Rút gọn :
a) \(\left(a-\frac{a^2+b^2}{a-b}\right).\left(\frac{1}{b}+\frac{2}{a+b}\right)\)
b) \(\left(\frac{3a+1}{a^2-3a}+\frac{3a-1}{a^2+3a}\right).\frac{a^2-9}{a^2+1}\)
K
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Những câu hỏi liên quan
11 tháng 8 2017
Bài 1:
Ta có:
\(\left(a-b+c\right)^3=a^3-b^3+c^3-3a^2b+3a^2c+3ab^2+3b^2c+3ac^2-3bc^2-6abc\)
\(\Rightarrow\left(\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}\right)^3=\frac{1}{9}-\frac{2}{9}+\frac{4}{9}-\frac{1}{3}.\sqrt[3]{2}+\frac{1}{3}.\sqrt[3]{4}+\frac{1}{3}.\sqrt[3]{4}+\frac{2}{3}.\sqrt[3]{2}\)
\(+\frac{2}{3}.\sqrt[3]{2}-\frac{2}{3}.\sqrt[3]{4}-\frac{4}{3}=\sqrt[3]{2}-1\)
\(\Rightarrow\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}\)
3 tháng 2 2022
\(=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{1}{a-1}\right]:\dfrac{2a}{3}\)
\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}\)
\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}=\dfrac{3}{2a}\)
KN
27 tháng 9 2020
a) \(ĐK:a\ne1;a\ne0\)
\(A=\left[\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\right]:\frac{a^3+4a}{4a^2}=\left[\frac{a^2-2a+1}{a^2+a+1}-\frac{1-2a^2+4a}{a^3-1}+\frac{a^2+a+1}{a^3-1}\right].\frac{4a^2}{a^3+4a}\)\(=\left[\frac{a^3-3a^2+3a-1}{a^3-1}-\frac{1-2a^2+4a}{a^3-1}+\frac{a^2+a+1}{a^3-1}\right].\frac{4a^2}{a^3+4a}=\frac{a^3-1}{a^3-1}.\frac{4a}{a^2+4}=\frac{4a}{a^2+4}\)
b) Ta có: \(a^2+4\ge4a\)(*)
Thật vậy: (*)\(\Leftrightarrow\left(a-2\right)^2\ge0\)
Khi đó \(\frac{4a}{a^2+4}\le1\)
Vậy MaxA = 1 khi x = 2
23 tháng 6 2021
a, \(M=\left[\frac{2}{3a}-\frac{2}{a+1}\left(\frac{a+1}{3a}-a-1\right)\right]:\frac{a-1}{a}\)ĐK : \(a\ne\pm1;0\)
\(=\left[\frac{2}{3a}-\frac{2}{a+1}\left(\frac{a+1-3a^2-3a}{3a}\right)\right]:\frac{a-1}{a}\)
\(=\left[\frac{2}{3a}-\frac{2}{a+1}\left(\frac{-3a^2-2a+1}{3a}\right)\right]:\left(\frac{a-1}{a}\right)\)
\(=\left[\frac{2}{3a}+\frac{2}{a+1}.\frac{\left(a+1\right)\left(3a-1\right)}{3a}\right]:\left(\frac{a-1}{a}\right)\)
\(=\left(\frac{2}{3a}+\frac{2\left(3a-1\right)}{3a}\right):\left(\frac{a-1}{a}\right)=\frac{2a}{a-1}\)
b, Để P nguyên \(\frac{2a}{a-1}=\frac{2\left(a-1\right)+2}{a-1}=2+\frac{2}{a-1}\)
Vì 2 nguyên nên \(\frac{2}{a-1}\)cũng phải nguyên
\(\Rightarrow a-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| a - 1 | 1 | -1 | 2 | -2 |
| a | 2 ( tm ) | 0 ( tm ) | 3 (tm ) | -1 (tm) |
c, Ta có : \(P\le1\Rightarrow\frac{2a}{a-1}\le1\Leftrightarrow\frac{2a}{a-1}-1\le0\)
\(\Leftrightarrow\frac{a+1}{a-1}\le0\)mà a + 1 > a - 1
\(\hept{\begin{cases}a+1\ge0\\a-1\le0\end{cases}\Leftrightarrow\hept{\begin{cases}a\ge-1\\a\le1\end{cases}\Leftrightarrow-1\le}a\le1}\)
