Rút gọn phân thức
\(N=\dfrac{\left(a-1\right)^4-11\left(a-1\right)^2+30}{3\left(a-1\right)^4-18\left(a^2-2a\right)-3}\)
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AH
Akai Haruma
Giáo viên
25 tháng 11 2018
Lời giải:
\(N=\frac{(a-1)^4-11(a-1)^2+30}{3(a-1)^4-18(a^2-2a+1)+15}=\frac{(a-1)^4-11(a-1)^2+30}{3(a-1)^4-18(a-1)^2+15}\)
Đặt \((a-1)^2=t\Rightarrow N=\frac{t^2-11t+30}{3t^2-18t+15}\)
\(=\frac{t^2-11t+30}{3(t^2-6t+5)}=\frac{(t-5)(t-6)}{3(t-1)(t-5)}\)
\(=\frac{t-6}{3(t-1)}=\frac{(a-1)^2-6}{3(a-1)^2-3}\)
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
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 : \(a-3< 0\)
\(\Rightarrow K=\dfrac{2\left(a^2-9\right)}{4\left|a-2\right|}=\dfrac{\left(a-3\right)\left(a+3\right)}{\left|2a-4\right|}\)
T
11 tháng 11 2023
\(N=\dfrac{\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)+1}{x^2+7x+11}\)
\(=\dfrac{\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]+1}{x^2+7x+11}\)
\(=\dfrac{\left(x^2+7x+10\right)\left(x^2+7x+12\right)+1}{x^2+7x+11}\)
Đặt \(x^2+7x+11=y\), thay vào \(N\) ta được:
\(N=\dfrac{\left(y-1\right)\left(y+1\right)+1}{y}\)
\(=\dfrac{y^2-1+1}{y}\)
\(=\dfrac{y^2}{y}\)
\(=y\)
\(=x^2+7x+11\)
Vậy \(N=x^2+7x+11\).
\(\text{#}Toru\)
Đặt \(\left(a-1\right)^2=t\)
Ta có: \(\left(a-1\right)^4-11\left(a-1\right)^2+30\)
\(=t^2-11t+30\)
\(=t\left(t-5\right)-6\left(t-5\right)=\left(t-5\right)\left(t-6\right)\)
\(=\left[\left(a-1\right)^2-5\right]\left[\left(a-1\right)^2-6\right]\)
\(=\left(a^2-2a-4\right)\left(a^2-2a-5\right)\)
Đặt \(a^2-2a=k\)
Ta có: \(3\left(a-1\right)^4-18\left(a^2-2a\right)-3\)
\(=3\left(a^2-2a+1\right)^2-18\left(a^2-2a\right)-3\)
\(=3\left(k+1\right)^2-18k-3\)
\(=3k^2+6k+3-18k-3\)
\(=3k^2-12k=3k\left(k-4\right)\)
\(=3\left(a^2-2a\right)\left(a^2-2a-4\right)\)(Ở đây bạn ghi thêm điều kiện nhé)
Khi đó: \(N=\frac{\left(a^2-2a-4\right)\left(a^2-2a-5\right)}{3\left(a^2-2a\right)\left(a^2-2a-4\right)}=\frac{a^2-2a-5}{3\left(a^2-2a\right)}\)