K
Khách

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

5 tháng 11 2020

a) x3 - 16x = 0

=> x(x2 - 16) = 0

=> x(x - 4)(x + 4) = 0

=> x = 0 hoặc x - 4 = 0 hoặc x + 4 = 0

=> x = 0 hoặc x = 4 hoặc x = -4

b) x4 - 2x+ 10x2 - 20x = 0

=> x3(x - 2) + 10x(x - 2) = 0

=> (x - 2)(x3 + 10x) = 0

=> x(x - 2)(x2 + 10) = 0 (1)

Vì x2 + 10 \(x^2+10\ge10>0\forall x\)

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

Vậy \(x\in\left\{0;2\right\}\)là giá trị cần tìm

5 tháng 11 2020

a) x3 - 16x = 0

⇔ x( x2 - 16 ) = 0

⇔ x( x - 4 )( x + 4 ) = 0

⇔ x = 0 hoặc x - 4 = 0 hoặc x + 4 = 0

⇔ x = 0 hoặc x = ±4

b) x4 - 2x3 + 10x2 - 20x = 0

⇔ x( x3 - 2x2 + 10x - 20 ) = 0

⇔ x[ ( x3 - 2x2 ) + ( 10x - 20 ) ] = 0

⇔ x[ x2( x - 2 ) + 10( x - 2 ) ] = 0

⇔ x( x - 2 )( x2 + 10 ) = 0

⇔ x = 0 hoặc x - 2 = 0 hoặc x2 + 10 = 0

⇔ x = 0 hoặc x = 2 ( x2 + 10 ≥ 10 > 0 ∀ x )

17 tháng 6 2018

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

17 tháng 6 2018

* \(x^3-16x=0\)

\(\Rightarrow x\left(x^2-16\right)=0\)

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

\(\Rightarrow x\left(x-4\right)\left(x+4\right)=0\)

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

6 tháng 8 2016

1, x(x - 5) - 4x + 20 = 0

=> x(x - 5) - 4(x - 5) = 0

=> (x - 4)(x - 5) = 0

=> x - 4 = 0 hoặc x - 5 = 0

=> x = 4 hoặc x = 5

=> x thuộc {4; 5}

2, 3(x + 1) + x(x + 1) 

= (3 + x)(x + 1)

3, 2x3 + x = 0

=> x(2x2 + 1) = 0

=> x = 0 hoặc 2x2 + 1 = 0

=> x = 0 hoặc 2x2 = -1

=> x = 0 hoặc x2 = -1/2 (vô lí vì x2 > hoặc = 0 với mọi x)

=> x = 0

4, x3 - 16x = 0

=> x(x2 - 16) = 0

=> x = 0 hoặc x2 - 16 = 0

=> x = 0 hoặc x2 = 16

=> x = 0 hoặc x = 4 hoặc x = -4

=> x thuộc {-4; 0; 4}

5, x2 + 6x = -9

=> x2 + 6x + 9 = 0

=> x2 + 2.3.x + 32 = 0

=> (x + 3)2 = 0

=> x + 3 = 0

=> x = -3

6, x4 - 2x3 + 10x2 - 20x = 0

=> x2(x2 + 10) - 2x(x2 + 10) = 0

=> (x2 + 2x)(x2 + 10) = 0

=> x(x +2)(x2 + 10) = 0

-TH1: x = 0

-TH2: x + 2 = 0 => x = -2

-TH3: x2 + 10 = 0 => x2 = -10 (vô lí vì x2 > hoặc = 0 với mọi x)

=> x thuộc {0; -2}

7, (2x - 3)2 = (x + 5)2

-TH1: 2x - 3 = x + 5

=> x = 8

- TH2: - 2x + 3 = x + 5

=> -3x = 2

=> x = \(\frac{-2}{3}\)

- TH3: 2x - 3 = - x - 5

=> 3x = -2

=> x = \(\frac{-2}{3}\)

- TH4: - 2x + 3 = - x - 5

=> -x = -8

=> x = 8`

=> x thuộc {\(\frac{-2}{3}\); 8}

22 tháng 7 2018

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

Vậy....

\(2x\left(x+6\right)=7x+42\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)

Vậy......

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

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

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

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

\(\Leftrightarrow\)\(x=5\)

\(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

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

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

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

Vậy...

16 tháng 9 2019

a) x3 - 16x = 0

x(x2 - 16) = 0

=> x = 0 hoặc x2 - 16 = 0

x = 4

Vậy x = 0 hoặc x = 4

b) x4 -2x3 + 10x2 - 20x = 0

x3 (x - 2) + 10x(x - 2) = 0

(x - 2)(x3 + 10x) = 0

=> x - 2 = 0 hoặc x3 + 10x = 0

x = 2 x(x2 + 10) = 0

+ TH1: x = 0

+ TH2: x2 + 10 = 0

x2 = -10 (vô lí)

Vậy x = 2 hoặc x = 0

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

(2x)2 + 2 . 2x . 3 + 32 = x2 + 2.x.5 + 52

4x2 + 12x + 9 = x2 + 10x + 25

4x2 + 12x - x2 - 10x = 25 - 9

3x2 + 2x = 16

x(3x + 2) = 16

Đến đây bạn làm nốt câu c nhé!

