\(2x\left(x^2-25\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x=0\\x^2-25=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

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

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

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x-5=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{3}\end{cases}}\)

\(9\left(3x-2\right)-x\left(2-3x\right)=0\)

\(9\left(3x-2\right)+x\left(3x-2\right)=0\)

\(\left(9+x\right)\left(3x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}9+x=0\\3x-2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-9\\x=\frac{2}{3}\end{cases}}\)

\(\left(2x-1\right)^2=25\)

\(\Rightarrow\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

\(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x^2-25\right)=0\)

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

\(\Leftrightarrow x^2=25\)

\(\Leftrightarrow x=\pm5\)

a)  3x – 15 = 25 – 5x 

=> 3x + 5x = 25 + 15

=> 8x = 40

=> x = 5

 b) 3x - 17 = 2x – 7     

=> 3x - 2x = -7 + 17

=> x = 10

 c) 2x – 17 =  – (3x – 18)

=> 2x - 17 = -3x + 18

=> 2x + 3x = 18 + 17

=> 5x = 35

=> x = 7

d) 3x – 14 = 2(x – 9) + 1

=> 3x - 14 = 2x - 18 + 1

=> 3x - 2x = -18 + 1 + 14

=> x = -3

f) (x – 5)² = 9          

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

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

a) Ta có: \(3x-15=25-5x\)

\(\Leftrightarrow3x-15-25+5x=0\)

\(\Leftrightarrow8x-40=0\)

\(\Leftrightarrow8x=40\)

hay x=5

Vậy: x=5

b) Ta có: \(3x-17=2x-7\)

\(\Leftrightarrow3x-17-2x+7=0\)

\(\Leftrightarrow x-10=0\)

hay x=10

Vậy: x=10

c) Ta có: \(2x-17=-\left(3x-18\right)\)

\(\Leftrightarrow2x-17=-3x+18\)

\(\Leftrightarrow2x-17+3x-18=0\)

\(\Leftrightarrow5x-35=0\)

\(\Leftrightarrow5x=35\)

hay x=7

Vậy: x=7

d) Ta có: \(3x-14=2\left(x-9\right)+1\)

\(\Leftrightarrow3x-14=2x-18+1\)

\(\Leftrightarrow3x-14-2x+18-1=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Vậy: x=-3

f) Ta có: \(\left(x-5\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

Vậy: \(x\in\left\{2;8\right\}\)

Bài 2. Tìm x, biết :                                

a) \(3x-15=25-5x\)

\(\Leftrightarrow8x=40\)

\(\Leftrightarrow x=5\)

Vậy x = 5

b) \(3x-17=2x-7\)

\(\Leftrightarrow x=10\)

Vậy x = 10

c) \(2x-17=-\left(3x-18\right)\)

\(\Leftrightarrow2x-17=18-3x\)

\(\Leftrightarrow5x=35\)

\(\Leftrightarrow x=7\)

Vậy x = 7

d) \(3x-14=2\left(x-9\right)+1\)

\(\Leftrightarrow3x-14=2x-18+1\)

\(\Leftrightarrow3x-14=2x-17\)

\(\Leftrightarrow x=-3\)

Vậy x = -3

e) \(\left(x-5\right)^2=9\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

Vậy x = {8; 2}

b: \(\left(17x-25\right):8+65=9^5:9^3\)

\(\Leftrightarrow\left(17x-25\right):8=9^2-65=81-65=16\)

\(\Leftrightarrow17x-25=128\)

hay x=9

b: 

hay x=9

\(a,\Leftrightarrow6x^2-6x^2-11x+10=-12\\ \Leftrightarrow-11x=-22\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^3+27-x^3-2x=12-5x\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\\ c,\Leftrightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x-8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

a: ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)

\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)

\(\Leftrightarrow-11x=-22\)

hay x=2

b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=12-5x\)

\(\Leftrightarrow x^3+27-x^3-2x+5x=12\)

\(\Leftrightarrow x=-5\)

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

3x+5 =0 hoặc 4-3x=0

3x=-5 hoặc 3x=-4

x=-5/3 hoặc x=-4/3

9(3x-2)=x(2-3x)

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

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

3x-2=0 hoặc 9-x=0

3x=2 hoặc x= -9

x =2/3 hoặc x=-9 

vậy x =2/3 ; x= -9

25x^2 - 2=0

(5x)^2 -√2^2=0

(5x-√2)(5x+√2)=0

5x=√2 hoặc 5x = -√2

x=√2/5 hoặc x= -√2/5

vậy x=√2/5 ; x=-√2/5

tìm x biết:

(3x-1) [- 1/2x+5]=0

1/4+1/3:(2x-1)=-5

[2x+3/5]² - 9/25=0

-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6

[x+1/2]x [2/3-2x]=0

17/2-|2x-3/4|=-7/4

2/3x-1/2x =5/12

(x+1/5)²+17/25=26/25

[x.44/7+3/7].11/5-3/7=-2

3[3x-1/2]+1/9=0

Toán lớp 6Tìm x

 Trả lời  Câu hỏi tương tự

Chưa có ai trả lời câu hỏi này,bạn hãy là người đâu tiên giúp nguyenvanhoang giải bài toán này !

