\(4x\left(x+1\right)=8\left(x+1\right)\)

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4x(x+1)=8(x+1)

=>\(x\left(x+1\right)=2\left(x+1\right)\)

=>\(\left(x+1\right)\left(x-2\right)=0\)

=>\(\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

15 tháng 1

gtttụ

18 tháng 7 2016

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

\(3x=5\)

\(x=\frac{5}{3}\)

b) \(\left(3x-8\right)^2=\left(2x-7\right)^2\)

\(3x-8=2x-7\)

\(x=1\)

c) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)

\(\left(4x^2-3x-18\right)^2=\left(4x^2+3x\right)^2\)

\(4x^2-3x-18=4x^2+3x\)

\(6x=-18\)

\(x=-3\)

d) Sai đề

e) ko bt

1, \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)

\(\Leftrightarrow-4x^2+28x+4x^3-20x=28x^2-13\)

\(\Leftrightarrow-32x^2+8x+4x^3-13=0\)( vô nghiệm )

2, \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)

\(\Leftrightarrow12x^3-7x^2-10x-7x^2-35x=-2x^2+11x-12+12x^3+2x^2\)

\(\Leftrightarrow12x^3-14x^2-45x=11x-12+12x^3\)

\(\Leftrightarrow-14x^2-56x-12=0\)( vô nghiệm )

20 tháng 8 2020

Mình làm riêng ra nhá , chứ nhiều quá nên thông cảm cho mình :))

1. \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)

=> \(-4x^2+28x+4x^3-20x=28x^2-13\)

=> \(-4x^2+4x^3+\left(28x-20x\right)=28x^2-13\)

=> \(-4x^2+4x^3+8x-28x^2+13=0\)

=> \(\left(-4x^2-28x^2\right)+4x^3+8x+13=0\)

=> \(-32x^2+4x^3+8x+13=0\)

=> vô nghiệm

2. \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)

=> \(4x^2\left(3x+2\right)-5x\left(3x+2\right)-7x\left(x+5\right)=-4\left(-2x+3\right)+x\left(-2x+3\right)+12x^3+2x^2\)

=> \(12x^3+8x^2-15x^2-10x-7x^2-35x=8x-12-2x^2+3x+12x^3+2x^2\)

=> \(12x^3+8x^2-15x^2-10x-7x^2-35x-8x+12+2x^2-3x-12x^3-2x^2=0\)

=> \(\left(12x^3-12x^3\right)+\left(8x^2-15x^2-7x^2+2x^2-2x^2\right)+\left(-10x-35x-8x-3x\right)+12=0\)

=> \(-14x^2-56x+12=0\)

=> .... tự tìm

Câu c dấu bằng chỗ nào ?

28 tháng 5 2017

a) (x-1)(5x+3)=(3x-8)(x-1)

= (x-1)(5x+3)-(3x-8)(x-1)=0

=(x-1)[(5x+3)-(3x-8)]=0

=(x-1)(5x+3-3x+8)=0

=(x-1)(2x+11)=0

\(\Leftrightarrow\) x-1=0 hoặc 2x+11=0

\(\Leftrightarrow\) x=1 hoặc x=\(\dfrac{-11}{2}\)

Vậy S={1;\(\dfrac{-11}{2}\)}

b) 3x(25x+15)-35(5x+3)=0

=3x.5(5x+3)-35(5x+3)=0

=15x(5x+3)-35(5x+3)=0

=(5x+3)(15x-35)=0

\(\Leftrightarrow\) 5x+3=0 hoặc 15x-35=0

\(\Leftrightarrow\) x=\(\dfrac{-3}{5}\) hoặc x=\(\dfrac{7}{3}\)

Vậy S={\(\dfrac{-3}{5};\dfrac{7}{3}\)}

c) (2-3x)(x+11)=(3x-2)(2-5x)

