\(a,3x^3-3x=0\)

\(b,x\left(x-2\right)+x-...">

K
Khách

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

11 tháng 12 2018

a) \(3x^3-3x=0\)

\(\Rightarrow3x\left(x^2-1\right)=0\)

\(\Rightarrow3x\left(x-1\right)\left(x+1\right)=0\)

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

Vậy \(x\in\left\{0;\pm1\right\}\)

b) \(x\left(x-2\right)+x-2=0\)

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

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

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

Vậy \(x\in\left\{-1;2\right\}\)

c) \(5x\left(x-2000\right)-x+2000=0\)

\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)

\(\Rightarrow\left(5x-1\right)\left(x-2000\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x-2000=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2000\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{1}{5};2000\right\}\)

11 tháng 12 2018

a, 3x 3 - 3x = 0

=> 3x ( x 2 - 1 ) = 0

=> \(\orbr{\begin{cases}3x=0\\x^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}\Rightarrow[}\begin{cases}x=0\\x=1\\x=-1\end{cases}}\)

b, x ( x - 2 ) + ( x - 2 ) = 0

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

=> \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)

c, 5x ( x - 2000 ) - x + 2000 = 0

=> ( x - 2000 ) ( 5x - 1 ) = 0

=> \(\orbr{\begin{cases}x-2000=0\\5x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2000\\x=\frac{1}{5}\end{cases}}}\)

17 tháng 7 2018

Lần sau đăng thì chia thành nhiều câu hỏi nhé

\(16^2-9.\left(x+1\right)^2=0\)

\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)

\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)

\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)

\(\left[13-3x\right].\left[19+3x\right]=0\)

\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)

KL:..............................

25 tháng 7 2018

Nhiều câu hỏi mà bn ??

AH
Akai Haruma
Giáo viên
13 tháng 8 2018

a)

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

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

b)

\(x^3-0,36x=0\Leftrightarrow x(x^2-0,36)=0\)

\(\Leftrightarrow x(x-0,6)(x+0,6)=0\)

\(\Rightarrow \left[\begin{matrix} x=0\\ x-0,6=0\\ x+0,6=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=0\\ x=0,6\\ x=-0,6\end{matrix}\right.\)

c)

\((5x+2)^2-(3x-1)^2=0\)

\(\Leftrightarrow (5x+2-3x+1)(5x+2+3x-1)=0\)

\(\Leftrightarrow (2x+3)(8x+1)=0\)

\(\Rightarrow \left[\begin{matrix} 2x+3=0\\ 8x+1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-3}{2}\\ x=\frac{-1}{8}\end{matrix}\right.\)

AH
Akai Haruma
Giáo viên
13 tháng 8 2018

d)

\(x^2-10x=-25\)

\(\Leftrightarrow x^2-10x+25=0\)

\(\Leftrightarrow x^2-2.5x+5^2=0\Leftrightarrow (x-5)^2=0\)

\(\Rightarrow x=5\)

e)

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

\(\Leftrightarrow 3(x+5)-x(x+5)=0\)

\(\Leftrightarrow (3-x)(x+5)=0\)

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

f)

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

\(\Leftrightarrow [(x-1)-(3x+2)]^2=0\)

\(\Leftrightarrow (-2x-3)^2=0\Rightarrow -2x-3=0\Rightarrow x=\frac{-3}{2}\)

9 tháng 6 2017

a) \(4x^2-8x=0\)

\(\Rightarrow4x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-2=0\end{matrix}\right.\)

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

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

Vậy \(x_1=0;x_2=2\)

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

\(\Rightarrow-3x^2-14x+5=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}-3x+1=0\\x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-5\end{matrix}\right.\)

Vậy \(x_1=-5;x_2=\dfrac{1}{3}\)

9 tháng 6 2017

\(a,4x^2-8x=0\Rightarrow4x\left(x-8\right)=0\Rightarrow\left[{}\begin{matrix}4x=0\\x-8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)\(b,\left(x+5\right)-3x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(1-3x\right)=0\Rightarrow\left[{}\begin{matrix}x+5=0\\1-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\3x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{3}\end{matrix}\right.\)

