Bài 1.Cho biểu thức
A = (\(\dfrac{2-x}{x+3}-\dfrac{3-x}{x+2}+\dfrac{2-x}{x^2+5x+6}\)) : (1-\(\dfrac{x}{x-1}\))
(a) Rút gọn A.
(b) Tìm x để A > 2.
Bài 2.Cho x+y=a,\(x^2+y^2=b\).Tính \(x^3+y^3\)theo a và b
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f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)
g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)
h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)
n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)
p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)
k: \(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-6}{x^2-4}\)
m: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
b)
\(P=A-B=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x^2-9}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-9-x^2+9}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x\left(2-x\right)}{\left(x-3\right)\left(x-2\right)}\\ =-\dfrac{x\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =-\dfrac{x}{x-3}\)
c)
Để \(P\le1\) thì:
\(-\dfrac{x}{x-3}\le1\)
\(\Leftrightarrow\dfrac{x}{x-3}\ge1\\ \Leftrightarrow x-3-x\ge1\\ \Leftrightarrow-3\ge1\left(vô.lý\right)\)
Vậy không tồn tại giá trị x để \(P\le1\)
`HaNa♬D`
Làm lại nha cái này đúng, kia sai nha=)
b)
Với \(\left\{{}\begin{matrix}x\ne3\\x\ne2\end{matrix}\right.\)
\(P=A-B=(\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)})+\dfrac{2x-1}{x-3}\\ =\left(\dfrac{2x-9-x^2-9}{\left(x-3\right)\left(x-2\right)}\right)+\dfrac{\left(2x-1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2}{\left(x-3\right)\left(x-2\right)}+\dfrac{2x^2-4x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2+2x^2-4x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x^2-3x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x^2-2x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x\left(x-2\right)-\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}=\dfrac{x-1}{x-3}\)
c)
Để P\(\ge1\) thì:
\(\dfrac{x-1}{x-3}\ge1\\ \Leftrightarrow x-3-x+1-1\ge0\\ \Leftrightarrow-3\ge0\left(vô.lý\right)\)
Vậy không tồn tại giá trị x để \(P\ge1\)
`HaNa☘D`
a: TXĐ: D=[0;+\(\infty\))\{1}
\(B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}-\dfrac{\sqrt{x}}{x-1}\)
\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\cdot2}\)
\(=\dfrac{-1}{\sqrt{x}+1}\)
\(a,ĐK:x\ge0\\ x\ne1\\ B=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ B=\dfrac{2\left(1-\sqrt{x}\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-1}{\sqrt{x}+1}\\ b,x=3\Leftrightarrow B=\dfrac{-1}{\sqrt{3}+1}=\dfrac{1-\sqrt{3}}{2}\\ c,\left|B\right|=\dfrac{1}{2}\Leftrightarrow\left|\dfrac{-1}{\sqrt{x}+1}\right|=\dfrac{1}{2}\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{2}\left(\sqrt{x}+1\ge1>0\right)\\ \Leftrightarrow\sqrt{x}+1=2\Leftrightarrow x=1\left(tm\right)\)
Bài 1:
b) ĐKXĐ: \(x\ne3\)
Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=100\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{13;-7\right\}\)
\(a,\)Với \(x\ne-3,x\ne2\) ta có :
\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}-\dfrac{1}{x-2}\)
\(=\dfrac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x-4}{x-2}\)
\(b,\) \(A=-3\Leftrightarrow\dfrac{x-4}{x-2}=-3\)
\(\Leftrightarrow x-4=-3\left(x-2\right)\)
\(\Leftrightarrow x-4+3x-6=0\)
\(\Leftrightarrow4x=10\Rightarrow x=\dfrac{10}{4}=\dfrac{5}{2}\)