2D

6

\(x^3+125=\left(x+5\right)\left(x^2-5x+25\right)\)

A là đa thức bậc 1

=>A=x+5

=>B=x^2-5x+25

=>Chọn A

Câu 2. M có bậc 2 + 7 = 9

Chọn D

Câu 6. x³ + 125 = x³ + 5³ = (x + 5)(x² - 5x + 25)

Chọn A

\(=25-\left(a-b\right)^2=\left(25-a+b\right)\left(25+a-b\right)\)

\(25-a^2+2ab-b^2\)

\(=25-\left(a^2-2ab+b^2\right)=25-\left(a-b\right)^2\)

\(=\left(5-a+b\right)\left(5+a-b\right)\)

`25-a^2+2ab-b^2`

`=25-(a^2-2ab+b^2)`

`=5^2-(a-b)^2`

`=(5+a-b)(5-a+b)`

Bài 3:

\(C=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)

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

\(=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{-\left(x^2-3x+9\right)}\)

\(=\dfrac{-3}{x-3}\)

a: \(\dfrac{3x-2}{2x^2+7}=\dfrac{\left(3x-2\right)\left(x+1\right)}{\left(2x^2+7\right)\left(x+1\right)}=\dfrac{3x^2+x-2}{\left(2x^2+7\right)\left(x+1\right)}\)

b:

Sửa đề: Đathức cho trước là x^3-x+5x^2-5

 \(\dfrac{\left(x^2+7x+10\right)\left(x-1\right)}{\left(x^2+3x+2\right)}\)

\(=\dfrac{\left(x+5\right)\left(x+2\right)\left(x-1\right)}{\left(x+2\right)\left(x+1\right)}=\dfrac{\left(x+5\right)\left(x-1\right)}{\left(x+1\right)}\)

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

\(Sửa,đề:x^2-10x+25\\ =x^2-2x.5+5^2=\left(x-5\right)^2=\left(x-5\right)\left(x-5\right)\\---\\ b,x^3+125=x^3+5^3=\left(x+5\right)\left(x^2-5x+25\right)\\ ---\\ 8x^3-y^3=\left(2x\right)^3-y^3\\ =\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

a, Bạn xem lại đề vì không thể tách được.

b, \(x^3+125\\ =x^3+5^3=\left(x+5\right)\left(x^2-5x+25\right)\)

c, \(8x^3-y^3\\ =\left(2x\right)^3-y^3\\ =\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

a/

\(x^2-10x+25=x^2+2.x.5+5^2=\left(x-5\right)^2=\left(x-5\right)\left(x-5\right)\)

b/

\(x^3+125=x^3+5^3=\left(x+5\right)\left(x^2-5x+25\right)\)

c/

\(8x^3-y^3=\left(2x\right)^3-y^3=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

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\(\inƯ\left(-3\right)\)

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

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 \(A⋮B\)

\(a,x^2-25-\left(x+5\right)=0\)

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

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

\(\Rightarrow\left(x+5\right)\left(x-6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+5=0\\x-6=0\end{cases}}\)      \(\Rightarrow\orbr{\begin{cases}x=-5\\x=6\end{cases}}\)

Vậy x = - 5; 6