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16 tháng 5 2023
a: Khi x=1 thì\(P=\dfrac{1-2}{1+2}=\dfrac{-1}{2}\)
b: \(=\dfrac{3x+6+5x-6+2x^2-4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2+4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x}{x-2}\)
c: \(P=A\cdot B=\dfrac{2x}{x-2}\cdot\dfrac{x-2}{x+1}=\dfrac{2x}{x+1}\)
\(P-2=\dfrac{2x-2x-2}{x+1}=\dfrac{-2}{x+1}\)
P<=2
=>x+1>0
=>x>-1
26 tháng 7 2017
1)
\(M=\dfrac{1}{3}x^2+2x+10\)
\(=\dfrac{1}{3}.\left(x^2+6x+30\right)\)
\(=\dfrac{1}{3}\left(x^2+2.x.3+9\right)+7\)
\(=\dfrac{1}{3}.\left(x+3\right)^2+7\) \(\ge\) 7 với \(\forall\) x
=> M luôn dương
=> đpcm
2)
a) \(2x-x^2-15\)
\(=-\left(x^2-2x+15\right)\)
\(=-\left(x^2-2x+1\right)-14\)
\(=-\left(x-1\right)^2-14\) \(\le-14\) với \(\forall\) x
=> \(2x-x^2-15\) luôn âm
=> đpcm
b) \(-5-\left(x-1\right)\left(x+2\right)\)
\(=-5-x^2-2x+x+2\)
\(=-x^2-x-3\)
\(=-\left(x^2+x+3\right)\)
\(=-\left(x^2+2.\dfrac{1}{2}.x+\dfrac{1}{4}\right)-\dfrac{11}{4}\)
\(=-\left(x+\dfrac{1}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\) với \(\forall\) x
=> \(-5-\left(x-1\right)\left(x+2\right)\) luôn âm
=> đpcm
TN
26 tháng 7 2017
\(M=\dfrac{1}{3}x^2+2x+10=\dfrac{1}{3}\left(x^2+6x+9\right)+7\)
\(=\dfrac{1}{3}\left(x+3\right)^2+7\)
Ta có:
\(\dfrac{1}{3}\left(x+3\right)^2\ge\forall x\Rightarrow\dfrac{1}{3}\left(x+3\right)^2+7>0\)
=>đpcm
\(2,a,2x-x^2-15\)
\(=-\left(x^2-2x+1\right)-14\)
\(=-\left(x-1\right)^2-14\)
Ta có:
\(-\left(x-1\right)^2\le0\forall x\Rightarrow-\left(x-1\right)^2-14< 0\)
=> đpcm
\(b,-5-\left(x-1\right)\left(x+2\right)\)
\(=-5-\left(x^2+x-2\right)\)
\(=-5-x^2-x+2\)
\(=-\left(x^2+x+\dfrac{1}{4}\right)-\dfrac{11}{4}\)
\(=-\left(x+\dfrac{1}{2}\right)^2-\dfrac{11}{4}\)
Ta có:
\(-\left(x+\dfrac{1}{2}\right)^2\le0\forall x\Rightarrow-\left(x+\dfrac{1}{2}\right)-\dfrac{11}{4}< 0\)=> đpcm
31 tháng 12 2022
a: DKXĐ: x<>1; x<>-1
b: \(A=\dfrac{x^2+2x+1+6-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{x^2+2x+7-x^2+x-3x+3}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)
22 tháng 11 2022
a: ĐKXĐ: \(x\notin\left\{1;-\dfrac{1}{2};\dfrac{1}{2}\right\}\)
b: \(A=\dfrac{4x^2-4x+1-1}{\left(2x+1\right)\left(2x-1\right)}\cdot\dfrac{2x-1}{x-1}+\dfrac{2}{2x+1}\)
\(=\dfrac{4x\left(x-1\right)}{\left(2x+1\right)\left(x-1\right)}+\dfrac{2}{2x+1}\)
\(=\dfrac{4x+2}{2x+1}=2\)
ND
1
27 tháng 6 2022
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a^2-2a+4}{a^2+2a+4}>=\dfrac{1}{3}\\\dfrac{a^2-2a+4}{a^2+2a+4}< =3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a^2-6a+12-a^2-2a-4>=0\\a^2-2a+4-3a^2-6a-12< =0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2a^2-8a+8>=0\\-2a^2-8a-8< =0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\left(a-2\right)^2>=0\left(đúng\right)\\-2\left(a+2\right)^2< =0\left(đúng\right)\end{matrix}\right.\)
=>ĐPCM
8 tháng 4 2023
1: \(B=\dfrac{2x+1-x^2+2x^2-3x-1}{x\left(2x+1\right)}=\dfrac{x^2-x}{x\left(2x+1\right)}=\dfrac{x-1}{2x+1}\)
2: \(C=A:B\)
\(=\dfrac{x-1}{x^2}:\dfrac{x-1}{2x+1}=\dfrac{2x+1}{x^2}\)
\(C+1=\dfrac{2x+1+x^2}{x^2}=\dfrac{\left(x+1\right)^2}{x^2}>=0\)
=>C>=-1
9 tháng 1 2018
1 ) \(A=\left(\dfrac{2x^3+2}{x+1}-2x\right)\left(\dfrac{x^3-1}{x-1}+x\right)\)
\(\Leftrightarrow A=\left(\dfrac{2x^3+2-2x^2-2x}{x+1}\right)\left(x^2+2x+1\right)\)
\(\Leftrightarrow A=\left(\dfrac{\left(2x^2-2\right)\left(x-1\right)}{x+1}\right)\left(x+1\right)^2\)
\(\Leftrightarrow A=\left(\dfrac{2\left(x-1\right)\left(x+1\right)\left(x-1\right)}{x+1}\right)\left(x+1\right)^2\)
\(\Leftrightarrow A=2\left(x-1\right)^2\left(x+1\right)^2\ge0\forall x\)
Vì \(x^2+1\ne0\) nên ta có thể viết lại:
\(\left(x^2+1\right)Q=2x^2+2x+2\Leftrightarrow Qx^2+Q=2x^2+2x+2\)\(\Leftrightarrow Qx^2-2x^2-2x+Q-2=0\Leftrightarrow\left(Q-2\right)x^2-2x+Q-2=0\) (*)
pt (*) có nghiệm khi \(\Delta'=\left(-1\right)^2-\left(Q-2\right)\left(Q-2\right)=1-\left(Q-2\right)^2\ge0\)\(\Leftrightarrow\left(Q-2\right)^2\le1\)\(\Leftrightarrow-1\le Q-2\le1\)\(\Leftrightarrow1\le Q\le3\) (đpcm)
khó vl