a)(x+2).(x+3)-(x-2).(x+5)=10

  ( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10

 x^2 +3x+2x+6-x^2 -5x+2x+10-10=0

 2x+6=0

2x=-6

x=-3

a) 

<=> 3x - 3 + x - 2 = 2x - 2 - x + 1

<=> 3x + x - 2x + x = -2 + 1 + 3 + 2

<=>    3x               = 4

<=>   x                  = 4/3

Các câu sau làm tương tự

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

<=>   \(3x-3+x-2=2x-2-x+1\)

<=>  \(4x-5=x-1\)

<=> \(3x=4\)

<=>  \(x=\frac{4}{3}\)

Vậy....

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 

\(2x^2-7x+5=0\)

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

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

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

\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)

\(x\left(2x-5\right)-4x+10=0\)

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

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

\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)

\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)

\(x^2-25-x^2+2x=15\)

\(2x=15+25\)

\(2x=40\)

\(x=\frac{40}{2}\)

\(x=20\)

\(x^2\left(2x-3\right)-12+8x=0\)

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

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

\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))

\(2x=3\)

\(x=\frac{3}{2}\)

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

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

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

\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)

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

\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)

\(4x^2-12x+9-4x^2+4x=5\)

\(-8x=5-9\)

\(-8x=-4\)

\(x=\frac{4}{8}\)

\(x=\frac{1}{2}\)

\(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(5x-2x^2+2x^2-2x=13\)

\(3x=13\)

\(x=\frac{13}{3}\)

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

\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)

\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)

\(\left(2x-5\right)\left(x+11\right)=0\)

\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)

\(a,2x^2-7x+5=0\Leftrightarrow2x^2-2x-5x+5=0\Leftrightarrow2x\left(x-1\right)-5\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2,5\end{matrix}\right.\)\(b,x\left(2x-5\right)-4x+10=0\Rightarrow x\left(2x-5\right)-2\left(2x-5\right)=0\Leftrightarrow\left(x-2\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-2=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=2,5\end{matrix}\right.\)\(c,\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\Leftrightarrow x^2-25-x^2+2x-15=0\Leftrightarrow2x-40=0\Rightarrow2x=40\Rightarrow x=20\)\(d,x^2\left(2x-3\right)-12+8x=0\Rightarrow2x^3-3x^2-12+8x=0\Leftrightarrow2x^3+8x-3x^2-12=0\Leftrightarrow2x\left(x^2+4\right)-2\left(x^2+4\right)=0\Leftrightarrow\left(2x-2\right)\left(x^2+4\right)=0\Rightarrow\left[{}\begin{matrix}2x-2=0\\x^2+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=2\\x^2=-4\end{matrix}\right.\Rightarrow x=1\)

toàn là bài dễ, những phương trình này đơn giản mà bạn

`C(x)=`\(5-8x^4+2x^3+x+5x^4+x^2-4x^3\)

`C(x)= (-8x^4+5x^4)+(2x^3-4x^3)+x^2+x+5`

`C(x)= -3x^4-2x^3+x^2+x+5`

`D(x)=`\(\left(3x^5+x^4-4x\right)-\left(4x^3-7+2x^4+3x^5\right)\)

`D(x)= 3x^5+x^4-4x-4x^3+7-2x^4-3x^5`

`D(x)=(3x^5-3x^5)+(x^4-2x^4)-4x^3-4x+7`

`D(x)=-x^4-4x^3-4x+7`

`P(x)=C(x)+D(x)`

`P(x)=( -3x^4-2x^3+x^2+x+5)+(-x^4-4x^3-4x+7)`

`P(x)=-3x^4-2x^3+x^2+x+5-x^4-4x^3-4x+7`

`P(x)=(-3x^4-x^4)+(-2x^3-4x^3)+x^2+(x-4x)+(5+7)`

`P(x)=-4x^4-6x^3+x^2-3x+12`

`Q(x)=C(x)-D(x)`

`Q(x)=( -3x^4-2x^3+x^2+x+5)-(-x^4-4x^3-4x+7)`

`Q(x)=-3x^4-2x^3+x^2+x+5+x^4+4x^3+4x-7`

`Q(x)=(-3x^4+x^4)+(-2x^3+4x^3)+x^2+(x+4x)+(5-7)`

`Q(x)=-2x^4+2x^3+x^2+5x-2`

`F(x)=Q(x)-(-2x^4+2x^3+x^2-12)`

`F(x)=(-2x^4+2x^3+x^2+5x-2)-(-2x^4+2x^3+x^2-12)`

`F(x)=-2x^4+2x^3+x^2+5x-2+2x^4-2x^3-x^2+12`

`F(x)=(-2x^4+2x^4)+(2x^3-2x^3)+(x^2-x^2)+5x+(-2+12)`

`F(x)=5x+10`

Đặt `5x+10=0`

`\Leftrightarrow 5x=0-10`

`\Leftrightarrow 5x=-10`

`\Leftrightarrow x=-10 \div 5`

`\Leftrightarrow x=-2`

Vậy, nghiệm của đa thức là `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 )

3,26 + 4/5 =?

làm nhanh lên giúp mình nhé

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

K cho mình cái

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

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

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