TOÁN 8 NÂNG CAO
Rút gọn biểu thức A:
A=\(\left[\dfrac{3\left(a+2\right)}{a^3+a^2+a+1}+\dfrac{2a^2-a-10}{a^3-a^2+a-1}\right]:\left[\dfrac{5}{a^2+1}+\dfrac{3}{2a+2}-\dfrac{3}{2a-2}\right]\)
(Mọi người ơi giải giúp mình bài này với !!!!!!!!!)
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Những câu hỏi liên quan
2 tháng 7 2021
a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)
\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)
=-a-1
b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)
\(=\left|3a-5\right|-2a+4\)
\(=5-3a-2a+4\)
=9-5a
c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)
\(=4a-3-\left|2a-1\right|\)
\(=4a-3-2a+1\)
\(=2a-2\)
d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
28 tháng 6 2021
`M=sqrt{(3a-1)^2}+2a-3`
`=|3a-1|+2a-3`
`=3a-1+2a-3(do \ a>=1/3)`
`=5a-4`
`N=sqrt{(4-a)^2}-a+5`
`=|4-a|-a+5`
`=a-4-a+5(do \ a>4)`
`=1`
`I=sqrt{(3-2a)^2}+2-7`
`=|3-2a|-5`
`=3-2a-5(do \ a<3/2)`
`=-2-2a`
`K=(a^2-9)/4*sqrt{4/(a-2)^2}`
`=(a^2-9)/4*|2/(a-2)|`
`=(a^2-9)/(2|a-2|)`
Nếu `3>a>2=>|a-2|=a-2`
`=>K=(a^2-9)/(2(a-2))`
Nếu `a<2=>|a-2|=2-a`
`=>K=(a^2-9)/(2(2-a))`
28 tháng 6 2021
\(M=\left|3a-1\right|+2a-3\)
Mà \(a-\dfrac{1}{3}\ge0\)
\(\Rightarrow M=3a-1+2a-3=5a-4\)
\(N=\left|4-a\right|-a+5\)
Mà \(4-a< 0\)
\(\Rightarrow N=a-4-a+5=1\)
\(I=\left|3-2a\right|-5\)
Mà \(a-\dfrac{3}{2}< 0\)
\(\Rightarrow I=3-2a-5=-2a-2\)
K, Ta có : \(a-3< 0\)
\(\Rightarrow K=\dfrac{2\left(a^2-9\right)}{4\left|a-2\right|}=\dfrac{\left(a-3\right)\left(a+3\right)}{\left|2a-4\right|}\)
26 tháng 11 2021
\(B=\left(\dfrac{a-b}{a^2+ab}-\dfrac{a}{b^2+ab}\right):\left(\dfrac{b^3}{a^3-ab^2}+\dfrac{1}{a+b}\right)\)
\(=\left(\dfrac{a-b}{a\left(a+b\right)}-\dfrac{a}{b\left(a+b\right)}\right):\left(\dfrac{b^3}{a\left(a-b\right)\left(a+b\right)}+\dfrac{1}{a+b}\right)\)
\(=\dfrac{b\left(a-b\right)-a^2}{ab\left(a+b\right)}:\dfrac{b^3+a\left(a-b\right)}{a\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{ab-b^2-a^2}{ab\left(a+b\right)}\cdot\dfrac{a\left(a-b\right)\left(a+b\right)}{a^2-ab+b^3}\)
\(=\dfrac{\left(a-b\right)\left(ab-b^2-a^2\right)}{b\left(a^2-ab+b^3\right)}\)
\(=\dfrac{-\left(a-b\right)\left(a^2-ab+b^2\right)}{b\left(a^2-ab+b^3\right)}\)
Đề lỗi rồi chứ mình ko rút gọn đc nữa
6 tháng 7 2023
\(=\dfrac{a+1-1}{\sqrt{a+1}}\cdot\dfrac{a^2+3\sqrt{a+1}-2a+2a-a^2}{a}\)
\(=\dfrac{3\sqrt{a+1}}{\sqrt{a+1}}=3\)
22 tháng 5 2022
\(A=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{a^3-1}+\dfrac{1}{a-1}\right]\cdot\dfrac{a\left(a^2+1\right)}{2a}\)
\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}\)
\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}=\dfrac{a^2+1}{2}\)
Đặt: \(L=\dfrac{3\left(a+2\right)}{a^3+a^2+a+1}+\dfrac{2a^2-a-10}{a^3-a^2+a-1}\)
Ta có:
\(\dfrac{3\left(a+2\right)}{a^3+a^2+a+1}=\dfrac{3\left(a+2\right)}{a^2\left(a+1\right)+1\left(a+1\right)}=\dfrac{3\left(a+2\right)}{\left(a^2+1\right)\left(a+1\right)}\)
\(\dfrac{2a^2-a-10}{a^3-a^2+a-1}=\dfrac{a\left(2a-1\right)-10}{a^2\left(a-1\right)+1\left(a-1\right)}=\dfrac{a\left(2a-1\right)-10}{\left(a^2+1\right)\left(a-1\right)}\)
Như vậy \(L=\dfrac{3\left(a+2\right)}{\left(a^2+1\right)\left(a+1\right)}+\dfrac{a\left(2a-1\right)-10}{\left(a^2+1\right)\left(a-1\right)}\)
Đặt:
\(N=\dfrac{5}{a^2+1}+\dfrac{3}{2a+2}-\dfrac{3}{2a-2}\)
\(N=\dfrac{5}{a^2+1}+\dfrac{3\left(2a-2\right)}{\left(2a+2\right)\left(2a-2\right)}-\dfrac{3\left(2a+2\right)}{\left(2a+2\right)\left(2a-2\right)}\)
\(N=\dfrac{5}{a^2+1}+\dfrac{6a-6}{4a^2-4}-\dfrac{6a+6}{4a^2-4}\)
\(N=\dfrac{5}{a^2+1}+\dfrac{6a-6-6a-6}{4a^2-4}=\dfrac{5}{a^2+1}+\dfrac{-12}{4a^2-4}\)
\(N=\dfrac{5}{a^2+1}+\dfrac{-12}{4\left(a^2-1\right)}=\dfrac{5}{a^2+1}+\dfrac{-3}{a^2-1}\)
\(N=\dfrac{5\left(a^2-1\right)}{\left(a^2+1\right)\left(a^2-1\right)}+\dfrac{-3\left(a^2+1\right)}{\left(a^2-1\right)\left(a^2+1\right)}\)
\(N=\dfrac{5a^2-5-3a^2-3}{a^4-1}=\dfrac{2a^2-8}{a^4-1}\)
Thay M với N vào A Mình cạn sức rồi
Cảm ơn nhiều!!!!