b) x-2x√x+x

\(\Leftrightarrow x\left(1-2\sqrt{x}+1\right)\)

\(\Leftrightarrow x\left(2-2\sqrt{x}\right)\)

\(\Leftrightarrow2x\left(1-\sqrt{x}\right)\)

ồ cuk khó nhỉ

Nếu các bn thích thì ...........

cứ cho NTN này nhé !

\(x^{16}+x^8-2\)

\(=\left(x^8-1\right)\left(x^8+2\right)\)

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

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

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

= (x^16 -1)+(x^8 -1)

= (x^8 -1)(x^8 +1)+ (x^8 -1)

= (x^8 -1)(x^8 +2)

= (x^4 -1)(x^4 +1)(x^8 +2)

= (x^2 -1)(x^2 +1)(x^4 +1)(x^8 +2)

= (x-1)(x+1)(x^2 +1)(x^4 +1)(x^8 +2)

=(100^2-99^2)+(98^2-97^2)...+(2^2-1^1)

=100+99+98+97+...+2+1(áp dụng hdt)

Tới đây tính tổng dãy có quy luật của lớp 5 thôi

Cảm ơn

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

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

\(\Leftrightarrow ab+bc+ca=\frac{-\left(a^2+b^2+c^2\right)}{2}=\frac{-2}{2}=-1\)

\(\Leftrightarrow\left(ab+bc+ca\right)^2=1\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=1\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=1\) (vì a+b+c=0)

Từ \(a^2+b^2+c^2=2\Leftrightarrow\left(a^2+b^2+c^2\right)^2=4\)

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

\(\Leftrightarrow a^4+b^4+c^4=4-2\left(a^2b^2+b^2c^2+c^2a^2\right)=4-2.1=2\)

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

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

\(a^4-25=\left(a^2\right)^2-5^2=\left(a^2-5\right)\left(a^2+5\right)\)

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

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

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

\(64x^3+\frac{1}{27}=\left(4x\right)^3+\left(\frac{1}{3}\right)^3=\left(4x+\frac{1}{3}\right)\left(16x^2+\frac{4}{3}x+\frac{1}{9}\right)\)

Tham khảo~

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

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

\(a^4-25=\left(a^2\right)^2-5^2=\left(a^2+5\right)\left(a^2-5\right)\)

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

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

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

\(64x^3+\frac{1}{27}=\left(4x\right)^3+\left(\frac{1}{3}\right)^3=\left(4x+\frac{1}{3}\right)\left(16x^2-\frac{4}{3}x+\frac{1}{9}\right)\)

\(3x^2+48+24x-12y^2\)

\(=-3\left(-x^2-16-8x+4y^2\right)\)

\(=-3\left[\left(2y\right)^2-\left(x^2+8x+16\right)\right]\)

\(=-3\left[\left(2y\right)^2-\left(x+4\right)^2\right]\)

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