Thu gọn biểu thức:(x-1).(x²+x+1)-(x+1).(2-x)-x³
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7 tháng 3 2022
Bài 4:
b: \(=x^2z\left(-1+3-7\right)=-5x^2z=-5\cdot\left(-1\right)^2\cdot\left(-2\right)=10\)
c: \(=xy^2\left(5+0.5-3\right)=2.5xy^2=2.5\cdot2\cdot1^2=5\)
17 tháng 3 2021
TXĐ: \(\left\{{}\begin{matrix}x\in R\\x\notin\left\{0;-1\right\}\end{matrix}\right.\)
LN
1
H
30 tháng 12 2023
\(a,M=\left(\dfrac{\sqrt{x}+2}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{x+3}{x-1}\right)\\ =\left(\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}+1-\sqrt{x}\left(\sqrt{x}-1\right)+x+3}{x-1}\right)\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{\sqrt{x}+1-x+\sqrt{x}+x+3}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{2\sqrt{x}+4}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{2\left(\sqrt{x}+2\right)}\\ =\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
`b,` Để `M>1` Thì :
\(\dfrac{\sqrt{x}+1}{2\sqrt{x}}>1\\ \Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}}-1>0\\ \Leftrightarrow\dfrac{\sqrt{x}+1-2\sqrt{x}}{2\sqrt{x}}>0\\ \Leftrightarrow\dfrac{-\sqrt{x}+1}{2\sqrt{x}}>0\)
\(\Leftrightarrow-\sqrt{x}+1>0\) `(` Vì \(2\sqrt{x}>0\) do \(x>0\) `)`
\(\Leftrightarrow-\sqrt{x}>-1\\ \Rightarrow x< 1\)
12 tháng 3 2022
a, \(M=2x^3+xy^2-3xy+1\)
b, Thay x = -1 ; y = 2 ta được
M = -2 - 2 + 6 + 1 = 3
AT
9 tháng 6 2021
ĐKXĐ: \(x\ge0,x\ne1\)
\(\left(\dfrac{6\sqrt{x}+6}{x+2\sqrt{x}-3}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right):\dfrac{1}{\sqrt{x}+3}\)
\(=\left(\dfrac{6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right).\left(\sqrt{x}+3\right)\)
\(=\dfrac{6\sqrt{x}+6-\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}.\left(\sqrt{x}+3\right)\)
\(=\dfrac{-x+\sqrt{x}}{\sqrt{x}-1}=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=-\sqrt{x}\)
AT
28 tháng 7 2021
\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(x\ge0,x\ne1\right)\)
\(=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right).\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\)
b) Ta có: \(x\ge0\Rightarrow x+\sqrt{x}+1\ge1\Rightarrow\dfrac{2}{x+\sqrt{x}+1}\le2\)
\(\Rightarrow max=2\) khi \(x=0\)
28 tháng 7 2021
Ta có: \(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\)
NT
2
17 tháng 8 2021
\(\left(x+2\right)^2-\left(x+4\right)^2+x^2-3x+1\)
\(=x^2+4x+4-x^2-8x-16+x^2-3x+1\)
\(=x^2-7x-11\)
17 tháng 8 2021
\(\left(x+2\right)^2-\left(x+4\right)^2+x^2-3x+1\)
\(=x^2+4x+4-x^2-8x-16+x^2-3x+1=x^2-7x-11\)
NN
0
\(=x^3-1+x^2-x-2-x^3=x^2-x-3\)