a) 3x3-2x2+2 chia x+1= 3x2-5x+5 dư -3 b) -3 chia hết x+1 vậy chon x =2

1)

a) \(-7x\left(3x-2\right)\)

\(=-21x^2+14x\)

b) \(87^2+26.87+13^2\)

\(=87^2+2.87.13+13^2\)

\(=\left(87+13\right)^2\)

\(=100^2\)

\(=10000\)

2)

a) \(x^2-25\)

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

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

b) \(3x\left(x+5\right)-2x-10=0\)

\(\Leftrightarrow3x\left(x+5\right)-\left(2x-10\right)=0\)

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

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

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

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

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

Vậy..........

3)

a) \(A:B=\left(3x^3-2x^2+2\right):\left(x+1\right)\)

Vậy \(\left(3x^3-2x^2+2\right):\left(x+1\right)=\left(3x^2-5x-5\right)+7\)

b)

Để \(A⋮B\Rightarrow7⋮\left(x+1\right)\)

\(\Rightarrow\left(x+1\right)\in U\left(7\right)=\left\{-1;1-7;7\right\}\)

Vì x là số nguyên nên x=0 ; x=6 thì \(A⋮B\)

C1

a) -7x(3x-2)=-21x^2+14x

b) 87^2+26.87+13^2=87^2+2.13.87+13^2=(87+13)^2=100^2

C2

a) (x-5)(x+5)

b)3x(x+5)-2(x+5)=(3x-2)(x+5)=0

\(\Rightarrow\left[\begin{array}{nghiempt}3x-2=0\\x+5=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{2}{3}\\x=-5\end{array}\right.\)

Vậy S={-5;2/3}

C3:

a)3x^3-2x^2+2=(x+1)(3x^2-5x-5)-3

b) Để A chia hết cho B=> x+1\(\inƯ\left(-3\right)\)

\(\Rightarrow\begin{cases}x+1=3\\x+1=-3\\x+1=1\\x+1=-1\end{cases}\)\(\Rightarrow\begin{cases}x=2\\x=-4\\x=0\\x=-2\end{cases}\)

`@` `\text {Ans}`

`\downarrow`

`1,`

`a)`

\(A(x) = 5x^5 + 2 - 7x - 4x^2 - 2x^5\)

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

`= 3x^5 - 4x^2 - 7x + 2`

`b)`

`A(x)+B(x)`

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

`= 3x^5 - 4x^2 - 7x + 2-3x^5 + 4x^2 + 3x - 7`

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

`= -4x - 5`

`b)`

`A(x) - B(x)`

`= 3x^5 - 4x^2 - 7x + 2 + 3x^5 - 4x^2 - 3x + 7`

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

`= 6x^5 - 8x^2 - 10x + 9`

`c)`

Thay `x=-1` vào đa thức `A(x)`

` 3*(-1)^5 - 4*(-1)^2 - 7*(-1) + 2`

`= 3*(-1) - 4*1 + 7 + 2`

`= -3 - 4 + 7 + 2`

`= -7+7 + 2`

`= 2`

Bạn xem lại đề ;-;.

`2,`

`M =` \(( 3 x - 2 )( 2 x + 1 )-( 3 x + 1 )( 2 x - 1 )\)

`= 3x(2x+1) - 2(2x+1) - [3x(2x-1) + 2x - 1]`

`= 6x^2 + 3x - 4x - 2 - (6x^2 - 3x + 2x - 1)`

`= 6x^2 - x - 2 - (6x^2 - x - 1)`

`= 6x^2 - x - 2 - 6x^2 + x + 1`

`= (6x^2 - 6x^2) + (-x+x) + (-2+1)`

`= -1`

Vậy, giá trị của biểu thức không phụ thuộc vào giá trị của biến.

2:

M=6x^2+3x-4x-2-6x^2+3x-2x+1

=-1

1;

a: A(x)=3x^5-4x^2-7x+2

b: B(x)=-3x^5+4x^2+3x-7

B(x)+A(x)

=-3x^5-4x^2-7x+2+3x^5+4x^2+3x-7

=-4x-5

A(x)-B(x)

=-3x^5-4x^2-7x+2-3x^5-4x^2-3x+7

=-6x^5-8x^2-10x+9

Bài `1`

\(a,A\left(x\right)=5x^5+2-7x-4x^2-2x^5\\ =\left(5x^5-2x^5\right)-4x^2-7x+2\\ =3x^5-4x^2-7x+2\)

\(b,A\left(x\right)+B\left(x\right)=3x^5-4x^2-7x+2+\left(-3x^5+4x^2+3x-7\right)\\ =3x^5-4x^2-7x+2-3x^5+4x^2+3x-7\\ =\left(3x^5-3x^5\right)+\left(-4x^2+4x^2\right)+\left(-7x+3x\right)+\left(2-7\right)\\ =-4x-5\)

\(A\left(x\right)-B\left(x\right)=\left(3x^5-4x^2-7x+2\right)-\left(-3x^5+4x^2+3x-7\right)\\ =3x^5-4x^2-7x+2+3x^5-4x^2-3x+7\\ =\left(3x^5+3x^5\right)+\left(-4x^2-4x^2\right)+\left(-7x-3x\right)+\left(2+7\right)\\ =6x^5-8x^2-10x+9\)

`c,` Thay `x=-1` Vào từng biểu thức ta có :

\(A\left(x\right)=3x^5-4x^2-7x+2\\=3\left(-1\right)^5-4.\left(-1\right)^2-7.\left(-1\right)+2\\ =3.\left(-1\right)-4.1-\left(-7\right)+2\\ =-3-4+7+2\\ =2\)

Cậu xem lại đề ạa

\(2,\\ M=\left(3x-2\right)\left(2x+1\right)-\left(3x+1\right)\left(2x-1\right)\\ =6x^2+3x-4x-2-\left(6x^2-3x+2x-1\right)\\=6x^2-x-2-6x^2+3x-2x+1\\ =\left(6x^2-6x^2\right)+\left(-x+3x-2x\right)+\left(-2+1\right)\\ =-1\)

`->` Vậy biểu thức không phụ thuộc vào biến `x`

\(x^2\left(y-1\right)-4\left(y-1\right)\\ =\left(y-1\right)\left(x^2-4\right)=\left(y-1\right)\left(x-2\right)\left(x+2\right)\)

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

\(x^2\left(y-1\right)-4\left(y-1\right)\\ =\left(x^2-4\right)\left(y-1\right)\\ =\left(x-2\right)\left(x+2\right)\left(y-1\right)\)

Câu a bạn haphuong01 làm sai kìa

2xy(x-3y) mới đúng

Hoc24h vẫn cho đúng là sao

Hy vọng GV hoc24h chấm cẩn thận hơn nha

a) x2 - 7x + 5 = ( x2 - 2 . 7/2 . x + 49 / 4 ) + 5 - 49 / 4 
= (x - 7/2)^2 - 29/4
= (x - 7/2)^2 - (√ 29 / 2 )^2
= ( x - ( 7 + √ 29 / 2 )). ( x + ( 7 - √ 29 / 2 ))