23 tháng 7 2017

\(a,x^3-16x=0\)

\(\Leftrightarrow x\left(x^2-16\right)=0\)

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

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

\(b,x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x=0\\x^2+10=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\\\left[{}\begin{matrix}x^2=10\\x^2=-10\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\\x=\sqrt{10}\\x=-\sqrt{10}\end{matrix}\right.\)\(c,\left(2x-1\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow4x^2-4x+1=x^2+6x+9\)

\(\Leftrightarrow4x^2-4x+1-x^2-6x-9=0\)

\(\Leftrightarrow3x^2-10x-8=0\)

\(\Leftrightarrow3x^2-12x+2x-8=0\)

\(\Leftrightarrow3x\left(x-4\right)+2\left(x-4\right)=0\)

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

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

Phần d tương tự

23 tháng 7 2017

Câu a :

\(x^3-16x=0\)

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

\(\Leftrightarrow x\left[\left(x+4\right)\left(x-4\right)\right]=0\)

\(\Rightarrow\) \(x=0\)

\(\Rightarrow\) \(x+4=0\Rightarrow x=-4\)

\(\Rightarrow x-4=0\Rightarrow x=4\)

Câu b :

\(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)\) \(=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)

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

\(\Rightarrow x=0\)

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

\(x^2+10=0\) \(\Rightarrow\) x ( loại )

a: \(\left(x+5\right)^2>=0\forall x\)

\(\left(2y-8\right)^2>=0\forall y\)

Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)

b: \(\left(x+3\right)\left(2y-1\right)=5\)

=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)

Bài 2 :

a ) \(x^3-16x=0\)

\(\Leftrightarrow x\left(x^2-16\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-16=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm4\end{matrix}\right.\)

Vậy..........

b ) \(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\Rightarrow x=2\\x^2+10=0\left(loại\right)\end{matrix}\right.\)

Vậy .......................

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

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

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

\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\)

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

Vậy.............

d ) \(x^2\left(x-2\right)-2x^2+8x-8=0\)

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

\(\Leftrightarrow x^3-4x^2+8x-8=0\)

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

\(\Rightarrow x=2\)

25 tháng 7 2018

Bài 2 :

a ) x3−16x=0x3−16x=0

⇔x(x2−16)=0⇔x(x2−16)=0

⇔[x=0x2−16=0⇒[x=0x=±4⇔[x=0x2−16=0⇒[x=0x=±4

Vậy..........

b ) x4−2x3+10x2−20x=0x4−2x3+10x2−20x=0

⇔x3(x−2)+10x(x−2)=0⇔x3(x−2)+10x(x−2)=0

⇔(x−2)(x3+10x)=0⇔(x−2)(x3+10x)=0

⇔x(x−2)(x2+10)=0⇔x(x−2)(x2+10)=0

⇔⎡⎢⎣x=0x−2=0⇒x=2x2+10=0(loại)⇔[x=0x−2=0⇒x=2x2+10=0(loại)

Vậy .......................

c ) (2x−1)2=(x+3)2(2x−1)2=(x+3)2

⇔(2x−1)2−(x+3)2=0⇔(2x−1)2−(x+3)2=0

⇔(2x−1−x−3)(2x−1+x+3)=0⇔(2x−1−x−3)(2x−1+x+3)=0

⇔(x−4)(3x+2)=0⇔(x−4)(3x+2)=0

⇔[x−4=03x+2=0⇒⎡⎣x=4x=−23⇔[x−4=03x+2=0⇒[x=4x=−23

Vậy.............

d ) x2(x−2)−2x2+8x−8=0x2(x−2)−2x2+8x−8=0

⇔x3−2x2−2x2+8x−8=0⇔x3−2x2−2x2+8x−8=0

⇔x3−4x2+8x−8=0⇔x3−4x2+8x−8=0

⇔⇔ (x−2)3=0(x−2)3=0

⇒x=2

27 tháng 6 2017

a) ... \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\hept{\begin{cases}x=1\\x=2\\x=-2\end{cases}}\)Vậy.....

b) ... \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\Rightarrow x\in\theta\end{cases}}\)(\(\theta\)là rỗng) Vậy.........

c) ... \(\Leftrightarrow2x-3=x+5\Leftrightarrow x=8\)Vậy.......

d) ... \(\Leftrightarrow x\left(x^2-16\right)=0\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=4\\x=-4\end{cases}}\)Vậy......

a: \(\Leftrightarrow\left(x+12-3x\right)\left(x+12+3x\right)=0\)

=>(-2x+12)(4x+12)=0

=>x=-3 hoặc x=6

b: \(\Leftrightarrow20x^3-15x^2+45x-45=0\)

=>\(x\simeq0.93\)

d: =>-4x+28+11x=-x+3x+15

=>7x+28=2x+15

=>5x=-13

=>x=-13/5

e: \(\Leftrightarrow4x^3-12x+x=4x^3-3x+5\)

=>-9x=-3x+5

=>-6x=5

=>x=-5/6