Tính gtrị bthức: 2/√3-1 - 2/√3+1
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VH
2
HB
0
TT
10 tháng 6 2018
a, \(Đkxđ:x\ne-3;x\ne2\)
b,\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)\(=\dfrac{x-4}{x-2}\)
c,\(A=-\dfrac{3}{4}\) khi \(\dfrac{x-4}{x-2}=-\dfrac{3}{4}\)
\(\Leftrightarrow\left(x-4\right).4=-3\left(x-2\right)\)
\(\Leftrightarrow4x-16=-3x+6\)
\(\Leftrightarrow7x=22\)
\(\Leftrightarrow x=\dfrac{22}{7}\)
Vậy khi \(x=\dfrac{22}{7}\) thì \(A=-\dfrac{3}{4}\)
10 tháng 6 2018
a) ĐKXĐ : \(\left\{{}\begin{matrix}x+3\ne0\\2-x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)
b) \(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}-\dfrac{1}{x-2}\)
\(A=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(A=\dfrac{-x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\)
c) Để \(A=\dfrac{-3}{4}\) thì :
\(A=\dfrac{x-4}{x-2}=\dfrac{-3}{4}\)
\(\Rightarrow\dfrac{x-4}{x-2}+\dfrac{3}{4}=0\)
\(\Rightarrow\dfrac{4\left(x-4\right)}{4\left(x-2\right)}+\dfrac{3\left(x-2\right)}{4\left(x-2\right)}=0\)
\(\Rightarrow4x-16+3x-6=0\)
\(\Rightarrow7x+22=0\)
\(\Rightarrow x=\dfrac{-22}{7}\)
d) Ta có : \(A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)
Vì \(1\in Z\) để \(A\in Z\) thì \(\dfrac{2}{x-2}\in Z\)
\(\Rightarrow x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Có : \(\left\{{}\begin{matrix}x-2=1=>x=3\\x-2=-1=>x=1\\x-2=2=>x=4\\x-2=-2=>0\end{matrix}\right.\)
Vậy để A nhận gt nguyên thì x \(\in\left\{3;1;4;0\right\}\)
e) \(x^2-9=0\)
\(\Rightarrow\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=3\end{matrix}\right.\)
Thay vào A ta có :
\(A=\dfrac{x-4}{x-2}=\dfrac{3-4}{3-2}=-1\)
HJ
0
T
1
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)