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19 tháng 8 2021

\(9x^2-\left(2x-1\right)^2=\left(5x-1\right)\left(x-3\right)\)

\(\Leftrightarrow\left(3x-2x+1\right)\left(3x+2x-1\right)=\left(5x-1\right)\left(x-3\right)\)

\(\Leftrightarrow\left(x+1\right)\left(5x-1\right)-\left(5x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x+1-x+3\right)=0\Leftrightarrow x=\frac{1}{5}\)

24 tháng 8 2019

\(a,-5x\left(x-3\right)\left(2x+4\right)-\left(x+3\right)\left(x-3\right)+\left(5x-2\right)\left(3x+4\right)\)

\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)

\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)

\(=-10x^3+19x^2+74x+1\)

\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)

\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)

\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)

\(=-5x^4-11x^3+24x^2+12x+7\)

\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)

\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)

\(=-2x^2-27x+57\)

24 tháng 8 2019

\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)

\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)

\(=-x^3+4x^2+22x+5\)

\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)

\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)

\(=-9x^3-55x^2+4x+35\)

\(g,\left(x-1\right)^2-\left(x+2\right)^2\)

\(=x^2-2x+1-x^2-4x-4\)

\(=-6x-3\)

10 tháng 4 2020

Khó quá, ai giúp em đi ạ

1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)

\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)

\(=x^2+4x+3-\dfrac{12}{2x+3}\)

 

1 tháng 10 2020

1) \(x^3+2x-3\)

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

\(=x^2\left(x-1\right)+x\left(x-1\right)+3\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x+3\right)\)

2) \(x^3-6x+4\)

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

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

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

1 tháng 10 2020

3) \(x^3-2x^2+1\)

\(=\left(x^3-x^2\right)-\left(x^2-x\right)-\left(x-1\right)\)

\(=x^2\left(x-1\right)-x\left(x-1\right)-\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2-x-1\right)\)

4) \(x^3+5x^2-12\)

\(=\left(x^3+2x^2\right)+\left(3x^2+6x\right)-\left(6x+12\right)\)

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

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

6 tháng 7 2017

a) (x + 2)(x + 3) - (x - 2)(x + 5) = 6
<=> x2 + 3x + 2x + 6 - x2 - 5x + 2x + 10 = 6
<=> 2x + 16 = 6
<=> 2x = -10
<=> x = -5
Vậy x = {-5}

Còn b với c mình đang tính :D

7 tháng 7 2017

cảm ơn bạn

3 tháng 9 2019

-3x^3+5x^2-9x+15 -3x-5 x^2 -3x^3-5x^2 - 10x^2-9x+15 -(10/3)x 10x^2+(50/3)x - -(23/3)x+15 +23/9 -(23/3)x-115/9 - 250/9

Chả biết có sai ko @@

3 tháng 9 2019

x^4-2x^3 +2x-1 x^2-1 x^2-2x x^4 -x^2 - -2x^3+x^2+2x-1 -2x^3 +2x - x^2-1 +1 x^2-1 - 0

a: Đặt x-3=a; x+1=b

Theo đề, ta có: \(a^3+b^3=\left(a+b\right)^3\)

\(\Leftrightarrow3ab\left(a+b\right)=0\)

=>(x-3)(x+1)(2x-2)=0

hay \(x\in\left\{3;-1;1\right\}\)

b: \(\Leftrightarrow\left(2x^2+1\right)^2+2x\left(2x^2+1\right)-15x^2-9x^2=0\)

\(\Leftrightarrow\left(2x^2+1\right)^2+2x\left(2x^2+1\right)-24x^2=0\)

\(\Leftrightarrow\left(2x^2+1\right)^2+6x\left(2x^2+1\right)-4x\left(2x^2+1\right)-24x^2=0\)

\(\Leftrightarrow\left(2x^2+1\right)\left(2x^2+6x+1\right)-4x\left(2x^2+6x+1\right)=0\)

\(\Leftrightarrow\left(2x^2-4x+1\right)\left(2x^2+6x+1\right)=0\)

\(\Leftrightarrow x^2+3x+\dfrac{1}{2}=0\)

\(\Leftrightarrow x^2+3x+\dfrac{9}{4}=\dfrac{7}{4}\)

\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=\dfrac{7}{4}\)

hay \(x\in\left\{\dfrac{\sqrt{7}-3}{2};\dfrac{-\sqrt{7}-3}{2}\right\}\)