a) \(2x\left(3x+1\right)+3x\left(4-2x\right)=7\)

\(\Rightarrow6x^2+2x+12x-6x^2=7\)

\(\Rightarrow14x=7\Rightarrow x=\frac{1}{2}\)

b) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)

\(72-20x-36x+84=30x-240-6x-84\)

\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)

\(-80x=-480\)

x = 6

c) \(\left(3x+2\right).\left(2x+9\right)-\left(x+2\right).\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)

\(\Rightarrow6x^2+4x+27x+18-6x^2-12x-x-2=x+1-x+6\) ( chỗ này bn tự phân tích ik nha, mk chỉ đưa ra kp sau khi phân tích thôi, ko thì viết ra dài lắm)

\(\Rightarrow18x+16=7\)

18x = -9

x = -2

18x = 

a: \(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

=>14x=7

hay x=1/2

b: \(\Leftrightarrow72-20x-36x+84=30x-240-6x-84\)

=>-56x+156=24x-324

=>-80x=-480

hay x=6

c: \(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6=7\)

=>18x+16=7

=>18x=-9

hay x=-1/2

Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1 

=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1 

=> (15x - 8x) + (25 + 12) = 11x + 37 

=> 7x + 37 = 11x + 37 

=> 11x - 7x = 0 

=>  x = 0 

a, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)

\(\Rightarrow72-20x-36x+84=30x-240-6x-84\)

\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)

\(\Rightarrow-80x=-480\Rightarrow x=6\)

b, \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x+12\right)+1\)

\(\Rightarrow15x+25-8x+12=5x+6x+36+1\)

\(\Rightarrow15x-8x-5x-6x=36+1-25-12\)

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

c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow10x-12x-12x=-16+11+16-15\)

\(\Rightarrow-14x=-4\Rightarrow x=\dfrac{2}{7}\)

d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)

\(\Rightarrow5x-3\left[4x-2\left(4x-15x+6\right)\right]=182\)

\(\Rightarrow5x-3\left(4x-8x+30x-12\right)=182\)

\(\Rightarrow5x-12x+24x-90x+36=182\)

\(\Rightarrow-73x=182-36\)

\(\Rightarrow-73x=146\Rightarrow x=-2\)

Chúc bạn học tốt!!!

a, \(\frac{x}{12}=-\frac{1}{24}-\frac{1}{8}\Leftrightarrow\frac{x}{12}=-\frac{1}{6}\Leftrightarrow6x=-12\Leftrightarrow x=-2\)

b, \(\frac{2}{3}x=\frac{1}{2}x+\frac{15}{12}\Leftrightarrow\frac{2x}{3}=\frac{x}{2}+\frac{15}{12}\Leftrightarrow\frac{4x}{6}-\frac{3x}{6}=\frac{15}{12}\Leftrightarrow\frac{x}{6}=\frac{15}{12}\Leftrightarrow x=\frac{15}{2}\)

c, \(\left|\frac{3}{4}-2x\right|+\frac{1}{6}=\frac{9}{2}\Leftrightarrow\left|\frac{3}{4}-2x\right|=\frac{13}{3}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{4}-2x=\frac{13}{3}\\\frac{3}{4}-2x=-\frac{13}{3}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-\frac{43}{12}\\2x=\frac{61}{12}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{43}{24}\\x=\frac{61}{24}\end{cases}}}\)

1) Ta có: 3x-12=5x(x-4)

\(\Leftrightarrow3x-12-5x\left(x-4\right)=0\)

\(\Leftrightarrow3x-12-5x^2+20x=0\)

\(\Leftrightarrow-5x^2+23x-12=0\)

\(\Leftrightarrow-5x^2+20x+3x-12=0\)

\(\Leftrightarrow\left(-5x^2+20x\right)+\left(3x-12\right)=0\)

\(\Leftrightarrow5x\left(-x+4\right)+3\left(x-4\right)=0\)

\(\Leftrightarrow5x\left(4-x\right)-3\left(4-x\right)=0\)

\(\Leftrightarrow\left(4-x\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\5x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{3}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{4;\frac{3}{5}\right\}\)

2) Ta có: 3x-15=2x(x-5)

\(\Leftrightarrow3x-15-2x\left(x-5\right)=0\)

\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)

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

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

Vậy: \(x\in\left\{5;\frac{3}{2}\right\}\)

3) Ta có: 3x(2x-3)+2(2x-3)=0

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

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

Vậy: \(x\in\left\{\frac{3}{2};-\frac{2}{3}\right\}\)

4) Ta có: (4x-6)(3-3x)=0

\(\Leftrightarrow\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=6\\3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{6}{4}=\frac{3}{2}\\x=1\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{3}{2};1\right\}\)

