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DH
9 tháng 12 2019
a, \(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=x^2+2xy+y^2-\left(x^2-2xy+y^2\right)\)
\(=x^2+2xy+y^2-x^2+2xy-y^2\)
\(=4xy\)
b, \(\left(a+b\right)^3+\left(a-b\right)^3-2a^3\)
\(=a^3+3a^2b+3ab^2+b^3+\left(a^3-3a^2b+3ab^2-b^3\right)-2a^3\)
\(=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-2a^3\)
\(=6ab^2\)
c, \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(=\left(9.2\right)^8-\left[\left(18^4\right)^2-1^2\right]\)
\(=18^8-\left(18^8-1\right)\)
\(=18^8-18^8+1\)
\(=1\)
PN
4 tháng 8 2021
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12 tháng 4 2020
ĐKXĐ: \(\left\{{}\begin{matrix}a-1\ne0\\a^2-1\ne0\\a-a^3\ne0\\a+a^3\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\ne1\\a\ne\left\{-1;1\right\}\\a\left(1-a^2\right)\ne0\\a\left(1+a^2\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\ne1\\a\ne\left\{1;-1\right\}\\a\ne\left\{-1;0;1\right\}\\a\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\ne0\\a\ne-1\\a\ne1\end{matrix}\right.\)
\(M=\frac{a^2}{a-1}+\left(\frac{a}{a^2-1}+\frac{1}{a-a^3}\right):\frac{1-a}{a+a^3}\)
\(=\frac{a^2}{a-1}+\left(\frac{a}{\left(a-1\right)\left(a+1\right)}+\frac{1}{a\left(1-a^2\right)}\right):\frac{1-a}{a\left(1+a^2\right)}\)
\(=\frac{a^2}{a-1}+\left(\frac{a^2}{a\left(a-1\right)\left(a+1\right)}-\frac{1}{a\left(a+1\right)\left(a-1\right)}\right):\frac{1-a}{a\left(1+a^2\right)}\)
\(=\frac{a^2}{a-1}+\frac{\left(a-1\right)\left(a+1\right)}{a\left(a-1\right)\left(a+1\right)}.\frac{a\left(1+a^2\right)}{1-a}\)
\(=\frac{a^2}{a-1}-\frac{1+a^2}{a-1}=\frac{a^2-1-a^2}{a-1}=-\frac{1}{a-1}\)
b/ Thay $a=\frac{1}{2}$ vào M ta được \(M=-\frac{1}{-\frac{1}{2}-1}=-\frac{1}{-\frac{3}{2}}=\frac{1}{\frac{3}{2}}=\frac{2}{3}\)
MN
15 tháng 8 2020
Bài 1 :
a) \(ĐKXĐ:x\ne1\)
\(A=\left(\frac{3}{x^2-1}+\frac{1}{x+1}\right):\frac{1}{x+1}\)
\(\Leftrightarrow A=\frac{3+x-1}{\left(x-1\right)\left(x+1\right)}\cdot\left(x+1\right)\)
\(\Leftrightarrow A=\frac{x+2}{x-1}\)
b) Thay x = \(\frac{2}{5}\)vào A ta được :
\(A=\frac{\frac{2}{5}+2}{\frac{2}{5}-1}=\frac{\frac{12}{5}}{-\frac{3}{5}}=-4\)
c) Để \(A=\frac{5}{4}\)
\(\Leftrightarrow\frac{x+2}{x-1}=\frac{5}{4}\)
\(\Leftrightarrow4x+8=5x-5\)
\(\Leftrightarrow x=13\)
d) Để \(A>\frac{1}{2}\)
\(\Leftrightarrow\frac{x+2}{x-1}>\frac{1}{2}\)
\(\Leftrightarrow\frac{x+2}{x-1}-\frac{1}{2}>0\)
\(\Leftrightarrow2x+4-x+1>0\)
\(\Leftrightarrow x+5>0\)
\(\Leftrightarrow x>-5\)
Bài 2 :
a) \(ĐKXĐ:\hept{\begin{cases}x\ne-1\\x\ne0\end{cases}}\)
\(A=\frac{x^2}{x^2+x}-\frac{1-x}{x+1}\)
\(A=\frac{x}{x+1}+\frac{x-1}{x+1}\)
\(\Leftrightarrow A=\frac{2x-1}{x+1}\)
b) Để \(A=1\)
\(\Leftrightarrow\frac{2x-1}{x+1}=1\)
\(\Leftrightarrow2x-1=x+1\)
\(\Leftrightarrow x=2\)
b) Để \(A< 2\)
\(\Leftrightarrow\frac{2x-1}{x+1}< 2\)
\(\Leftrightarrow\frac{2x-1}{x+1}-2< 0\)
\(\Leftrightarrow2x-1-2x-1< 0\)
\(\Leftrightarrow-2< 0\)(luôn đúng)
Vậy A < 2 <=> mọi x
SA
1
4 tháng 8 2018
mk nghỉ bài này đề sai
a) điều kiện : \(x\ne0;x\ne-1;x\ne2\)
ta có : \(A=1+\left(\dfrac{x+1}{x^3+1}-\dfrac{1}{x-x^2-1}+\dfrac{2}{x+1}\right):\dfrac{x^3-2x^2}{x^3-x^2+x}\)
\(\Leftrightarrow A=1+\left(\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{1}{x^2-x+1}+\dfrac{2}{x+1}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\) \(\Leftrightarrow A=1+\left(\dfrac{x+1+x+1+2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\) \(\Leftrightarrow A=1+\left(\dfrac{2x^2+4}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x^2-x+1}{x\left(x-2\right)}\) \(\Leftrightarrow A=1+\dfrac{2x^2+4}{x\left(x+1\right)\left(x-2\right)}=\dfrac{2x^2+4+x\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}\)\(\Leftrightarrow A=\dfrac{x^3+x^2-2x+4}{x\left(x+1\right)\left(x-2\right)}\)
b) ta có : \(\left|x-\dfrac{3}{4}\right|=\dfrac{5}{4}\) \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{5}{4}\\x-\dfrac{3}{4}=\dfrac{-5}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\left(L\right)\\x=\dfrac{-1}{2}\end{matrix}\right.\)
thế vào \(A\) ta có : \(A=\dfrac{41}{5}\)
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