\(a,\left(a+b\right)^3+\left(a-b\right)^3\)

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

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

\(=2a\left(a^2+3b^2\right)\)

\(b,9x^2+6xy+y^2\)

\(=\left(3x\right)^2+2.3x.y+y^2\)

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

\(c,4x^2-25\)

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

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

yiouoiyy

\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)

\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)

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

Vì \(\hept{\begin{cases}\left(x+y+z\right)^2\ge0\\\left(x+5\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)\(\Rightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y+z=0\\x+5=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y+z=0\\x=-5\\y=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=-3\\z=8\end{cases}}}\)

A = \(\frac{x}{x-1}+\frac{x}{x+1}+\frac{2-x^2}{1-x^2}\)

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

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

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

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

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

cậu xem lại đề nha

Đã bao giờ có 2 đường trung tuyến vuông góc bao giờ chưa?????

\(\left(0,3x^3-1,2xy^2\right).\left(0,2x^2-3y^2\right)\)

\(=0,06x^5-0,9x^3y^2-0,24x^3y^2+3,6xy^4\)

\(=0,06x^5-1,14x^3y^2+3,6xy^4\)

\(\frac{1}{\left(a+1\right)^2}+\frac{1}{\left(b+1\right)^2}\ge\frac{1}{ab+1}\)

\(\Leftrightarrow ab^3-a^2b^2+a^3b-2ab+1\ge0\)

\(\Leftrightarrow ab\left(a-b\right)^2+\left(ab-1\right)^2\ge0\)đúng

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