\(B=\frac{a+1}{a^2-a+1}-\frac{1}{a+1}\)\(-\frac{a-2}{a^3+1}\)
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H9
HT.Phong (9A5)
CTVHS
18 tháng 8 2023
a) \(a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}\cdot a^{\dfrac{7}{6}}=a^{\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{7}{6}}=a^2\)
b) \(a^{\dfrac{2}{3}}\cdot a^{\dfrac{1}{4}}:a^{\dfrac{1}{6}}=a^{\dfrac{2}{3}+\dfrac{1}{4}-\dfrac{1}{6}}=a^{\dfrac{3}{4}}\)
c) \(\left(\dfrac{3}{2}a^{-\dfrac{3}{2}}\cdot b^{-\dfrac{1}{2}}\right)\left(-\dfrac{1}{3}a^{\dfrac{1}{2}}b^{\dfrac{2}{3}}\right)=\left(\dfrac{3}{2}\cdot-\dfrac{1}{3}\right)\left(a^{-\dfrac{3}{2}}\cdot a^{\dfrac{1}{2}}\right)\left(b^{-\dfrac{1}{2}}\cdot b^{\dfrac{2}{3}}\right)\)
\(=-\dfrac{1}{2}a^{-1}b^{-\dfrac{1}{3}}\)
L
2
17 tháng 7 2018
\(B=\frac{a+1}{a^2-a+1}-\frac{1}{a+1}+\)\(\frac{a-2}{a^3+1}\)
\(B=\frac{\left(a+1\right)^2}{a^3+1}-\frac{a^2-a+1}{a^3+1}+\)\(\frac{a-2}{a^3+1}\)
\(B=\frac{a^2+2a+1-a^2+a-1-a+2}{a^3+1}\)
\(B=\frac{2a+2}{a^3+1}\)
\(B=\frac{2\left(a+1\right)}{\left(a+1\right)\left(a^2+a+1\right)}\)
\(B=\frac{2}{a+1}\)
17 tháng 7 2018
\(B=\frac{a+1}{a^2-a+1}-\frac{1}{a+1}-\frac{a-2}{a^3+1}\) ĐKXĐ : \(x\ne-1\)
\(=\frac{\left(a+1\right)^2}{\left(a+1\right)\left(a^2-a+1\right)}-\frac{a^2-a+1}{\left(a+1\right)\left(a^2-a+1\right)}\)\(-\frac{a-2}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\frac{a^2+2a+1-a^2+a-1-a+2}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\frac{\left(a^2-a^2\right)+\left(2a+a-a\right)+\left(1-1+2\right)}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\frac{2a+2}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\frac{2\left(a+1\right)}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\frac{2}{a^2-a+1}\)
PN
13 tháng 5 2021
1,
\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)
\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a^2-a}.\frac{a-2}{a+2}\)
\(=\frac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{a\left(a-1\right)}.\frac{a-2}{a+2}\)
\(=\frac{2a\left(a-1\right)\left(a-2\right)}{a\left(a-1\right)\left(a+2\right)}=\frac{2\left(a-2\right)}{a+2}\)
Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)
PN
14 tháng 5 2021
2,
a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)
b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)
\(=\frac{2\sqrt{x}+2+2}{x-4}=\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{2}{\sqrt{x}-2}\)
c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
18 tháng 5 2021
a, Với \(a\ge0;a\ne9\)
\(A=\left(\frac{1}{\sqrt{a}-3}+\frac{1}{\sqrt{a}+3}\right)\left(1-\frac{3}{\sqrt{a}}\right)\)
\(=\left(\frac{2\sqrt{a}}{a-9}\right)\left(\frac{\sqrt{a}-3}{\sqrt{a}}\right)=\frac{2}{\sqrt{a}+3}\)
b, Ta có : \(\frac{2}{\sqrt{a}+3}>\frac{1}{2}\Rightarrow\frac{2}{\sqrt{a}+3}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{1-\sqrt{a}}{2\sqrt{a}+6}>0\Rightarrow1-\sqrt{a}>0\)vì \(2\sqrt{a}+6>0\)
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
\(B=\frac{a+1}{a^2-a+1}-\frac{1}{a+1}-\frac{a-2}{a^3+1}=\frac{\left(a+1\right)^2}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a^2-a+1}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a-2}{a^3+1}\\ \)
\(=\frac{a^2+2a+1}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a^2-a+1}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a-2}{\left(a+1\right).\left(a^2-a+1\right)}\)
\(=\frac{a^2+2a+1-\left(a^2-a+1\right)-\left(a-2\right)}{\left(a+1\right).\left(a^2-a+1\right)}=\frac{2a+2}{\left(a+1\right).\left(a^2-a+1\right)}=\frac{2}{a^2-a+1}\)
ơ bài này là bài lớp 2 mà , sao lại ghi là lớp 8 vậy