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Bài 3:
Áp dụng BĐT Bunhiacopxky ta có:
\((2x+3y)^2\leq (2x^2+3y^2)(2+3)\)
\(\Leftrightarrow A^2\leq 5(2x^2+3y^2)\leq 5.5\)
\(\Leftrightarrow A^2\leq 25\Leftrightarrow A^2-25\leq 0\)
\(\Leftrightarrow (A-5)(A+5)\leq 0\Leftrightarrow -5\leq A\leq 5\)
Vậy \(A_{\min}=-5\Leftrightarrow (x,y)=(-1;-1)\)
\(A_{\max}=5\Leftrightarrow x=y=1\)
Bài 4:
Lời giải:
\(B=\sqrt{x-1}+\sqrt{5-x}\)
\(\Rightarrow B^2=(\sqrt{x-1}+\sqrt{5-x})^2=4+2\sqrt{(x-1)(5-x)}\)
Vì \(\sqrt{(x-1)(5-x)}\geq 0\Rightarrow B^2\geq 4\)
Mặt khác \(B\geq 0\)
Kết hợp cả hai điều trên suy ra \(B\geq 2\)
Vậy \(B_{\min}=2\).
Dấu bằng xảy ra khi \((x-1)(5-x)=0\Leftrightarrow x\in\left\{1;5\right\}\)
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\(A=\sqrt{x^2+x+1}+\sqrt{x^2-x+1}\)
\(\Rightarrow A^2=2x^2+2+2\sqrt{(x^2+x+1)(x^2-x+1)}\)
\(\Leftrightarrow A^2=2x^2+2+2\sqrt{(x^2+1)^2-x^2}=2x^2+2+2\sqrt{x^4+1+x^2}\)
Vì \(x^2\geq 0\forall x\in\mathbb{R}\)
\(\Rightarrow A^2\geq 2+2\sqrt{1}\Leftrightarrow A^2\geq 4\)
Mà $A$ là một số không âm nên từ \(A^2\geq 4\Rightarrow A\geq 2\)
Vậy \(A_{\min}=2\Leftrightarrow x=0\)
a: \(P=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\left(\sqrt{x}+1\right)\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
b: \(P=x-\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\)
Dấu '=' xảy ra khi x=1/4
\(a.P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2x-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\left(x\ne1;x>0\right)\)
\(b.P=x-\sqrt{x}+1=x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+1-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow P_{MIN}=\dfrac{3}{4}."="\Leftrightarrow x=\dfrac{1}{4}\)
Để em làm câu c cho 2 chị :3
\(Q=\dfrac{2\sqrt{x}}{P}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=\dfrac{2\sqrt{x}}{x}-2+2\sqrt{x}\)
Để \(Q\in Z\Leftrightarrow\) \(\dfrac{2\sqrt{x}}{x}-2+2\sqrt{x}\in Z\) . Do đó ta cần 2 điều kiện sau :
ĐK1 : \(2\sqrt{x}\) chia hết cho \(x\)
ĐK2 : \(x\) thuộc số chính phương : \(\left(0;1;4;9;.......\right)\)
Xét ĐK1 : Ta có : \(2\sqrt{x}\le x^2\)
Do vậy nên \(2\sqrt{x}\) chia hết cho \(x^2\) khi và chỉ khi \(2\sqrt{x}=x^2\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\) ( Thỏa mãn )
Vậy \(x=0\) hoặc \(x=1\) thì \(Q\in Z\)
\(\left(\dfrac{x-4}{2x-4}+\dfrac{2}{x^2-2x}\right):\dfrac{x-2}{x+1}\)
\(=\left(\dfrac{x-4}{2\left(x-2\right)}+\dfrac{2}{x\left(x-2\right)}\right).\dfrac{x+1}{x-2}\)
\(=\dfrac{x\left(x-4\right)+4}{2x\left(x-2\right)}.\dfrac{x+1}{x-2}\)
\(=\dfrac{x^2-4x+4}{2x\left(x-2\right)}.\dfrac{x+1}{x-2}\)
\(=\dfrac{\left(x-2\right)^2\left(x+1\right)}{2x\left(x-2\right)\left(x-2\right)}\)
\(=\dfrac{x+1}{2x}\)
Mình làm nốt bài 2 nhé :
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\)
⇔ \(\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c\)
⇔ \(\dfrac{a^2+a\left(b+c\right)}{b+c}+\dfrac{b^2+b\left(c+a\right)}{c+a}+\dfrac{c^2+c\left(a+b\right)}{a+b}=a+b+c\)