4) (4x - 6 ) ( 3 - 3x ) = 0

<=> \(\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}4x=6\\3x=3\end{matrix}\right.\)

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

Bài 1 :

a, Ta có : \(3x-12=5x\left(x-4\right)\)

=> \(3x-12=5x^2-20x\)

=> \(3x-12-5x^2+20x=0\)

=> \(5x^2-23x+12=0\)

=> \(5x^2-20x-3x+12=0\)

=> \(5x\left(x-4\right)-3\left(x-4\right)=0\)

=> \(\left(5x-3\right)\left(x-4\right)=0\)

=> \(\left[{}\begin{matrix}5x-3=0\\x-4=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{3}{5}\\x=4\end{matrix}\right.\)

Vậy phương trình có nghiệm là x = \(\frac{3}{5}\) và x = 4 .

b, Ta có : \(3x-15=2x\left(x-5\right)\)

=> \(3x-15-2x\left(x-5\right)=0\)

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

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

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

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

Vậy phương trình có nghiệm là x = \(\frac{3}{2}\) và x = 5 .

c, Ta có : \(3x\left(2x-3\right)+2\left(2x-3\right)=0\)

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

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

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

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

Vậy phương trình có nghiệm là x = \(-\frac{2}{3}\) và x = \(\frac{3}{2}\) .

d, Ta có : \(\left(4x-6\right)\left(3-3x\right)=0\)

=> \(\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}4x=6\\-3x=-3\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{6}{4}\\x=1\end{matrix}\right.\)

Vậy phương trình có nghiệm là x = 1 và x = \(\frac{6}{4}\) .

3,26 + 4/5 =?

làm nhanh lên giúp mình nhé

3,26+4/5=3,26+0,8=3,34

K cho mình cái

\(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x+16+11\)

\(\Leftrightarrow-2x-1=12x+27\Leftrightarrow-14x-28=0\Leftrightarrow x=-2\)

Bài 1.

1) ( 2x + 1 )³ - ( 2x + 1 )( 4x² - 2x + 1 ) - 3( 2x - 1 ) = 15

<=> 8x³ + 12x² + 6x + 1 - [ ( 2x )³ - 1³ ] - 6x + 3 = 15

<=> 8x³ + 12x² + 4 - 8x³ + 1 = 15

<=> 12x² + 15 = 15

<=> 12x² = 0

<=> x = 0

2) x( x - 4 )( x + 4 ) - ( x - 5 )( x² + 5x + 25 ) = 13

<=> x( x² - 16 ) - ( x³ - 5³ ) = 13

<=> x³ - 16x - x³ + 125 = 13

<=> 125 - 16x = 13

<=> 16x = 112

<=> x = 7

Bài 2.

A = ( x + 5 )( x² - 5x + 25 ) - ( 2x + 1 )³ - 28x³ + 3x( -11x + 5 )

= x³ + 5³ - ( 8x³ + 12x² + 6x + 1 ) - 28x³ - 33x² + 15x

= -27x³ + 125 - 8x³ - 12x² - 6x - 1 - 33x² + 15x

= -33x³ - 45x² + 9x + 124 ( có phụ thuộc vào biến )

B = ( 3x + 2 )³ - 18x( 3x + 2 ) + ( x - 1 )³ - 28x³ + 3x( x - 1 )

= 27x³ + 54x² + 36x + 8 - 54x² - 36x + x³ - 3x² + 3x - 1 - 28x³ + 3x² - 3x

= 7 ( đpcm )

C = ( 4x - 1 )( 16x² + 4x + 1 ) - ( 4x + 1 )³ + 12( 4x + 1 )³ + 12( 4x + 1 ) - 15

= ( 4x )³ - 1³ - [ ( 4x + 1 )³ - 12( 4x + 1 )³ - 12( 4x + 1 ) ] - 15

= 64x³ - 1 - ( 4x + 1 )[ ( 4x + 1 )² - 12( 4x + 1 )² - 12 ] - 15

= 64x³ - 16 - ( 4x + 1 )[ 16x² + 8x + 1 - 12( 16x² + 8x + 1 ) - 12 ]

= 64x³ - 16 - ( 4x + 1 )( 16x² + 8x - 11 - 192x² - 96x - 12 )

= 64x³ - 16 - ( 4x + 1 )( -176x² - 88x - 23 )

= 64x³ - 16 - ( -704x³ - 528x² - 180x - 23 )

= 64x³ - 16 + 704x³ + 528x² + 180x + 23 

= 768x³ + 528x² + 180x + 7 ( có phụ thuộc vào biến )