3(3x-1/2)^3+1/9=0

a) = (x+3).(x-3)^2-(x-3)(x+3)^2

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

=(x^2-9)(x-3-x-3)

=-6(x^2-9)

các câu còn lại tương tự

\(a,\left(x+3\right)\left(x^2-3x+9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)

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

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

\(=6\)

\(b,\left(x-5\right)\left(x^2+5x+25\right)-\left(x+5\right)\left(x^2-5x+25\right)\)

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

\(=-125-125\)

\(=-250\)

1: \(\left(3x-\dfrac{1}{5}\right)^2=\left(-\dfrac{3}{25}\right)^2\)

=>3x-1/5=3/25 hoặc 3x-1/5=-3/25

=>3x=8/25 hoặc 3x=2/25

=>x=8/75 hoặc x=2/75

2: \(\left(2x-\dfrac{1}{3}\right)^2=\left(-\dfrac{2}{9}\right)^2\)

=>2x-1/3=2/9 hoặc 2x-1/3=-2/9

=>2x=5/9 hoặc 2x=1/9

=>x=5/18 hoặc x=1/18

a) ( x² - 25 )² - ( x - 5 )²

= [ ( x - 5 )( x + 5 ) ]² - ( x - 5 )²

= [ ( x - 5 )( x + 5 ) - ( x - 5 ) ][ ( x - 5 )( x + 5 ) + ( x - 5 ) ]

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

= ( x - 5 )²( x + 4 )( x + 6 )

b) ( 4x² - 25 )² - 9( 2x - 5 )²

= ( 4x² - 25 )² - 3²( 2x - 5 )²

= ( 4x² - 25 )² - ( 6x - 15 )² 

= [ ( 4x² - 25 ) - ( 6x - 15 ) ][ ( 4x² - 25 ) + ( 6x - 15 ) ]

= ( 4x² - 25 - 6x + 15 )( 4x² - 25 + 6x - 15 )

= ( 4x² - 6x - 10 )( 4x² + 6x - 40 )

= ( 4x² + 4x - 10x - 10 )( 4x² + 16x - 10x - 40 )

= [ 4x( x + 1 ) - 10( x + 1 ) ][ 4x( x + 4 ) - 10( x + 4 ) ]

= ( x + 1 )( 4x - 10 )( x + 4 )( 4x - 10 )

= ( 4x - 10 )²( x + 1 )( x + 4 )

c) 4( 2x - 3 )² - 9( 4x² - 9 )²

= 2²( 2x - 3 )² - 3²( 4x² - 9 )²

= ( 4x - 6 )² - ( 12x² - 27 )²

= [ ( 4x - 6 ) - ( 12x² - 27 ) ][ ( 4x - 6 ) + ( 12x² - 27 ) ]

= ( 4x - 6 - 12x² + 27 )( 4x - 6 + 12x² - 27 )

= ( -12x² + 4x + 21 )( 12x² + 4x - 33 )

= ( -12x² + 18x - 14x + 21 )( 12x² - 18x + 22x - 33 )

= [ -12x( x - 3/2 ) - 14( x - 3/2 ) ][ 12x( x - 3/2 ) + 22( x - 3/2 ) ]

= ( x - 3/2 )( -12x - 14 )( x - 3/2 )( 12x + 22 )

= ( x - 3/2 )²( -12x - 14 )( 12x + 22 )

d) a⁶ - a⁴ + 2a³ + 2a²

= a²( a⁴ - a² + 2a + 2 )

= a²( a⁴ - 2a³ + 2a³ + 2a² - 4a² + a² + 4a - 2a + 2 )

= a²[ ( a⁴ - 2a³ + 2a² ) + ( 2a³ - 4a² + 4a ) + ( a² - 2a + 2 ) ]

= a²[ a²( a² - 2a + 2 ) + 2a( a² - 2a + 2 ) + 1( a² - 2a + 2 ) ]

= a²( a² + 2a + 1 )( a² - 2a + 2 )

= a²( a + 1 )²( a² - 2a + 2 )

e) ( 3x² + 3x + 2 )² - ( 3x² + 3x - 2 )²

= [ ( 3x² + 3x + 2 ) - ( 3x² + 3x - 2 ) ][ ( 3x² + 3x + 2 ) + ( 3x² + 3x - 2 ) ]

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

= 4( 6x² + 6x ) 

= 4.6x( x + 1 )

= 24( x + 1 )

e) là 24x( x + 1 ) nhé mình đánh thiếu