a) \(x^{12}-3x^6+1\)

\(=\left(x^6\right)^2-2\cdot x^6\cdot1+1^2-x^6\)

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

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

b) \(x^4+6x^3+7x^2-6x+1\)

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

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

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

Đặt \(2x^2+x-2018=a;x^2-5x-2017=b\) ta có : 

\(a^2+4b^2=4ab\)

\(\Leftrightarrow\)\(a^2-4ab+4b^2=0\)

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

\(\Leftrightarrow\)\(a-2b=0\)

\(\Leftrightarrow\)\(2x^2+x-2018-2\left(x^2-5x-2017\right)=0\)

\(\Leftrightarrow\)\(2x^2+x-2018-2x^2+10x+4034=0\)

\(\Leftrightarrow\)\(11x+2016=0\)

\(\Leftrightarrow\)\(x=\frac{-2016}{11}\)

Vậy \(x=\frac{-2016}{11}\)

Chúc bạn học tốt ~ 

MN = 1 / 2 BC 

ó gợi ý đấy mình hơi bận nên không ghi roc ược sorry

ukm, k sao đ

xét tứ giác MNPQ có góc M+ góc N+ góc P + góc Q=360⁰

=>góc M+85⁰+120⁰+70⁰=360⁰

=>góc M=360⁰-275⁰=85⁰

vậy...

Bài 1:

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

\(x^2+6x+9+x^2-2x-2x^2=0\)

\(4x+9=0\)

\(x=\frac{-9}{4}\)

b) \(5x\left(x-4\right)-x+4=0\)

\(5x\left(x-4\right)-\left(x-4\right)=0\)

\(\left(x-4\right)\left(5x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\5x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{1}{5}\end{cases}}}\)

Bài 2:

a) \(x^2-4x=x\left(x-4\right)\)

b) \(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=\left(x+5\right)^2\)

c) \(x^2-y^2+2y-1\)

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

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

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

d) \(x^2-11x+18\)

\(=x^2-2x-9x+18\)

\(=x\left(x-2\right)-9\left(x-2\right)\)

\(=\left(x-2\right)\left(x-9\right)\)

(x + 3)² + x(x - 2) = 2x² 

x² + 6x + 9 + x² - 2x = 2x² 

<=> 2x² + 4x + 9 = 2x² 

<=> 4x = -9

<=> x = -9/4