Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

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

c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)

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

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chiều mai bn nộp thì làm luôn đi còn hỏi đáp nữa !!!!!!

mình làm bài 2 trước nha:

a) y.(a-b)+a.(y-b)=a.y-b.y+a.y-b.y

                        =(a.y+a.y)-(b.y+b.y)

                         =2.a.y-2.b.y

                        =2.y.(a-b)

b)x².(x+y)-y.(x²-y²)=x³+x².y-x²y+y³=x³+y³

1 a

2c

3b

4d

5c

6c

Đề bài là gì sao không ghi rõ?? 

Bài 1:

a) \((x-5)(3x+3)-3x(x-3)+3x+7\)

\(=3(x-5)(x+1)-3x(x-3)+3x+7\)

\(=3(x^2+x-5x-5)-(3x^2-9x)+3x+7\)

\(=3(x^2-4x-5)-(3x^2-9x)+3x+7\)

\(=-8\)

b) \((x-3)(x^2+3x+9)-(54+x^3)\)

\(=(x-3)(x^2-3.x+3^2)-(54+x^3)\)

\(=x^3-3^3-(54+x^3)=-81\)

c) Sửa đề:

\((3x+y)(9x^2-3xy+y^2)-(3x-y)(9x^2+3xy+y^2)\)

\(=(3x+y)[(3x)^2-3x.y+y^2]-(3x-y)[(3x)^2+3x.y+y^2]\)

\(=(3x)^3+y^3-[(3x)^3-y^3]\)

\(=2y^3\)

Câu 2:

\(a)14x^2y^2-21xy^2+28x^2y\)

\(=7xy(2xy-3y+4x)\)

b) \((x+y)^2-4x^2\)\(=(x+y)^2-(2x)^2=(x+y-2x)(x+y+2x)\)

\(=(y-x)(3x+y)\)

c) \(2x^2-2xy-5x+5y\)

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

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

d) \(2xy-x^2-y^2+16\)

\(=16-(x^2+y^2-2xy)=4^2-(x-y)^2\)

\(=[4-(x-y)][4+(x-y)]=(4-x+y)(4+x-y)\)

Bài 3:
a)

\(x^2(x+1)+2x(x+1)=0\)

\(\Leftrightarrow (x+1)(x^2+2x)=0\)

\(\Leftrightarrow (x+1)x(x+2)=0\)

\(\Rightarrow \left[\begin{matrix} x+1=0\\ x=0\\ x+2=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-1\\ x=0\\ x=-2\end{matrix}\right.\)

b)

\(x(3x-2)-5(2-3x)=0\)

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

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

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

c)

\(16-25x^2=0\)

\(\Leftrightarrow 4^2-(5x)^2=0\)

\(\Leftrightarrow (4-5x)(4+5x)=0\)

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

Bài 2:

\(M=x^2-2xy+y^2=\left(x-y\right)^2=\left(-3\right)^2=9\)

\(N=x^2+y^2=\left(x-y\right)^2+2xy=9+2.10=29\)

\(P=x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3=\left(-3\right)^3=-27\)

\(Q=x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=\left(-3\right)^3+3.10.\left(-3\right)=-117\)

Bài 1:

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

b)  \(B=x^2+y^2=\left(x+y\right)^2-2xy=\left(-1\right)^2-2.\left(-12\right)=25\)

c)  \(C=x^3+3x^2y+3xy^2+y^3=\left(x+y\right)^3=\left(-1\right)^3=-1\)

d)  \(D=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=\left(-1\right)^3-3.\left(-12\right).\left(-1\right)=-37\)

a: \(\sqrt{x}=\dfrac{1}{3}\) nên x=1/9

\(\sqrt{y}=1\) nên y=1

\(D=3\cdot\dfrac{1}{81}-2\cdot\dfrac{1}{9}\cdot1+1^2=\dfrac{1}{27}-\dfrac{2}{9}+1=\dfrac{22}{27}\)

b: a/b=1/3

nên b=3a

\(E=\dfrac{3a+2\cdot3a}{4a-3\cdot3a}=\dfrac{9a}{-5a}=\dfrac{-9}{5}\)