=(2-3x)(x+11)-(3x-2)(2-5x)=0

=(3x-2)[(x+11)-(2-5x)]=0

=(3x-2)(x+11-2+5x)=0

=(3x-2)(6x+9)=0

\(\Leftrightarrow\) 3x-2=0 hoặc 6x+9=0

\(\Leftrightarrow\) x=\(\dfrac{2}{3}\) hoặc x=\(\dfrac{-3}{2}\)

Vậy S={\(\dfrac{2}{3};\dfrac{-3}{2}\)}

d) (2x2+1)(4x-3)=(2x2+1)(x-12)

=(2x2+1)(4x-3)-(2x2+1)(x-12)=0

=(2x2+1)[(4x-3)-(x-12)=0

=(2x2+1)(4x-3-x+12)=0

=(2x2+1)(3x+9)=0

\(\Leftrightarrow\)2x2+1=0 hoặc 3x+9=0

\(\Leftrightarrow\)x=\(\dfrac{1}{2}\)hoặc x=\(\dfrac{-1}{2}\) hoặc x=-3

Vậy S={\(\dfrac{1}{2};\dfrac{-1}{2};-3\)}

e) (2x-1)2+(2-x)(2x-1)=0

=(2x-1)[(2x-1)+(2-x)=0

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

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

\(\Leftrightarrow\) 2x-1=0 hoặc x+1=0

\(\Leftrightarrow\) x=\(\dfrac{-1}{2}\) hoặc x=-1

Vậy S={\(\dfrac{-1}{2}\);-1}

f)(x+2)(3-4x)=x2+4x+4

=(x+2)(3-4x)=(x+2)2

=(x+2)(3-4x)-(x+2)2=0

=(x+2)[(3-4x)-(x+2)]=0

=(x+2)(3-4x-x-2)=0

=(x+2)(-5x+1)=0

\(\Leftrightarrow\) x+2=0 hoặc -5x+1=0

\(\Leftrightarrow\) x=-2 hoặc x=\(\dfrac{1}{5}\)

Vậy S={-2;\(\dfrac{1}{5}\)}

AH
Akai Haruma
Giáo viên
15 tháng 7 2018

a)

\((x+2)(x+4)(x+6)(x+8)+16\)

\(=[(x+2)(x+8)][(x+4)(x+6)]+16\)

\(=(x^2+10x+16)(x^2+10x+24)+16\)

\(=a(a+8)+16\) (Đặt \(x^2+10x+16=a\) )

\(=a^2+2.4.a+4^2=(a+4)^2\)

\(=(x^2+10x+16+4)^2\)

\(=(x^2+10x+20)^2\)

AH
Akai Haruma
Giáo viên
15 tháng 7 2018

b) \((x^2+x)(x^2+x+1)-6\)

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

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

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

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

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

\(=[x(x-1)+2(x-1)](x^2+x+3)\)

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

c)

\((x^2-4x)^2-8(x^2-4x)+15\)

\(=(x^2-4x)^2-3(x^2-4x)-5(x^2-4x)+15\)

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

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

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

\(=(x^2-4x-3)[x(x+1)-5(x+1)]=(x^2-4x-3)(x+1)(x-5)\)

28 tháng 9 2019

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

\(\Leftrightarrow x^3-3x^2+3x-1+3\left(x^2+2x+1\right)=x^3+8\)

\(\Leftrightarrow-3x^2+3x+3x^2+6x+3=9\)

\(\Leftrightarrow9x=6\Leftrightarrow x=\frac{2}{3}\)

b) \(x^2-4=8\left(x-2\right)\)

\(\Leftrightarrow x^2-4=8x-16\)

\(\Leftrightarrow x^2-8x+12=0\)

\(\Leftrightarrow x^2-2x-6x+12=0\)

\(\Leftrightarrow x\left(x-2\right)-6\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=2\end{cases}}\)

28 tháng 9 2019

c) \(x^2-4x+4=9\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)^2=9\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)^2-9\left(x-2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-11=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=11\end{cases}}\)

d) \(4x^2-12x+9=\left(5-x\right)^2\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x-3=5-x\\2x-3=x-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=-2\end{cases}}\)