K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

4 tháng 12 2018

Câu e) là: 2x3 + 6x2 = x2 + 3x nhé

4 tháng 12 2018

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\)

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

b) \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)

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

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

\(\Rightarrow\left(x-2\right)\left(3x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

c) \(\left(2x+5\right)^2=\left(x+2\right)^2\)

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

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

\(\Rightarrow\left(x+3\right)\left(3x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\3x+7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\3x=-7\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)

d) \(x^2-5x+6=0\)

\(\Rightarrow x^2-2x-3x+6=0\)

\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

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

\(\Rightarrow2x^3+6x^2-x^2-3x=0\)

\(\Rightarrow2x^3+5x^2-3x=0\)

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

\(\Rightarrow2x^2+5x-3=0\)

\(\Rightarrow2x^2-6x+x-3=0\)

\(\Rightarrow2x\left(x-3\right)+\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(2x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

f) \(\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)-2x^2\)

\(\Rightarrow\left(x^2-1\right)\left(x+2\right)-\left(x^3-8\right)-2x^2=0\)

\(\Rightarrow x^3+2x^2-x+2-x^3+8-2x^2=0\)

\(\Rightarrow-x+10=0\)

\(\Rightarrow x=10\)

a: (2x+1)(3-x)(4-2x)=0

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

hay \(x\in\left\{-\dfrac{1}{2};3;2\right\}\)

b: 2x(x-3)+5(x-3)=0

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

=>x=3 hoặc x=-5/2

c: =>(x-2)(x+2)+(x-2)(2x-3)=0

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

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

=>x=2 hoặc x=1/3

d: =>(x-2)(x-3)=0

=>x=2 hoặc x=3

e: =>(2x+5+x+2)(2x+5-x-2)=0

=>(3x+7)(x+3)=0

=>x=-7/3 hoặc x=-3

f: \(\Leftrightarrow2x^3+5x^2-3x=0\)

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

\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)

hay \(x\in\left\{0;-3;\dfrac{1}{2}\right\}\)

25 tháng 3 2020

Bài 1:

a) (3x - 2)(4x + 5) = 0

<=> 3x - 2 = 0 hoặc 4x + 5 = 0

<=> 3x = 2 hoặc 4x = -5

<=> x = 2/3 hoặc x = -5/4

b) (2,3x - 6,9)(0,1x + 2) = 0

<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0

<=> 2,3x = 6,9 hoặc 0,1x = -2

<=> x = 3 hoặc x = -20

c) (4x + 2)(x^2 + 1) = 0

<=> 4x + 2 = 0 hoặc x^2 + 1 # 0

<=> 4x = -2

<=> x = -2/4 = -1/2

d) (2x + 7)(x - 5)(5x + 1) = 0

<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0

<=> 2x = -7 hoặc x = 5 hoặc 5x = -1

<=> x = -7/2 hoặc x = 5 hoặc x = -1/5

13 tháng 12 2020

bài 2:

a, (3x+2)(x^2-1)=(9x^2-4)(x+1)

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

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

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

b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0

x(x^2-9)-(x^3+8)=0

x^3-9x-x^3-8=0

-9x-8=0

tự tìm x nha

10 tháng 12 2021

\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)

\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)

1) Ta có: \(x^2-4x+4=0\)

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

\(\Leftrightarrow x-2=0\)

hay x=2

Vậy: S={2}

12 tháng 12 2020

Bạn chú ý đăng lẻ câu hỏi! 1/

a/ \(=x^3-2x^5\)

b/\(=5x^2+5-x^3-x\)

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)

12 tháng 12 2020

cảm ơn bạn đã nhắc

 

29 tháng 10 2021

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)

d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

2 tháng 1 2022

Giúp mk vs

 

a: \(=x^3-2x^5\)

e: \(=x^4+2x^3-x^2-2x\)

29 tháng 8 2021

a) \(2x\left(x^2-7x-3\right)=2x.x^2-2x.7x-2x.3=2x^3-14x^2-6x\)

b) \(\left(-2x^3+y^2-7xy\right)4xy^2=\left(-2x^3\right)4xy^2+y^24xy^2-7xy.4xy^2=-8x^4y^2+4xy^4-28x^2y^3\)

c) \(\left(-5x^3\right)\left(2x^2+3x-5\right)=-5x^32x^2-5x^33x-5x^3.-5=-10x^5-15x^4+25x^3\)

d) \(\left(2x^2-xy+y^2\right)\left(-3x^3\right)=-3x^32x^2-3x^3.-xy-3x^3y^2=-6x^5+3x^4y-3x^3y^2\)

29 tháng 8 2021

e) \(\left(x^2-2x+3\right)\left(x-4\right)=x\left(x^2-2x+3\right)-4\left(x^2-2x+3\right)=x^3-2x^2+3x-4x^2+8x-12=x^3-6x^2+11x-12\)

f) \(\left(2x^3-3x-1\right)\left(5x+2\right)=5x\left(2x^3-3x-1\right)+2\left(2x^3-3x-1\right)=10x^4-15x^2-5x+4x^3-6x-2=10x^4+4x^3-15x^2-11x-2\)

23 tháng 2 2018

a,\(4x^2-14x^2-6x=-10x^2-6x\)

các câu còn lại làm tg tuj

2 tháng 9 2018

a) 2x.(x2 - 7x - 3)

= 2xx2 + 2x(-7x) + 2x(-3)

= 2x2x - 2.7xx - 2.3x

= 2x3 - 14x2 - 6x