7 tháng 6 2017

n) \(\left|3-x\right|+x^2-x\left(x+4\right)=0\)

\(\Rightarrow\left|3-x\right|+x^2-x^2-4x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3-x-4x=0\left(đk:3-x\ge0\right)\\-\left(3-x\right)-4x=0\left(đk:3-x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(đk:x\le3\right)\\x=-1\left(đk:x>3\right)\end{matrix}\right.\)

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

Vậy \(x=\dfrac{3}{5}\)

m) \(\left(x-1\right)^2+\left|x+21\right|-x^2-13=0\)

\(\Rightarrow x^2-2x+1+\left|x+21\right|-x^2-13=0\)

\(\Leftrightarrow-2x-12+\left|x+21\right|=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x-12+x+21=0\left(đk:x+21\ge0\right)\\-2x-12-\left(x+21\right)=0\left(đk:x+21< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\left(đk:x\ge-21\right)\\x=-11\left(đk:x< -21\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=9\\x\in\varnothing\end{matrix}\right.\)

Vậy \(x=9\)

7 tháng 6 2017

e) \(\left|5x\right|=3x-2\)

\(\Rightarrow5\cdot\left|x\right|=3x-2\)

\(\Leftrightarrow5\cdot\left|x\right|-3x=-2\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-3x=-2\left(đk:x\ge0\right)\\5\cdot\left(-x\right)-3x=-2\left(đk:x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(đk:x\ge0\right)\\x=\dfrac{1}{4}\left(đk:x< 0\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)

Vậy \(x\in\varnothing\)

g) \(\left|-2,5x\right|=x-12\)

\(\Rightarrow2,5\cdot\left|x\right|=x-12\)

\(\Leftrightarrow2x5\cdot\left|x\right|-x=-12\)

\(\Leftrightarrow\left[{}\begin{matrix}2,5x-x=-12\left(đk:x\ge0\right)\\2,5\cdot\left(-x\right)-x=-12\left(đk:x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-8\left(đk:x\ge0\right)\\x=\dfrac{24}{7}\left(đk:x< 0\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)

Vậy \(x\in\varnothing\)

25 tháng 6 2019

Tìm x,biết:

a/ x + 5x2 =0

⇔x ( 1 + 5x ) = 0

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

1) x = 0

2) 1+ 5x = 0 \(\Leftrightarrow\) x = \(\frac{-1}{5}\)

Vậy: S = \(\left\{0;\frac{-1}{5}\right\}\)

b/x+1=(x+1)2

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

\(\Leftrightarrow\) ( x+ 1)(1-x-1) = 0

\(\Leftrightarrow\) (x+1).(-x) = 0

\(\Leftrightarrow\) x+1 = 0 hoặc x = 0

\(\Leftrightarrow\) x= -1 ; 0

Vậy: S=\(\left\{-1;0\right\}\)

c/ x3+x=0

\(\Leftrightarrow\) x(x2 + 1) = 0

\(\Leftrightarrow\) x = 0 hoặc x2 + 1 = 0

Ta có : x2 + 1 \(\ge\) 0 vs mọi x

Vậy: S = \(\left\{0\right\}\)


d/5x(x2)(2x)=0

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

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

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

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

Vậy: S = \(\left\{\frac{-1}{5};2\right\}\)

g/ x(x4)+(x4)2=0

⇔ (x - 4)( x+x-4) = 0

\(\Leftrightarrow\) x - 4 = 0 hoặc 2x-4=0

\(\Leftrightarrow\) x = 4 hoặc x = 2

Vậy: S= \(\left\{2;4\right\}\)

h/ x23x=0

⇔x (x-3) = 0

\(\Leftrightarrow\) x = 0 hoặc x = 3

Vậy: S = \(\left\{0;3\right\}\)

Vậy: S= \(\left\{0;3\right\}\)
i/4x(x+1)=8(x+1)

4x(x+1)-8(x+1) = 0

\(\Leftrightarrow\) 4(x+1) (x - 2) = 0

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

\(\Leftrightarrow\) x= -1 hoặc x = 2

Vậy: S=\(\left\{-1;2\right\}\)