e lớp 7 nên sai thì thôi ạ

\(P=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right).\frac{x+2007}{x}\left(ĐK:x\ne\pm1;0\right)\)

\(=\left(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{x^2-1}+\frac{x^2-4x-1}{x^2-1}\right).\frac{x+2007}{x}\)

\(=\left[\frac{\left(x+1+x-1\right)\left(x+1-x-1\right)}{x^2-1}+\frac{x^2-4x-1}{x^2-1}\right].\frac{x+2007}{x}\)

\(=\left(\frac{2x.0}{x^2-1}+\frac{x^2-4x-1}{x^2-1}\right).\frac{2007}{x}+\frac{x^2-4x-1}{x^2-1}\)

\(=\frac{2007\left(x^2-4x-1\right)}{x^3-x}+\frac{x^2-4x-1}{x^2-1}\)

\(=\frac{2007x^2-8028x-2007}{x^3-x}+\frac{x^3-4x^2-x}{x^3-x}\)

\(=\frac{x^3+2003x^2-8029x-2007}{x^3-x}\)( số to vch )

ừm , sai thật em ạ, tìm x mà số to quá

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

\(\Leftrightarrow\left(4x^2+x-2\right)\sqrt{x+4}=4-3x-8x^2\)

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

\(\Leftrightarrow\left(4x^2+x-2\right)^2\left(x+4\right)-\left(4-3x-8x^2\right)^2=0\)

ĐKXĐ: \(x\ge-4\)

PT \(\Leftrightarrow\left(\sqrt{x+4}+2\right)\left(2x+1-\sqrt{x+4}\right)\left(2x+\sqrt{x+4}\right)=0\)

\(x\in\left(2;+\infty\right)\)

ĐNÁ: Đông Nam Á.còn câu còn lại mình không biết đâu.

Ta thấy : \(\sqrt{5}>\sqrt{4}=2\)

Nên \(2-\sqrt{5}< 0\)

Mà \(\sqrt{3}>0\)

Nên dẫn đến \(\sqrt{3}>2-\sqrt{5}\)

\(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+2\sqrt{7}}=\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\frac{\sqrt{2}}{2}\)

Hải Ngọc bạn làm mỗi 1 vế mà lại làm vế dễ nhát nữa thì chịu rồi

\(\frac{2}{\sqrt{3}}+\frac{\sqrt{2}}{3}+\frac{2}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\frac{2\sqrt{3}}{3}+\frac{\sqrt{2}}{3}+\frac{2\sqrt{3}}{3}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)

\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\sqrt{12}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\sqrt{12\left(\frac{5}{12}-\frac{1}{\sqrt{6}}\right)}\)

\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\sqrt{5-2\sqrt{6}}=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\cdot\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\left|\sqrt{3}-\sqrt{2}\right|=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\left(\sqrt{3}-\sqrt{2}\right)\)(vì \(\sqrt{3}-\sqrt{2}>0\))

\(=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{3}=\sqrt{3}\)

xét hiệu a³+b³+3abc=(a+b+c)(a²+b²+c²-ab-ac-bc)=(a+b+c)\(\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}\ge0\)

đẳng thức xảy ra khi a=b=c

Ta có: \(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)^3-3.\left(a+b\right).c.\left(a+b+c\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right).\left[\left(a+b+c\right)^2-3\left(a+b\right).c-3ab\right]\)

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

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

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

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

\(=\frac{1}{2}.\left(a+b+c\right).\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)( Vì a, b, c không âm )

\(\Rightarrow a^3+b^3+c^3\ge3abc\)( đpcm )