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

\(1,x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\\ =\left(x^2+10x\right)\left(x^2+10x+24\right)+128\\ =\left(x^2+10x+12-12\right)\left(x^2+10x+12+12\right)+128\\ =\left(x^2+10x+12\right)-12^2+128=\left(x^2+10x+12\right)^2-16\\ =\left(x^2+10x+12-16\right)\left(x^2+10x+12+16\right)\\ =\left(x^2+10-4\right)\left(x^2+10x+28\right)\)

Có số số hạng là: (100-1)+1=100(số hạng)

​Tổng trên là: (100+1):2•100=5050

​

​

\(45^2+40^2-15^2+80x45\\ =\left(45+40\right)^2-15x^2\\ =\left(45+40-15\right)\left(45+40+15\right)\\ =70.100=70000\)

\(A=ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\\ =a^2b-ab^2+b^2c-bc^2+ac^2-a^2c\\ =\left(a-c\right)\left(b-c\right)\left(a-b\right)\)

\(B=a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\\ =ab^2-ac^2+bc^2-a^2b+a^2c-b^2c\\ =\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(C=\left(a+b+c\right)^3-a^3-b^3-c^3\\ =3\left(a^2b+ab^2+bc^2+b^2c+a^2c+ac^2+2abc\right)\\ =3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(a,\left(1+x^2\right)^2-4x\left(1-x^2\right)\\ =\left(1+x^2\right)^2-\left(\sqrt{4x\left(1-x^2\right)}\right)^2\\ =\left(1+x^2-\sqrt{4x\left(1-x^2\right)}\right)\left(1+x^2+\sqrt{4x\left(1-x^2\right)}\right)\)

\(b,\left(x^2-8\right)^2-36\\ =\left(x^2-8-6\right)\left(x^2-8+6\right)\\ =\left(x^2-14\right)\left(x^2-2\right)\)

Theo mk là trừ 36 nhé

\(3^{x+1}\)\(+3^{x+3}+2^{x+3}+2^{x+2}\)

= \(3^x.3+3^x.9+2^x.8+2^x.4\)

=\(3^x.12+2^x.12\)

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

 vì 12 chia hét cho 6=> bt trên cx chia hét cho 6

\(A=2x^2+6x+\dfrac{9}{2}-\dfrac{9}{2}\\ =2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)

Với mọi x thì \(2\left(x+\dfrac{3}{2}\right)^2\ge0\\ \Rightarrow2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

Hay \(A\ge-\dfrac{9}{2}\)

Để \(A=-\dfrac{9}{2}\) thì \(\left(x+\dfrac{3}{2}\right)^2=0\)

=>\(x+\dfrac{3}{2}=0\)

=>\(x=-\dfrac{3}{2}\)

Vậy...

\(C1:x^2-2x+xy-2y\\ =x\left(x-2\right)+y\left(x-2\right)\\ =\left(x-2\right)\left(x+y\right)\\ C2:x^2-2x+xy-2y\\ =x\left(x+y\right)-2\left(x+y\right)\\ =\left(x+y\right)\left(x-2\right)\)

\(a,x+5x^2=0\\ \Leftrightarrow x\left(5x+1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\\\5x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\\\x=-\dfrac{1}{5}\end{matrix}\right.\)

Vậy...

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

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

Vậy...

\(c,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow x=0\left(do...x^2+1>0\right)\)

Vậy...