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Bài 1:
Ta có:
\(A=(x^2-x)(x^2+3x+2)=x(x-1)(x+1)(x+2)\)
\(=[x(x+1)][(x-1)(x+2)]=(x^2+x)(x^2+x-2)\)
\(=(x^2+x)^2-2(x^2+x)=(x^2+x)^2-2(x^2+x)+1-1\)
\(=(x^2+x-1)^2-1\geq -1\)
Vậy GTNN của $A$ là $-1$. Dấu "=" xảy ra khi \((x^2+x-1)^2=0\Leftrightarrow x^2+x-1=0\Leftrightarrow x=\frac{-1\pm \sqrt{5}}{2}\)
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\(B=x^4+(x-2)^4+6x^2(x-2)^2=x^4+(x-2)^4-2x^2(x-2)^2+8x^2(x-2)^2\)
\(=[x^2-(x-2)^2]^2+8x^2(x-2)^2\)
\(=16(x-1)^2+8[x(x-2)]^2=16(x^2-2x+1)+8(x^2-2x)^2\)
\(=8[(x^2-2x)^2+2(x^2-2x+1)]=8[(x^2-2x)^2+2(x^2-2x)+1+1]\)
\(=8[(x^2-2x+1)^2+1]=8(x^2-2x+1)^2+8\geq 8\)
Vậy GTNN của biểu thức là $8$ khi \((x^2-2x+1)^2=0\Leftrightarrow (x-1)^4=0\Leftrightarrow x=1\)
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\(C=4x^2+4x-6|2x+1|+6=(4x^2+4x+1)-6|2x+1|+5\)
\(=|2x+1|^2-6|2x+1|+5\)
\(=|2x+1|^2-6|2x+1|+9-4=(|2x+1|-3)^2-4\geq -4\)
Vậy GTNN của biểu thức là $-4$ khi \(|2x+1|=3\Leftrightarrow x=1\) hoặc $x=-2$
Bài 2:
ĐKXĐ: \(x\geq 0\)
Áp dụng BĐT Cô-si cho các số không âm ta có:
\(x+1\geq 2\sqrt{x}\Rightarrow x+1+\sqrt{x}\geq 3\sqrt{x}\)
\(\Rightarrow E=\frac{\sqrt{x}}{x+1+\sqrt{x}}\leq \frac{\sqrt{x}}{3\sqrt{x}}=\frac{1}{3}\)
Vậy GTLN của $E$ là $\frac{1}{3}$ khi $x=1$
a) \(x+3+\sqrt{x^2-6x+9}\left(x\le3\right)\)
\(=x+3+\sqrt{\left(x-3\right)^2}\)
\(=x+3+\left|x-3\right|\)
\(=x+3-\left(x-3\right)\)
\(=x+3-x+3\)
\(=6\)
b) \(\sqrt{x^2+4x+4}-\sqrt{x^2}\left(-2\le x\le0\right)\)
\(=\sqrt{\left(x+2\right)^2}-\sqrt{x^2}\)
\(=\left|x+2\right|-\left|x\right|\)
\(=x+2-\left(-x\right)\)
\(=x+2+x\)
\(=2x+2=2\left(x+1\right)\)
c) \(\frac{\sqrt{x^2-2x+1}}{x-1}\left(x>1\right)\)
\(=\frac{\sqrt{\left(x-1\right)^2}}{x-1}\)
\(=\frac{\left|x-1\right|}{x-1}\)
\(=\frac{x-1}{x-1}=1\)
d) \(\left|x-2\right|+\frac{\sqrt{x^2-4x+4}}{x-2}\)
\(=\left|x-2\right|+\frac{\sqrt{\left(x-2\right)^2}}{x-2}\)
\(=\left|x-2\right|+\frac{\left|x-2\right|}{x-2}\)
\(=\left|x-2\right|+\frac{-\left(x-2\right)}{x-2}\)
\(=\left|x-2\right|-1\)
\(=-\left(x-2\right)-1\)
\(=-x+2-1\)
\(=-x+1=-\left(x-1\right)\)
2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
a) \(A=\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-3\right)^2}\)
\(=\left|x-1\right|+\left|x-3\right|\ge\left|\left(x-1\right)+\left(3-x\right)\right|=2\)
Vậy\(A_{min}=2\Leftrightarrow\left(x-1\right)\left(3-x\right)\ge0\)
\(TH1:\hept{\begin{cases}x-1\ge0\\3-x\ge0\end{cases}}\Leftrightarrow1\le x\le3\)
\(TH1:\hept{\begin{cases}x-1\le0\\3-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge3\end{cases}}\left(L\right)\)
Vậy \(A_{min}=2\Leftrightarrow1\le x\le3\)
\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\sqrt{3+x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}{\sqrt{3-x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}\)
\(=\frac{\sqrt{3+x}}{\sqrt{3-x}}\)
\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right)\sqrt{x^2-6x+8}}\)
\(=\frac{\left(x-3\right)\left(x-2\right)+3\sqrt{\left(x-4\right)\left(x-2\right)}}{3\left(x-4\right)+\left(x-3\right)\sqrt{\left(x-4\right)\left(x-2\right)}}\)
\(=\frac{\sqrt{x-2}\left(\left(x-3\right)\sqrt{x-2}+3\sqrt{x-4}\right)}{\sqrt{x-4}\left(3\sqrt{x-4}+\left(x-3\right)\sqrt{x-2}\right)}\)
\(=\frac{\sqrt{x-2}}{\sqrt{x-4}}\)