⇔ \(\dfrac{a^2}{b+c}+a+\dfrac{b^2}{c+a}+b+\dfrac{c^2}{a+b}+c=a+b+c\)
⇔ \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\)
A)
Đặt \(\sqrt{1+2x}=a; \sqrt{1-2x}=b\) (\(a,b>0\) )
\(\Rightarrow \left\{\begin{matrix} a^2+b^2=2\\ a^2-b^2=4x=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} 2a^2=2+\sqrt{3}\rightarrow 4a^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\\ 2b^2=2-\sqrt{3}\rightarrow 4b^2=4-2\sqrt{3}=(\sqrt{3}-1)^2\end{matrix}\right.\)
\(\Rightarrow a=\frac{\sqrt{3}+1}{2}; b=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow ab=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}=\frac{1}{2}; a-b=1\)
Có:
\(A=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+ab^2}{(1+a)(1-b)}\)
\(=\frac{2-ab(a-b)}{1+(a-b)-ab}=\frac{2-\frac{1}{2}.1}{1+1-\frac{1}{2}}=1\)
B)
\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)
\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\)
\(\rightarrow 4(x^2-1)=\frac{a}{b}+\frac{b}{a}-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)
\(\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\) do $a>b$
T có: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)
\(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\)
Đề bài sai: Khi \(x=4\) thì \(A=\dfrac{1}{2};B=\dfrac{28}{9};\dfrac{A}{B}=\dfrac{9}{56};\dfrac{x-2}{4\sqrt{x}}=\dfrac{1}{4}\Rightarrow\dfrac{A}{B}\ne\dfrac{x-2}{4\sqrt{x}}\)
a. \(y=\sqrt{x^2-6x+10}=\sqrt{x^2-6x+9+1}=\sqrt{\left(x-3\right)^2+1}\ge\sqrt{0+1}=1\)
\(\Rightarrow Min_y=1\Leftrightarrow x=3\)
b. \(y=\sqrt{\dfrac{x^2}{9}-\dfrac{2x}{15}+1}=\sqrt{\left(\dfrac{x}{3}\right)^2-2.\dfrac{x}{3}.\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{24}{25}}=\sqrt{\left(\dfrac{x}{3}-\dfrac{1}{5}\right)^2+\dfrac{24}{25}}\ge\sqrt{0+\dfrac{24}{25}}=\sqrt{\dfrac{24}{25}}\)
\(\Rightarrow Min_y=\sqrt{\dfrac{24}{25}}\Leftrightarrow x=\dfrac{3}{5}\)
Giải:
a) \(y=\sqrt{x^2-6x+10}\)
\(\Leftrightarrow y=\sqrt{x^2-6x+9+1}\)
\(\Leftrightarrow y=\sqrt{\left(x^2-6x+9\right)+1}\)
\(\Leftrightarrow y=\sqrt{\left(x-3\right)^2+1}\ge1\)
\(\Leftrightarrow y_{Min}=1\)
\("="\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy ...
b) \(y=\sqrt{\dfrac{x^2}{9}-\dfrac{2x}{15}+1}\)
\(\Leftrightarrow y=\sqrt{\dfrac{x^2}{9}-\dfrac{2x}{15}+\dfrac{1}{25}+\dfrac{24}{25}}\)
\(\Leftrightarrow y=\sqrt{\left(\dfrac{x^2}{9}-\dfrac{2x}{15}+\dfrac{1}{25}\right)+\dfrac{24}{25}}\)
\(\Leftrightarrow y=\sqrt{\left(\dfrac{x}{3}-\dfrac{1}{5}\right)^2+\dfrac{24}{25}}\ge\dfrac{24}{25}\)
\(\Leftrightarrow y_{Min}=\dfrac{24}{25}\)
\("="\Leftrightarrow\dfrac{x}{3}-\dfrac{1}{5}=0\Leftrightarrow x=\dfrac{3}{5}\)
Vậy ...
\(B=\dfrac{x^2+x}{x^2+x+1}=\dfrac{3x^2+3x}{3\left(x^2+x+1\right)}=\dfrac{-\left(x^2+x+1\right)+4x^2+4x+1}{3\left(x^2+x+1\right)}\)
\(=-\dfrac{1}{3}+\dfrac{\left(2x+1\right)^2}{3\left(x+\dfrac{1}{2}\right)^2+\dfrac{9}{4}}\ge-\dfrac{1}{3}\)
\(B_{min}=-\dfrac{1}{3}\) khi \(x=-\dfrac{1}{2}\)