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\(A=\frac{1}{x^2-x}+\frac{1}{x^2+x+1}+\frac{2x}{1-x^3}\)
\(A=\frac{1}{x.\left(x-1\right)}+\frac{1}{x^2+x+1}+\frac{2x}{\left(1-x\right)\left(x^2+x+1\right)}\)
\(A=\frac{x^2+x+1}{x.\left(x-1\right)\left(x^2+x+1\right)}+\frac{x\left(x-1\right)}{x.\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2}{x.\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{x^2+x+1}{x.\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2-x}{x.\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2}{x.\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{x^2+x+1+x^2-x-2x^2}{x.\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{1}{x.\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{1}{x.\left(x^3-1\right)}\)
Với x=10
\(\Rightarrow A=\frac{1}{10.\left(10^3-1\right)}\)
\(A=\frac{1}{10.999}\)
\(A=\frac{1}{9990}\)
Vậy \(A=\frac{1}{9990}\)tại x=10
a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)
a)\(\left(\frac{1-x^3+1-x-x}{1-x}\right):\frac{-\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)^2}=\left(\frac{-x^3-2x+2}{1-x}\right)\cdot\left(1-x\right)=-x^3-2x+2\)
b) \(-\left(-1\frac{2}{3}\right)-2\cdot\left(-1\frac{2}{3}\right)+2=\frac{5}{3}+\frac{10}{3}+2=7\)
Bài 1:
a: \(A=\dfrac{x+1+x}{x+1}:\dfrac{3x^2+x^2-1}{x^2-1}\)
\(=\dfrac{2x+1}{x+1}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{x-1}{2x-1}\)
b: Thay x=1/3 vào A, ta được:
\(A=\left(\dfrac{1}{3}-1\right):\left(\dfrac{2}{3}-1\right)=\dfrac{-2}{3}:\dfrac{-1}{3}=2\)
a) A = \(\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
A = \(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{x+2}\right]:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
A = \(\left[\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right]:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
A = \(-\frac{6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
A = \(-\frac{6\left(x+2\right)}{6\left(x-2\right)\left(x+2\right)}\)
A = \(-\frac{6}{6\left(x-2\right)}\)
A = \(-\frac{1}{x-2}\)
b) |x| = \(\hept{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
+) với x = 1/2, ta có:
A = \(-\frac{1}{\frac{1}{2}-2}=\frac{2}{3}\)
+) với x = -1/2, ta có:
A = \(-\frac{1}{\left(-\frac{1}{2}\right)-2}=\frac{2}{5}\)
\(ĐKXĐ:x\ne1\)
a) \(A=\left(1+\frac{x^2}{x^2+1}\right):\left(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\right)\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\left[\frac{1}{x-1}-\frac{2x}{x\left(x^2+1\right)-\left(x^2+1\right)}\right]\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\left[\frac{1}{x-1}-\frac{2x}{\left(x^2+1\right)\left(x-1\right)}\right]\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\frac{x^2+1-2x}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\frac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\frac{x-1}{x^2+1}\)
\(\Leftrightarrow A=\frac{\left(2x^2+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{2x^2+1}{x-1}\)
b) Thay \(x=-\frac{1}{2}\)vào A, ta được :
\(A=\frac{2\left(-\frac{1}{2}\right)^2+1}{-\frac{1}{2}-1}\)
\(\Leftrightarrow A=\frac{\frac{3}{2}}{-\frac{3}{2}}\)
\(\Leftrightarrow A=-1\)
c) Để A < 1
\(\Leftrightarrow2x^2+1< x-1\)
\(\Leftrightarrow2x^2-x+2< 0\)
\(\Leftrightarrow2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right)+\frac{15}{8}< 0\)
\(\Leftrightarrow2\left(x-\frac{1}{4}\right)^2+\frac{15}{8}< 0\)
\(\Leftrightarrow x\in\varnothing\)
Vậy để \(A< 1\Leftrightarrow x\in\varnothing\)
d) Để A có giá trị nguyên
\(\Leftrightarrow2x^2+1⋮x-1\)
\(\Leftrightarrow2x^2-2x+2x-2+3⋮x-1\)
\(\Leftrightarrow2x\left(x-1\right)+2\left(x-1\right)+3⋮x-1\)
\(\Leftrightarrow2\left(x+1\right)\left(x-1\right)+3⋮x-1\)
\(\Leftrightarrow3⋮x-1\)
\(\Leftrightarrow x-1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow x\in\left\{2;0;4;-2\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{2;0;4;-2\right\}\)
\(A=\left(x+1\right)\left(x^2-x+1\right)+x-\left(x-1\right)\left(x^2+x+1\right)+2021\)
\(=x^3+1+x-x^3+1+2021\)
\(=x+2023\)
=-2022+2023
=1
\(A=\left(x+1\right)\left(x^2-x+1\right)+x-\left(x-1\right)\left(x^2+x+1\right)+2021=x^3+1+x-x^2+1+2021=x+2023=-2022+2023=1\)