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VV
3
3 tháng 7 2018
\(B=\frac{a+1}{a^2-a+1}-\frac{1}{a+1}-\frac{a-2}{a^3+1}=\frac{\left(a+1\right)^2}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a^2-a+1}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a-2}{a^3+1}\\ \)
\(=\frac{a^2+2a+1}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a^2-a+1}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a-2}{\left(a+1\right).\left(a^2-a+1\right)}\)
\(=\frac{a^2+2a+1-\left(a^2-a+1\right)-\left(a-2\right)}{\left(a+1\right).\left(a^2-a+1\right)}=\frac{2a+2}{\left(a+1\right).\left(a^2-a+1\right)}=\frac{2}{a^2-a+1}\)
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
HM
2 tháng 2 2017
\(\left(\frac{a}{a-1}-\frac{1}{a^2-a}\right)=\frac{a^2-1}{a^2-a}=\frac{a+1}{a}\)
ở phàn a+/a thiếu số 1 nhé
\(\frac{1}{a+1}+\frac{2}{a^2-1}=\frac{a-1+2}{a^2-1}=\frac{1}{a-1}\)
=> K =\(\frac{a^2-1}{a}\)
đkxđ: a khác +-1
b, thay vào mà tình
2 tháng 2 2017
a/ \(K=\left(\frac{a}{a-1}-\frac{1}{a^2-a}\right):\left(\frac{1}{a+1}+\frac{2}{a^2-1}\right)\)
\(=\left(\frac{a}{a-1}-\frac{1}{a\left(a-1\right)}\right):\left(\frac{1}{a+1}+\frac{2}{\left(a-1\right)\left(a+1\right)}\right)\)
\(=\frac{a^2-1}{a\left(a-1\right)}:\frac{a-1+2}{\left(a-1\right)\left(a+1\right)}\)
\(=\frac{\left(a-1\right)\left(a+1\right)}{a\left(a-1\right)}.\frac{\left(a-1\right)\left(a+1\right)}{a-1}\)
\(=\frac{a+1}{a}.a+1\)
\(=\frac{\left(a+1\right)^2}{a}\)
b, Thay a=1/2
\(\Rightarrow\frac{\left(\frac{1}{2}+1\right)^2}{\frac{1}{2}}=\frac{\frac{9}{4}}{\frac{1}{2}}=\frac{9}{2}\)
10 tháng 8 2021
Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
10 tháng 8 2021
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)
11 tháng 8 2016
Điều kiện : \(a\ne1\)
\(A=\left(1+\frac{a}{a^2+1}\right):\left(\frac{1}{a-1}+\frac{2a}{a^2+1-a^3-a}\right)-1\)
\(=\frac{a^2+a+1}{a^2+1}:\left(\frac{-a^2-1}{\left(1+a^2\right)\left(1-a\right)}+\frac{2a}{\left(1+a^2\right)\left(1-a\right)}\right)-1\)
\(=\frac{a^2+a+1}{a^2+1}.\frac{\left(a-1\right)\left(1+a^2\right)}{\left(a-1\right)^2}-1=\frac{a^2+a+1}{a-1}-1=\frac{a^2+2}{a-1}\)
b) A < 2 \(\Rightarrow\frac{a^2+2}{a-1}< 2\Leftrightarrow\frac{\left(a^2-2a+1\right)+2\left(a-1\right)+3}{a-1}< 2\)
\(\Leftrightarrow a-1+2+\frac{3}{a-1}< 2\Leftrightarrow a-1+\frac{3}{a-1}< 0\)
Đặt t = a-1 , xét :
Nếu t > 0 thì \(t+\frac{3}{t}< 0\Leftrightarrow t^2+3< 0\) không thỏa mãn vì \(t^2+3>3>0\)
Nếu t < 0 thì \(t+\frac{3}{t}< 0\Leftrightarrow t^2+3>0\) thỏa mãn
Vậy a - 1 < 0 => a < 1 thỏa mãn đề bài
28 tháng 12 2016
\(\left(\frac{a}{a-1}-\frac{1}{a^2-a}\right):\left(\frac{1}{a+1}+\frac{2}{a^2-1}\right)=\left(\frac{a}{a-1}-\frac{1}{a\left(a-1\right)}\right):\left(\frac{1}{a+1}+\frac{2}{\left(a-1\right)\left(a+1\right)}\right)\)
\(=\frac{a^2-1}{a\left(a-1\right)}:\frac{a-1+2}{\left(a-1\right)\left(a+1\right)}=\frac{\left(a^2-1\right)\left(a-1\right)\left(a+1\right)}{a\left(a-1\right)\left(a+1\right)}=\frac{a^2-1}{a}=a-\frac{1}{a}\)
\(B=\frac{a+1}{a^2-a+1}-\frac{1}{a+1}+\)\(\frac{a-2}{a^3+1}\)
\(B=\frac{\left(a+1\right)^2}{a^3+1}-\frac{a^2-a+1}{a^3+1}+\)\(\frac{a-2}{a^3+1}\)
\(B=\frac{a^2+2a+1-a^2+a-1-a+2}{a^3+1}\)
\(B=\frac{2a+2}{a^3+1}\)
\(B=\frac{2\left(a+1\right)}{\left(a+1\right)\left(a^2+a+1\right)}\)
\(B=\frac{2}{a+1}\)
\(B=\frac{a+1}{a^2-a+1}-\frac{1}{a+1}-\frac{a-2}{a^3+1}\) ĐKXĐ : \(x\ne-1\)
\(=\frac{\left(a+1\right)^2}{\left(a+1\right)\left(a^2-a+1\right)}-\frac{a^2-a+1}{\left(a+1\right)\left(a^2-a+1\right)}\)\(-\frac{a-2}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\frac{a^2+2a+1-a^2+a-1-a+2}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\frac{\left(a^2-a^2\right)+\left(2a+a-a\right)+\left(1-1+2\right)}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\frac{2a+2}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\frac{2\left(a+1\right)}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\frac{2}{a^2-a+1}\)