Kết hợp với đk vậy \(-1< a< 1;a\ne0\)thì \(P\le1\)
3 tháng 4 2020
a) P = \(\left(\frac{3\sqrt{a}}{a+\sqrt{a}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right):\frac{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}{\left(2.a+2.\sqrt{ab}+2.b\right)}\)
= \(\left(\frac{3\sqrt{a}.\left(\sqrt{a}-\sqrt{b}\right)-3.a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right).\left(a+\sqrt{ab}+b\right)}\right).\frac{2.\left(a+\sqrt{ab}+b\right)}{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\frac{a-2.\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}.\frac{2}{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\frac{2}{a-1}\)
b) P nguyên <=> \(\frac{2}{a-1}\)nguyên => 2 \(⋮\)a - 1
=> ( a- 1 ) = { \(\pm\)1 ; \(\pm\) 2} => a = { -1 ; 0 ; 2 ;3 }
DT
1
DT
26 tháng 6 2016
\(\left(\frac{3a+1}{a^2-3a}+\frac{3a-1}{a^2+3a}\right)\):\(\frac{a^2+1}{a^2-9}\)
=\(\left[\frac{3a+1}{a\left(a-3\right)}+\frac{3a-1}{a\left(a+3\right)}\right]\): \(\frac{a^2+1}{\left(a-3\right)\left(a+3\right)}\)
=\(\left[\frac{\left(3a+1\right)\left(a+3\right)}{a\left(a-3\right)\left(a+3\right)}+\frac{\left(3a-1\right)\left(a-3\right)}{a\left(a+3\right)\left(a-3\right)}\right]\): \(\frac{a^2+1}{\left(a-3\right)\left(a+3\right)}\)
=\(\frac{3a^2+9a+a+3+3a^2-9a-a+3}{a\left(a-3\right)\left(a+3\right)}\): \(\frac{a^2+1}{\left(a-3\right)\left(a+3\right)}\)
=\(\frac{6a^2+6}{a\left(a-3\right)\left(a+3\right)}\): \(\frac{a^2+1}{\left(a-3\right)\left(a+3\right)}\)
=\(\frac{6\left(a^2+1\right)}{a\left(a-3\right)\left(a+3\right)}\).\(\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)
=\(\frac{6}{a}\)
a/ đk: a\(\ne b\), b\(\ne0,a\ne-b\)
= \(\frac{a\left(a-b\right)-a^2-b^2}{a-b}.\frac{a+b+2b}{b\left(a+b\right)}\)
= \(\frac{a^2-ab-a^2-b^2}{a-b}.\frac{a+3b}{b\left(a+b\right)}\)
= \(\frac{-ab-b^2}{a-b}.\frac{a+3b}{b\left(a+b\right)}\)
= \(\frac{-b\left(a+b\right)\left(a+3b\right)}{b\left(a+b\right)\left(a-b\right)}\)
= \(\frac{-a-3b}{a-b}\)
b/ đk: a\(\ne0,a\ne\pm3\)
= \(\left[\frac{3a+1}{a\left(a-3\right)}+\frac{3a-1}{a\left(a+3\right)}\right].\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)
= \(\frac{\left(3a+1\right)\left(a+3\right)+\left(3a-1\right)\left(a-3\right)}{a\left(a-3\right)\left(a+3\right)}.\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)
= \(\frac{6a^2+6}{a\left(a-3\right)\left(a+3\right)}.\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)
= \(\frac{6\left(a^2+1\right)\left(a-3\right)\left(a+3\right)}{a\left(a^2+1\right)\left(a-3\right)\left(a+3\right)}\)
= \(\frac{6}{a}\)