Mình làm 1 bài thôi nhé

Bài 5 

\(a.1-2y+y^2=\left(1-y\right)^2\)

\(b.\left(x+1\right)^2-25=\left(x+1\right)^2-5^2=\left(x-4\right)\left(x+6\right)\)

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

\(d.27+27x+9x^2+x^3=3^3+3.3^3.x+3.3.x^2+x^3=\left(3+x\right)^3\)

\(f.8x^3-12x^2y+6xy-y^3=\left(2x\right)^3-3.\left(2x\right)^2.y+3.2x.y-y^3=\left(2x-y\right)^3\)

Bài 4 : 

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

b, bạn xem lại đề nhé 

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

d, \(5x+5-x^2+1=5\left(x+1\right)+\left(1-x\right)\left(x+1\right)=\left(x+1\right)\left(6-x\right)\)

Bài 3: 

a) \(\left(2-3x\right)^2-\left(3-x\right)^2=\left[\left(2-3x\right)-\left(3-x\right)\right]\left[\left(2-3x\right)+\left(3-x\right)\right]\)

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

b) \(49\left(x-3\right)^2-9\left(x+2\right)^2\)

\(=\left[7\left(x-3\right)\right]^2-\left[3\left(x+2\right)\right]^2\)

\(=\left[\left(7x-21\right)-\left(3x+6\right)\right]\left[\left(7x-21\right)+\left(3x+6\right)\right]\)

\(=\left(4x-27\right)\left(10x-15\right)\)

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

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

\(=\left(x-3\right)\left(3x+11\right)\)

e) \(16x^2-\left(x^2+4\right)^2=\left(4x-x^2-4\right)\left(4x+x^2+4\right)\)

\(=-\left(x-2\right)^2\left(x+2\right)^2\)

f) \(1-2x+2yz+x^2-y^2-z^2=\left(x-1\right)^2-\left(y-z\right)^2\)

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

Bài 5: 

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

b) \(2x^2-14x+20=2x^2-4x-10x+20=2x\left(x-2\right)-10x\left(x-2\right)=2\left(x-5\right)\left(x-2\right)\)

c) \(3x^2+8x+5=3x^2+3x+5x+5=3x\left(x+1\right)+5\left(x+1\right)=\left(3x+5\right)\left(x+1\right)\)

d) \(6x^2-xy-7y^2=6x^2+6xy-7xy-7y^2=6x\left(x+y\right)-7y\left(x+y\right)\)

\(=\left(6x-7y\right)\left(x+y\right)\)

Bài 4: 

a) \(x^3-6x^2+12x-8=x^3-2.3.x^2+3.2^2.x-2^3=\left(x-2\right)^3\)

b) \(\left(x-1\right)^3+\left(3-x\right)^3=\left(x-1+3-x\right)\left[\left(x-1\right)^2-\left(x-1\right)\left(3-x\right)+\left(3-x\right)^2\right]\)

\(=2\left(x^2-2x+1+x^2-4x+3+x^2-6x+9\right)\)

\(=2\left(3x^2-12x+13\right)\)

c) \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3xy-3yz-3zx\right]\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

Trả lời:

Bài 1:

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

b, \(x^3+27=x^3+3^3=\left(x+3\right)\left(x^2-3x+9\right)\)

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

d, \(x^4-81=\left(x^2\right)^2-9^2=\left(x^2-9\right)\left(x^2+9\right)\)\(=\left(x^2-3^2\right)\left(x^2+9\right)=\left(x-3\right)\left(x+3\right)\left(x^2+9\right)\)

e, \(64x^3-1=\left(4x\right)^3-1^3=\left(4x-1\right)\left(16x^2+4x+1\right)\)

f, \(x^6+8y^3=\left(x^2\right)^3+\left(2y\right)^3=\left(x^2+2y\right)\left(x^4-2x^2y+4y^2\right)\)

ai giải giúp em đi ạ em đang cần gấp lắm ạ 

Bài 9. 

d) \(a^2+b^2+c^2=ab+bc+ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\).

Bài 10.

\(x^2+y^2-2x+4y+5=0\)

\(\Leftrightarrow x^2-2x+1+y^2+4y+4=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\).

Bài 4 : 

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

\(=\left(2x-3y-1+2x\right)\left(2x-3y+1-2x\right)-9y^2+4+12xy-4x\)

\(=\left(4x-3y-1\right)\left(1-3y\right)-9y^2+4+12xy-4x\)

\(=4x-12xy-3y+9y^2-1+3y-9y^2+4+12xy-4x=3\)

Vậy biểu thức ko phụ thuộc giá trị biến x 

Bài 2 : 

a, \(\left(a-3b\right)^2=a^2-6ab+9b^2\)

b, \(x^2-16y^4=\left(x-4y^2\right)\left(x+4y^2\right)\)

c, \(25a^2-\frac{1}{4}b^2=\left(5a-\frac{1}{2}b\right)\left(5a+\frac{1}{2}b\right)\)

Bài 3 : 

a, \(9x^2-6x+1=\left(3x-1\right)^2\)

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

c, \(4\left(2x-y\right)^2-8x+4y+1=\left(4x-2y\right)^2-2\left(4x-2y\right)+1=\left(4x-2y-1\right)^2\)

\(\left(n^2-1\right)^{2016}:n\)

Ta có \(n^2⋮n\)

\(\Rightarrow\left(n^2\right)^{2016}⋮n\)

Mà \(\left(-1\right)^{2016}:n=a\left(dư1\right)\)

Vậy số dư khi chia \(\left(n^2-1\right)^{2016}\) cho \(n\) là 1.

Nếu answer thôi thì nè: n=3