Rút gon \(\frac{48c^2+27b^2+36a^2}{a^2+b^2+c^2}\)
K
Khách
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NK
2
16 tháng 10 2020
8a3 - 36a2b + 54ab2 - 27b3 - 8
= ( 8a3 - 36a2b + 54ab2 - 27b3 ) - 8
= ( 2a - 3b )3 - 23
= ( 2a - 3b - 2 )[ ( 2a - 3b )2 + 2( 2a - 3b ) + 4 ]
= ( 2a - 3b - 2 )( 4a2 - 12ab + 9b2 + 4a - 6b + 4 )
CH
Cô Hoàng Huyền
Admin
VIP
24 tháng 11 2017
\(R=\frac{3a^2-2ab-b^2}{2a^2+ab-b^2}:\frac{3a^2-4ab+b^2}{3a^2+2ab-b^2}\)
\(R=\frac{3a^2-2ab-b^2}{2a^2+ab-b^2}.\frac{3a^2+2ab-b^2}{3a^2-4ab+b^2}\)
\(R=\frac{\left(3a+b\right)\left(a-b\right)}{\left(a+b\right)\left(2a-b\right)}.\frac{\left(a+b\right)\left(3a-b\right)}{\left(a-b\right)\left(3a-b\right)}\)
\(R=\frac{3a+b}{2a-b}\)
29 tháng 12 2019
Hình như đề sai.Sửa đề luôn nha !
\(ĐKXĐ:x\ne\pm2\)
\(A=\left(\frac{x}{x^2-4}-\frac{2}{x-2}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right):\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\)
\(=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{x^2-4+10-x^2}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{-6}=\frac{1}{x-2}\)
b
Để \(A< 0\Rightarrow\frac{1}{x-2}< 0\Rightarrow x-2< 0\Rightarrow x< 2\)
c
Để A nguyên thì \(\frac{1}{x-2}\) nguyên
\(\Rightarrow1⋮x-2\)
\(\Rightarrow x-2\in\left\{1;-1\right\}\Rightarrow x\in\left\{3;1\right\}\)
29 tháng 2 2020
\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}\left(\frac{1}{a}+\frac{1}{b}\right)\right].\frac{ab}{\left(a+b\right)^2}\)
\(=\left(\frac{1}{a}+\frac{1}{b}\right)^2.\frac{ab}{\left(a+b\right)^2}\)
\(=\frac{1}{ab}\)
\(B=\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+14xy+y^2}{16x}\)
\(=\frac{\left(2x+y\right)^2+2\left(2x+y\right)\left(2x-y\right)+\left(2x-y\right)^2}{\left(2x+y\right)^2.\left(2x-y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)
\(=\frac{\left(2x+y+2x-y\right)^2}{\left(2x+y\right)^2.\left(2x-y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)
\(=\frac{x}{\left(2x-y\right)^2}\)
LT
29 tháng 2 2020
\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}.\left(\frac{1}{a}+\frac{1}{b}\right)\right].\frac{ab}{\left(a+b\right)^2}\)
ĐK: a, b khác 0, a khác -b
\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}.\left(\frac{a+b}{ab}\right)\right].\frac{ab}{\left(a+b\right)^2}\)
\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{ab}\right].\frac{ab}{\left(a+b\right)^2}=\left(\frac{1}{a}+\frac{1}{b}\right)^2.\frac{ab}{\left(a+b\right)^2}\)
\(A=\frac{\left(a+b\right)^2}{ab}.\frac{ab}{\left(a+b\right)^2}=1\)
\(B=\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{\left(4x^2-y^2\right)}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+4xy+y^2}{16xy}\)
ĐK: xy khác 0, y \(\ne\pm\)2x
\(B=\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{\left(2x-y\right).\left(2x+y\right)}+\frac{1}{\left(2x+y\right)^2}\right].\frac{\left(2x+y\right)^2}{16xy}\)
\(B=\left[\frac{1}{\left(2x-y\right)}+\frac{1}{\left(2x+y\right)}\right]^2.\frac{\left(2x+y\right)^2}{16xy}\)
\(B=\left(\frac{2x+y+2x-y}{\left(2x-y\right).\left(2x+y\right)}\right)^2.\frac{\left(2x+y\right)^2}{16xy}\)
\(B=\frac{16x^2}{\left(2x-y\right)^2.\left(2x+y\right)^2}.\frac{\left(2x+y\right)^2}{16xy}\)
\(B=\frac{x}{\left(2x-y\right)^2.y}\)
31 tháng 5 2017
- Đk : \(\hept{\begin{cases}x-3\ne0\\x-2\ne0\\x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-2\\x\ne2\end{cases}}}\)
- \(P=\frac{\left(2+x\right)^2+4x^2-\left(2-x\right)^2}{\left(x-2\right)\left(x+2\right)}.\frac{x^2\left(2-x\right)}{x\left(x-3\right)}\)\(\Rightarrow P=\frac{8x+4x^2}{\left(x-2\right)\left(x+2\right)}.\frac{x\left(2-x\right)}{x-3}\)\(\Rightarrow p=\frac{4x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}.\frac{x\left(x-2\right)}{3-x}=\frac{4x^2}{3-x}\)
- \(|x-5|=2\)
- nếu \(x\ge5\)=> x-5=2 =>x=7 (TM) => \(P=\frac{4.7^2}{-7+3}=-49\)
- Nếu \(x< 5\)=> x-5 = -2 => x = 3 Loại
\(\frac{48c^2+27b^2+36c^2}{a^2+b^2+c^2}=\frac{14\cdot3c^2+9\cdot3b^2+12\cdot3a^2}{a^2+b^2+c^2}=\frac{3\left(14c^2+9b^2+12a^2\right)}{a^2+b^2+c^2}\)