Cho biểu thức \(Q=\dfrac{2x^2+2x+x}{x^2+1}\). Chứng minh \(1\le Q\le3\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
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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)
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Đk:\(-1\le x\le3\) (chính là cái bài cho kia)
Nếu \(x=0\) thì \(A=\sqrt{3}\) ta sẽ chứng minh nó là GTNN của \(A\)
Tức là ta cần chứng minh
\(\sqrt{-x^2+2x+3}+\sqrt{3}\le\sqrt{-x^2+4x+12}\)
Sau khi bình phương 2 vế rồi rút gọn ta cần chứng minh
\(\sqrt{-3\left(x^2+2x+3\right)}\le x+3\)
Từ khi \(x+3>0\), ta cần chứng minh
\(3\left(-x^2+2x+3\right)\le\left(x+3\right)^2\Leftrightarrow x^2\ge0\) (Đúng)
Vậy \(A_{Min}=\sqrt{3}\Leftrightarrow x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
a: DKXĐ: x<>1; x<>-1
b: \(A=\dfrac{x^2+2x+1+6-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{x^2+2x+7-x^2+x-3x+3}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
-\(x+3+\sqrt{x^2-6x+9}\)
\(=x+3+\left|x\right|-6x+9\)
\(x< 0\)
\(--->x+3-x-6x+9\)
\(=\left(x-x\right)-6x+3+9\)
\(=-6x+\left(3+9\right)=-6x+12\)
\(x>0\)
\(--->3+x+x-6x+9\)
\(=\left(x+x-6x\right)+\left(3+9\right)\)
\(=\left(2x-6x\right)+12\)
\(=4x+12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(M=\dfrac{1}{3}x^2+2x+10=\dfrac{1}{3}\left(x^2+6x+9\right)+7\)
\(=\dfrac{1}{3}\left(x+3\right)^2+7\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ĐKXĐ: \(x\notin\left\{1;-\dfrac{1}{2};\dfrac{1}{2}\right\}\)
b: \(A=\dfrac{4x^2-4x+1-1}{\left(2x+1\right)\left(2x-1\right)}\cdot\dfrac{2x-1}{x-1}+\dfrac{2}{2x+1}\)
\(=\dfrac{4x\left(x-1\right)}{\left(2x+1\right)\left(x-1\right)}+\dfrac{2}{2x+1}\)
\(=\dfrac{4x+2}{2x+1}=2\)
Bạn có bị nhầm đề không? (sao lại \(2x+x\))