Sửa đề: CMR: \(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\)

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{a^2}+\frac{1}{b^2}\ge2.\sqrt{\frac{1}{a^2}.\frac{1}{b^2}}=2.\frac{1}{ab}=\frac{2}{ab}\)

                                                   đpcm

\(\left(18x^3-6x^2+2x\right):2x-9x\left(x-2\right)=9x^2-3x+1-9x^2+18x=-8;\)

\(\Leftrightarrow18x-3x+1=-8\Leftrightarrow15x+1=-8\Leftrightarrow15x=-8-1=-9\Leftrightarrow x=-9:15=\frac{-9}{15}\)

\(=\frac{-3}{5}\)

\(S=\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right).\left(x+2\right)}+....+\frac{1}{\left(x+99\right).\left(x+100\right)}\)

\(S=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+99}-\frac{1}{x+100}\)

\(S=\frac{1}{x}-\frac{1}{x+100}=\frac{x+100-x}{x.\left(x+100\right)}=\frac{100}{x^2+100x}\)

Sửa đề: \(P=\frac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)

\(P=\frac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)

\(P=\frac{\left(a+b\right)^3+c^3-3abc-3a^2b-3ab^2}{a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2}\)

\(P=\frac{\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)}{2.\left(a^2+b^2+c^2-ab-bc-ca\right)}\)

\(P=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc+3ab\right)}{2.\left(a^2+b^2+c^2-ab-bc-ca\right)}\)

\(P=\frac{5\left(a^2+b^2+c^2-ab-ac-bc\right)}{2.\left(a^2+b^2+c^2-ab-bc-ca\right)}\)( a+b+c=0)

\(P=\frac{5}{2}\left[\left(a^2+b^2+c^2-ab-bc-ca\right)\ne0\right]\)

\(\frac{x^4-8x}{4x-x^3}=\frac{x\left(x^3-2^3\right)}{x\left(4-x^2\right)}\)

\(=\frac{x\left(x-2\right)\left(x^2+2x+4\right)}{x\left(x+2\right)\left(x-2\right)}=\frac{\left(x^2+2x+4\right)}{\left(x-2\right)}\)

k hiểu gì cả viết có dấu với bạn

\(27x^3-8y^3=\left(3x\right)^3-\left(2y\right)^3=\left(3x-2y\right).\left[\left(3x\right)^2+3x3y+\left(2y\right)^2\right]\)

\(27x^3-8y^3=\left(3x\right)^3-\left(2y\right)^3\)

\(=\left(3x-2y\right)\left[\left(3x\right)^2+3x.2y+\left(2y^{ }\right)^2\right]\)

\(=\left(3x-2y\right)\left(9x+6xy+4y\right)\)