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25 tháng 8 2020
a) Ta có: \(5x\left(x+1\right)-5\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[5x-5\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(5x-5x+10\right)=0\)
\(\Leftrightarrow10\left(x+1\right)=0\)
mà \(10\ne0\)
nên x+1=0
hay x=-1
Vậy: x=-1
b) Ta có: \(\left(4x+1\right)\left(x-2\right)-\left(2x-3\right)=4\)
\(\Leftrightarrow4x^2-8x+x-2-2x+3-4=0\)
\(\Leftrightarrow4x^2-9x-3=0\)
\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot\frac{9}{4}+\frac{81}{16}-\frac{129}{16}=0\)
\(\Leftrightarrow\left(2x-\frac{9}{4}\right)^2=\frac{129}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{9}{4}=\frac{\sqrt{129}}{4}\\2x-\frac{9}{4}=-\frac{\sqrt{129}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{9+\sqrt{129}}{4}\\2x=\frac{9-\sqrt{129}}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{9+\sqrt{129}}{8}\\x=\frac{9-\sqrt{129}}{8}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{9+\sqrt{129}}{8};\frac{9-\sqrt{129}}{8}\right\}\)
c) Ta có: \(2x^3-18x=0\)
\(\Leftrightarrow2x\left(x^2-9\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)
mà \(2\ne0\)
nên \(\left[{}\begin{matrix}x=0\\x+3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=3\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-3;3\right\}\)
d) Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)
\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)
\(\Leftrightarrow-13x-2=11\)
\(\Leftrightarrow-13x=13\)
hay x=-1
Vậy: x=-1
e) Ta có: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-\left(x^3+8\right)=3-3x^2\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8-3+3x^2=0\)
\(\Leftrightarrow3x-12=0\)
\(\Leftrightarrow3x=12\)
hay x=4
Vậy: x=4
f) Ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-1\)
\(\Leftrightarrow6x^2-\left(6x^2-4x+15x-10\right)+1=0\)
\(\Leftrightarrow6x^2-6x^2+4x-15x+10+1=0\)
\(\Leftrightarrow-11x+11=0\)
\(\Leftrightarrow-11x=-11\)
hay x=1
Vậy: x=1
14 tháng 12 2021
\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)
NT
1
29 tháng 10 2018
a) x(x+1)+3(x+1)=0
⇌ (x+1)(x+3)=0
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
b)3x(12x-4)-2x(18x+3)=0
⇒36x2-12x-36x2+6x=0
⇒ -6x = 0
⇒ x=0
MC
Tìm x
a) (12x-5)(3x-1)-(18x-1)(2x+3)=5
b) (x+2)(x-3)-(x-2)(x+5)=2(x+3)
c) (2x+3)(2x-1)-(2x+5)-(2x-3)=12
0
30 tháng 9 2020
Bài 1.
1) ( 2x + 1 )3 - ( 2x + 1 )( 4x2 - 2x + 1 ) - 3( 2x - 1 ) = 15
<=> 8x3 + 12x2 + 6x + 1 - [ ( 2x )3 - 13 ] - 6x + 3 = 15
<=> 8x3 + 12x2 + 4 - 8x3 + 1 = 15
<=> 12x2 + 15 = 15
<=> 12x2 = 0
<=> x = 0
2) x( x - 4 )( x + 4 ) - ( x - 5 )( x2 + 5x + 25 ) = 13
<=> x( x2 - 16 ) - ( x3 - 53 ) = 13
<=> x3 - 16x - x3 + 125 = 13
<=> 125 - 16x = 13
<=> 16x = 112
<=> x = 7
Bài 2.
A = ( x + 5 )( x2 - 5x + 25 ) - ( 2x + 1 )3 - 28x3 + 3x( -11x + 5 )
= x3 + 53 - ( 8x3 + 12x2 + 6x + 1 ) - 28x3 - 33x2 + 15x
= -27x3 + 125 - 8x3 - 12x2 - 6x - 1 - 33x2 + 15x
= -33x3 - 45x2 + 9x + 124 ( có phụ thuộc vào biến )
B = ( 3x + 2 )3 - 18x( 3x + 2 ) + ( x - 1 )3 - 28x3 + 3x( x - 1 )
= 27x3 + 54x2 + 36x + 8 - 54x2 - 36x + x3 - 3x2 + 3x - 1 - 28x3 + 3x2 - 3x
= 7 ( đpcm )
C = ( 4x - 1 )( 16x2 + 4x + 1 ) - ( 4x + 1 )3 + 12( 4x + 1 )3 + 12( 4x + 1 ) - 15
= ( 4x )3 - 13 - [ ( 4x + 1 )3 - 12( 4x + 1 )3 - 12( 4x + 1 ) ] - 15
= 64x3 - 1 - ( 4x + 1 )[ ( 4x + 1 )2 - 12( 4x + 1 )2 - 12 ] - 15
= 64x3 - 16 - ( 4x + 1 )[ 16x2 + 8x + 1 - 12( 16x2 + 8x + 1 ) - 12 ]
= 64x3 - 16 - ( 4x + 1 )( 16x2 + 8x - 11 - 192x2 - 96x - 12 )
= 64x3 - 16 - ( 4x + 1 )( -176x2 - 88x - 23 )
= 64x3 - 16 - ( -704x3 - 528x2 - 180x - 23 )
= 64x3 - 16 + 704x3 + 528x2 + 180x + 23
= 768x3 + 528x2 + 180x + 7 ( có phụ thuộc vào biến )
\(a,2x^2-18x+28=0\)
\(\Leftrightarrow2\left(x^2-9x+14\right)=0\)
\(\Leftrightarrow x^2-9x+14=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=2\end{matrix}\right.\)
\(b,\dfrac{x-2}{x^2-9}+\dfrac{3x-1}{x+3}=\dfrac{2x+1}{x-3}+1\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{x-2}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(3x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(2x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-1=0\)
\(\Leftrightarrow\dfrac{x-2}{\left(x-3\right)\left(x+3\right)}+\dfrac{3x^2-10x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{2x^2+7x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=0\)\(\Rightarrow x-2+3x^2-10x+3-2x^2-7x-3-x^2+9=0\)
\(\Leftrightarrow-16x+7=0\)
\(\Leftrightarrow-16x=-7\)
\(\Leftrightarrow x=\dfrac{7}{16}\left(tm\right)\)
\(VậyS=\left\{\dfrac{7}{16}\right\}\)
a: =>x^2-9x+14=0
=>(x-2)(x-7)=0
=>x=2 hoặc x=7
b: =>x-2+(3x-1)(x-3)=(2x+1)(x+3)+x^2-9
=>x-2+3x^2-9x-x+3=2x^2+7x+3+x^2-9
=>3x^2-9x+1=3x^2+7x-6
=>-16x=-7
=>x=7/16