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a) Ta có: \(\left(x-2\right)\cdot x=2x\cdot\left(x+5\right)\)

\(\Leftrightarrow x\cdot\left(x-2\right)-2x\left(x+5\right)=0\)

\(\Leftrightarrow x\cdot\left[x-2-2\left(x+5\right)\right]=0\)

\(\Leftrightarrow x\left(x-2-2x-10\right)=0\)

\(\Leftrightarrow x\left(-x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vậy: S={0;-8}

b) Ta có: \(\left(2x-5\right)\left(x+11\right)=\left(5-2x\right)\left(2x+1\right)\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)-\left(5-2x\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11+2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(3x+12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\3x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{2};-4\right\}\)

c) Ta có: \(x^2+6x+9=4x^2\)

\(\Leftrightarrow\left(x+3\right)^2-\left(2x\right)^2=0\)

\(\Leftrightarrow\left(x+3-2x\right)\left(x+3+2x\right)=0\)

\(\Leftrightarrow\left(-x+3\right)\left(3x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x+3=0\\3x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-3\\3x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy: S={3;-1}

d) Ta có: \(\left(x+2\right)\left(5-4x\right)=x^2+4x+4\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x^2+4x+4\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x-x-2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(-5x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\-5x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{-2;\dfrac{3}{5}\right\}\)

a: \(\Leftrightarrow x\left(2x+10\right)-x\left(x-2\right)=0\)

=>x(2x+10-x+2)=0

=>x(x+12)=0

=>x=0 hoặc x=-12

b: \(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)

=>(2x-5)(3x+12)=0

=>x=5/2 hoặc x=-4

c: \(\Leftrightarrow\left(2x\right)^2-\left(x+3\right)^2=0\)

=>(x-3)(3x+3)=0

=>x=3 hoặc x=-1

d: \(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x-x-2\right)=0\)

=>(x+2)(-5x+3)=0

=>x=-2 hoặc x=3/5

6 tháng 2 2022

\(a,\left(x-2\right)x=2x\left(x+5\right)\)

\(\Leftrightarrow\left(x-2\right)x-2x\left(x+5\right)=0\)

\(\Leftrightarrow x.\left(x-2-2x-10\right)=0\)

\(\Leftrightarrow x\left(-x-12\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+12=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-12\end{matrix}\right.\)

Câu 5:B

Câu 4: C

Câu 3: D

Câu 2: A

Câu 1: A

31 tháng 8 2015

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0 

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52 

=> 19x = -4

=> x = -4/19

d/ 20x2 - 16x - 34 = 10x2 + 3x - 34

=> 10x2 - 19x = 0

=> x(10x - 19) = 0

=> x = 0 

hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10

Vậy x = 0 ; x = 19/10

2 tháng 1 2016

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52

=> 19x = -4

=> x = -4/19

d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34

=> 10x 2 - 19x = 0

=> x(10x - 19) = 0

=> x = 0 hoặc 10x - 19 = 0

=> 10x = 19

=> x = 19/10

Vậy x = 0 ; x = 19/10 

5 tháng 7 2018

1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)

\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)

\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)

\(=27x^3-4x^2+20x-1\)

b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)

\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)

\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)

\(=13x-28x^2-21-x^3\)

c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)

\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)

\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)

\(=16x^2-17+x^3\)

d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)

\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)

\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)

\(=-27x^2+63x-46\)

e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)

\(=12x^2-24x-6x^2-10x-4x^2\)

\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)

\(=2x^2-34x\)

f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)

\(=30x^2-25x-36x+30-3x^2-10x\)

\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)

\(=27x^2-71x+30\)

5 tháng 7 2018

2) a)\(x\left(x+3\right)-x^2=6\)

\(\Rightarrow x^2+3x-x^2=6\)

\(\Rightarrow\left(x^2-x^2\right)+3x=6\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

Vậy x=2

b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)

\(\Rightarrow2x^2-10x-2x^2-x=6\)

\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)

\(\Rightarrow-11x=6\)

\(\Rightarrow x=-\dfrac{6}{11}\)

\(\)Vậy \(x=-\dfrac{6}{11}\)

c) x(x+5)-(x+1)(x-2)=7

\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)

\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)

\(\Rightarrow6x=5\)

\(\Rightarrow x=\dfrac{5}{6}\)

Vậy x=\(\dfrac{5}{6}\)

d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)

\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)

\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)

\(\Rightarrow10x-10=10\)

\(\Rightarrow10x=20\)

\(\Rightarrow x=2\)

Vậy x=2

b: =x-2

d: \(=-x^3+\dfrac{3}{2}-2x\)

11 tháng 7 2021

\(\left(x^2+3\right)\left(3-x^2\right)\)

\(\left(x^2+3\right)\left(-x^2+3\right)\)

\(\left(-x^2+3\right).x^2+3\left(-x^2+3\right)\)

\(-x^2.x^2+3x^2+3\left(-x^2+3\right)\)

\(-x^2.x^2+3x^2-3x^2+9\)

\(-x^2.x^2+9\)

11 tháng 7 2021

\(\left(2x+5\right)\left(2x-5\right)\)

\(2x\left(2x-5\right)+5\left(2x-5\right)\)

\(4x^2-10x+5\left(2x-5\right)\)

\(4x^2-10x+10x-25\)

\(4x^2-25\)

8 tháng 9 2021

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

10 tháng 10 2020

Ko có j

20 tháng 12 2019

a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.