\(x^4-3x^2+9\)

\(=\left(x^2\right)^2+2.x^2.3+3^2-9x^2\)

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

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

       \(x^4+3x^2+4\)

\(=\left(x^2\right)^2+2.x^2.2+2^2-x^2\)

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

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

      \(118^2-118.36+18^2\)

\(=118^2-2.118.18+18^2\)

\(=\left(118-18\right)^2\)

\(=100^2=10000\)

\(118^2-118.36+18^2\)

=\(118^2-2.118.18+18^2\)

=\(\left(118-18\right)^2\)

=\(100^2\)

Bài 1:

a) \(2x^2y-xy=xy\left(2x-1\right)\)

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

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

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

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

Bài 2:

a)\(x^3-\frac{1}{9}x=0\)

\(\Leftrightarrow x\left(x^2-\frac{1}{9}\right)=0\)

\(\Leftrightarrow x\left(x-\frac{1}{3}\right)\left(x+\frac{1}{3}\right)=0\)

\(\Rightarrow x=0\text{ hoặc }x-\frac{1}{3}=0\Leftrightarrow x=\frac{1}{3}\text{ hoặc }x+\frac{1}{3}=0\Leftrightarrow x=-\frac{1}{3}\)

Vậy...

b)\(\left(x+1\right)^2=5x\left(x+1\right)\)

\(\Leftrightarrow\left(x+1\right)^2-5x\left(x+1\right)=0\)

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

\(\Leftrightarrow\left(x+1\right)\left(-4x+1\right)=0\)

\(\Leftrightarrow-\left(x+1\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\4x=1\Leftrightarrow x=\frac{1}{4}\end{cases}}}\)

Vậy...

a) \(2x^2-50\)

\(=2\left(x^2-25\right)\)

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

b) \(x^2z+4xyz+4y^2z\)

\(=z\left(x^2+4xy+4y^2\right)\)

\(=z\left(x+2y\right)^2\)

c) \(x^2-y^2+12y-36\)

\(=x^2-\left(y^2-2\cdot x\cdot6+6^2\right)\)

\(=x^2-\left(y-6\right)^2\)

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

d) Đặt \(D=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

\(D=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]+1\)

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

Đặt \(x^2+3x+1=a\)

\(D=\left(a-1\right)\left(a+1\right)+1\)

\(D=a^2-1^2+1\)

\(D=a^2\)

Thay \(x^2+3x+1=a\)vào D ta có :

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

a: \(2x^2-50=2\left(x^2-5^2\right)\)

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

b:\(x^2z+4xyz+4y^2z=z\left(x^2+4xy+4y^2\right)\)

\(=z\left(x+2y\right)^2\)

c:\(x^2-y^2+12y-36=x^2-\left(y^2-12y+36\right)\)

\(=x^2-\left(y-6\right)^2\)

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

d:\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1\)

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

Đặt \(t=x^2+3x+1\)Ta có:

\(=\left(t-1\right)\left(t+1\right)+1\)

\(=t^2-1+1=t^2\)

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

áp dụng định lí fecma nhé bạn

Theo định lí Fecma nhỏ,ta có:\(n^5-n\equiv0\left(mod5\right)\)

Do vậy \(n^5-n⋮5^{\left(đpcm\right)}\)

~ Học tốt nha bạn~

Mong mọi người giúp với! Mình đang cần gấp! Thanhs nhiều!