cho A=\(\left[\frac{1}{a^2}+\left(\frac{1}{a}+\frac{1}{b}\right):\frac{a+b}{2}+\frac{1}{b^2}\right].\frac{a^2b2}{a^3+b^3}:\left(a+b\right)\)
a, Rút gọn A.
b, Chứng minh A dương
Rút gọn hộ mình với đừng lướt qua thui nha, xin đó
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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) \(\left(\dfrac{3}{2}a^{-\dfrac{3}{2}}\cdot b^{-\dfrac{1}{2}}\right)\left(-\dfrac{1}{3}a^{\dfrac{1}{2}}b^{\dfrac{2}{3}}\right)=\left(\dfrac{3}{2}\cdot-\dfrac{1}{3}\right)\left(a^{-\dfrac{3}{2}}\cdot a^{\dfrac{1}{2}}\right)\left(b^{-\dfrac{1}{2}}\cdot b^{\dfrac{2}{3}}\right)\)
\(=-\dfrac{1}{2}a^{-1}b^{-\dfrac{1}{3}}\)
DT
6 tháng 5 2019
B= \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\)\(\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)
B= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}\)= \(\frac{1}{20}\)
vậy B= \(\frac{1}{20}\)
11 tháng 6 2017
a) Điều kiện : \(a\ne-b;b\ne1;a\ne-1\)
\(P=\frac{a^2\left(1+a\right)-b^2\left(1-b\right)-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{a^3+a^2+b^3-b^2-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a+b\right)\left(a-b\right)-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{\left(a+b\right)\left(a^2-ab+b^2+a-b-a^2b^2\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{a^2+b^2-a^2b^2+a-b-ab}{\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{a^2\left(1-b^2\right)-\left(1-b^2\right)+a\left(1-b\right)+\left(1-b\right)}{\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{\left(1-b\right)\left(a^2+a^2b-1-b+a+1\right)}{\left(1-b\right)\left(1+a\right)}\)
\(P=\frac{a^2+a^2b+a-b}{1+a}\)
\(P=\frac{a\left(a+1\right)+b\left(a-1\right)\left(a+1\right)}{1+a}\)
\(P=\frac{\left(a+1\right)\left(a+ab-b\right)}{1+a}\)
P = a + ab - b
b)
P = 3
<=> a + ab - b = 3
<=> a(b+1) - (b+1) +1 - 3 = 0
<=> (b+1)(a-1) = 2
Ta có bảng sau với a, b nguyên
| b+1 | 1 | 2 | -1 | -2 |
| a-1 | 2 | 1 | -2 | -1 |
| b | 0 | 1 | -2 | -3 |
| a | 3 | 2 | -1 | 0 |
| so với đk | loại | loại |
Vậy (a;b) \(\in\){ (3; 0) ; (0; -3)}
help me, hic
a, ĐK: \(a\ne0,b\ne0,a+b\ne0\)
\(A=\left[\frac{1}{a^2}+\left(\frac{1}{a}+\frac{1}{b}\right):\frac{a+b}{2}+\frac{1}{b^2}\right].\frac{a^2b^2}{a^3+b^3}:\left(a+b\right)\)
\(=\left[\frac{1}{a^2}+\frac{a+b}{ab}:\frac{a+b}{2}+\frac{1}{b^2}\right].\frac{a^2b^2}{a^3+b^3}:\left(a+b\right)\)
\(=\left[\frac{1}{a^2}+\frac{2}{ab}+\frac{1}{b^2}\right].\frac{a^2b^2}{a^3+b^3}:\left(a+b\right)\)
\(=\frac{\left(a+b\right)^2}{a^2b^2}.\frac{a^2b^2}{\left(a+b\right)\left(a^2-ab+b^2\right)}.\frac{1}{a+b}\)
\(=\frac{1}{a^2-ab+b^2}\)
b, \(a^2-ab+b^2=\left(a-\frac{1}{2}b\right)^2+\frac{3}{4}b^2>0\left(a,b\ne0\right)\)
\(\Rightarrow A=\frac{1}{a^2-ab+b^2}>0\forall a;b\)