x+49-x^2+3x=12\)
\(\Leftrightarrow17x=-37\)
\(\Leftrightarrow x=\frac{-37}{17}\)
Vậy x = -37/17
b) \(\left(x+1\right)^2-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow x^2+2x+1-\left(x^2-4\right)=0\)
\(\Leftrightarrow x^2+2x+1-x^2+4=0\)
\(\Leftrightarrow2x=-5\)
\(\Leftrightarrow x=\frac{-5}{2}\)
Vậy x = -5/2