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a) Đk: x > 0 và x khác +-1
Ta có: A = \(\left(\frac{x+1}{x}-\frac{1}{1-x}-\frac{x^2-2}{x^2-x}\right):\frac{x^2+x}{x^2-2x+1}\)
A = \(\left[\frac{\left(x-1\right)\left(x+1\right)+x-x^2+2}{x\left(x-1\right)}\right]:\frac{x\left(x+1\right)}{\left(x-1\right)^2}\)
A = \(\frac{x^2-1+x-x^2+2}{x\left(x-1\right)}\cdot\frac{\left(x-1\right)^2}{x\left(x+1\right)}\)
A = \(\frac{x+1}{x}\cdot\frac{x-1}{x\left(x+1\right)}=\frac{x-1}{x^2}\)
b) Ta có: A = \(\frac{x-1}{x^2}=\frac{1}{x}-\frac{1}{x^2}=-\left(\frac{1}{x^2}-\frac{1}{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\frac{1}{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
Dấu "=" xảy ra <=> 1/x - 1/2 = 0 <=> x = 2 (tm)
Vậy MaxA = 1/4 <=> x = 2
a) ĐKXĐ : x ≠ ±2
\(=\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\div\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(=\left[\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right]\div\left(\frac{x^2-4+10-x^2}{x+2}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\div\frac{6}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\times\frac{x+2}{6}=\frac{-1}{x-2}\)
b) Để A < 0 thì -1/x-2 < 0
=> x - 2 > 0 <=> x > 2
Vậy với x > 2 thì A < 0
Câu 3 :
\(a,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\) ĐKXđ : \(x\ne\pm1\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{5\left(x-1\right)}\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{10}{x+1}\)
\(B=\left(\frac{x}{3x-9}+\frac{2x-3}{3x-x^2}\right).\frac{3x^2-9x}{x^2-6x+9}.\)
ĐKXđ : \(x\ne0;x\ne3\)
\(B=\left(\frac{x}{3\left(x-3\right)}+\frac{2x-3}{x\left(3-x\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\left(\frac{x^2}{3x\left(x-3\right)}+\frac{9-6x}{3x\left(x-3\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\frac{x^2-6x+9}{3x\left(x-3\right)}.\frac{3x\left(x-3\right)}{x^2-6x+9}=1\)
a, \(A=\left(\frac{4}{2x+1}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4\left(x^2+1\right)}{\left(2x+1\right)\left(x^2+1\right)}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4x^2+4+4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\frac{\left(2x+1\right)^2}{\left(x^2+1\right)\left(2x+1\right)}\frac{x^2+1}{x^2+2}=\frac{2x+1}{x^2+2}\)
a) \(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\) (ĐKXĐ: \(x\ne\pm1\) )
\(=\left(\frac{x+1+2\left(1-x\right)-5+x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\left(\frac{x+1+2-2x-5+x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\left(\frac{-2}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\frac{2}{x^2-1}.\frac{x^2-1}{1-2x}=\frac{2}{1-2x}\)
b) Để x nhận giá trị nguyên <=> 2 chia hết cho 1 - 2x
<=> 1-2x thuộc Ư(2) = {1;2;-1;-2}
Nếu 1-2x = 1 thì 2x = 0 => x= 0
Nếu 1-2x = 2 thì 2x = -1 => x = -1/2
Nếu 1-2x = -1 thì 2x = 2 => x =1
Nếu 1-2x = -2 thì 2x = 3 => x = 3/2
Vậy ....
a) Với \(x\ne1\)ta có:
\(A=\left(\frac{x^2+2}{x^3-1}+\frac{x}{x^2+x+1}+\frac{1}{1-x}\right):\frac{x-1}{2}\)
\(=\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x}{x^2+x+1}-\frac{1}{x-1}\right].\frac{2}{x-1}\)
\(=\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right].\frac{2}{x-1}\)
\(=\frac{\left(x^2+2\right)+x\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{2}{x-1}\)
\(=\frac{x^2+2+x^2-x-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{2}{x-1}\)
\(=\frac{2\left(x^2-2x+1\right)}{\left(x-1\right)^2.\left(x^2+x+1\right)}=\frac{2\left(x-1\right)^2}{\left(x-1\right)^2.\left(x^2+x+1\right)}=\frac{2}{x^2+x+1}\)
b) \(A=\frac{2}{3}\)\(\Leftrightarrow\frac{2}{x^2+x+1}=\frac{2}{3}\)
\(\Leftrightarrow x^2+x+1=3\)\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow x^2-x+2x-2=0\)\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\left(ktmĐKXĐ\right)\\x=-2\left(tmĐKXĐ\right)\end{cases}}\)
Vậy \(A=\frac{2}{3}\)\(\Leftrightarrow x=-2\)
b) Ta có: \(x^2+x+1=x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
\(\Rightarrow x^2+x+1\ge\frac{3}{4}\forall x\)\(\Rightarrow\frac{1}{x^2+x+1}\le\frac{4}{3}\forall x\)
\(\Rightarrow\frac{2}{x^2+x+1}\le\frac{8}{3}\forall x\)\(\Rightarrow A\le\frac{8}{3}\)
Dấu " = " xảy ra \(\Leftrightarrow x+\frac{1}{2}=0\)\(\Leftrightarrow x=-\frac{1}{2}\)( thỏa mãn ĐKXĐ )
Vậy \(maxA=\frac{8}{3}\Leftrightarrow x=-\frac{1}{2}\)