a, \(100a^2-\left(a^2+25\right)^2=\left(10a\right)^2-\left(a^2+25\right)^2\)

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

b,\(-5\left(xy\right)^3-5\left(xy\right)^3=-10\left(xy\right)^3\)

c,\(16+2\left(xy\right)^3=2\left(2+xy\right)\left(4-2xy+x^2y^2\right)\)

\(\left|-5x\right|-16=3x\Leftrightarrow\left|-5x\right|=3x+16\)ĐK : x > = -16/3 

TH1 : \(-5x=3x+16\Leftrightarrow-8x=16\Leftrightarrow x=-2\)

TH2 : \(-5x=-3x-16\Leftrightarrow-2x=-16\Leftrightarrow x=8\)

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

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

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

4, \(4x^2-9+\left(2x+3\right)^2=\left(2x+3\right)\left(2x-3\right)+\left(2x+3\right)=2\left(2x+3\right)\left(x-1\right)\)

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

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

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

3,\(\left(4x^2-9\right)+\left(2x+3\right)^2=\left(2x+3\right)\left(2x-3\right)+\left(2x+3\right)^2=\left(2x+3\right)4x\)

Trả lời:

dư: a - 2 

Để phép chia hết thì a - 2 = 0 <=> a = 2

Vậy để A chia hết cho B thì a = 2

Đặt \(x+3=t\ne0\Rightarrow x=t-3\)

\(A=\frac{\left(t+2\right)\left(t-4\right)}{t^2}=\frac{t^2-2t-8}{t^2}\)

\(=-\frac{8}{t^2}-\frac{2}{t}+1=-8\left(\frac{1}{t}+\frac{1}{8}\right)^2+\frac{9}{8}\le\frac{9}{8}\)

\(Amax\) \(=\frac{9}{8}\)khi t = - 8 hay x = - 11

chiu ba bao tui lam lo lon ak

Bài 3B : 

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

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

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

\(\Leftrightarrow\left(x-3\right)\left(-8-3x\right)=0\Leftrightarrow x=-\frac{8}{3};x=3\)

b, \(x^3-8=2x^2-4x\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-2x\left(x-2\right)=0\)

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

c, \(x^3-6x^2+8x=0\Leftrightarrow x\left(x^2-6x+8\right)=0\)

\(\Leftrightarrow x\left(x^2-6x+9-1\right)=0\Leftrightarrow x\left[\left(x-3\right)^2-1\right]=0\)

\(\Leftrightarrow x\left(x-4\right)\left(x-2\right)=0\Leftrightarrow x=0;x=2;x=4\)

câu 1 

làm tính nhân 

2x . ( x^2 + 3 ) = 2x^3+6x

phân tích đa thức thành nhân tử 

a) x^2 - 4x 

= (x-4)x

b) đang nghĩ ạ 

Câu 1:

1,  \(2x\left(x^2+3\right)=2x^3+6x\)

2, \(73^2+27^2+54.73=73^2+2.73.27+27^2=\left(73+27\right)^2=100^2=10000\)

3, 

a, \(x^2-4x=x\left(x-4\right)\)

b) \(x^2-9y^2+6y-1\)