Thực hiện phép tính:

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

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

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

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

\(=x^4-x^2+6x-9\)

b)\(\left(5x-1\right)\left(x+3\right)-\left(x-2\right)\left(5x-4\right)\)

\(=\left(5x^2+15x-x-3\right)-\left(5x^2-4x-10x+8\right)\)

\(=5x^2+15x-x-3-5x^2+4x+10x-8\)

\(=28x-11\)

Phân tích đa thức thành nhân tử:

\(6xy+5x-5y-3x^2-3y^2\)

\(=-3x^2+6xy-3y^2+5x-5y\)

\(=-3\left(x^2-2xy+y^2\right)+5\left(x-y\right)\)

\(=-3\left(x-y\right)^2+5\left(x-y\right)\)

\(=\left(x-y\right)\left[-3\left(x-y\right)+5\right]\)

\(=\left(x-y\right)\left(-3x+3y+5\right)\)

Ta thấy cứ 2 số tự nhiên liên tiếp chắc chắn có một số chia hết cho 2

Nên tích của chúng phải chia hết cho 2

Tìm x, biết:

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

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

\(\Leftrightarrow\left(4x^2-20x+25\right)\left(8x^3-60x^2+150x-125\right)+2x-5=0\)

\(\Leftrightarrow32x^5-240x^4+600x^3-500x^2-160x^4+1200x^3+3000x^2+2500x+200x^3-1500x^2+3750x-3125+2x-5=0\)

\(\Leftrightarrow32x^5-400x^4+2000x^3+1000x^2+6252x-3130=0\)

h)\(a^6+a^4+a^2b^2+b^4-b^6\)

\(=\left(a^4+a^2b^2+b^4\right)+\left(a^6-b^6\right)\)

\(=\left(a^4+a^2b^2+b^4\right)+\left[\left(a^2\right)^3-\left(b^2\right)^3\right]\)

\(=\left(a^4+a^2b^2+b^4\right)+\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)\)

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

\(y^3\left(x-1\right)-7y^3+7xy^3\)

\(=y^3\left(x-1\right)-\left(7y^3-7xy^3\right)\)

\(=y^3\left(x-1\right)-7y^3\left(1-x\right)\)

\(=y^3\left(x-1\right)+7y^3\left(x-1\right)\)

\(=\left(y^3+7y^3\right)\left(x-1\right)\)

\(=y^3\left(1+7\right)\left(x-1\right)\)

Tìm x, biết:

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

\(\Leftrightarrow x^2-2x+3x-6=0\)

\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Vậy x=2 ; x=-3

\(x^2+x-20\)

\(=x^2+5x-4x-20\)

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

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