Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
Biểu thức 1:
\(y=\frac{2x^2-2x+2}{x^2+1}=\frac{2(x^2+1)-2x}{x^2+1}\)
\(\Leftrightarrow y=2-\frac{2x}{x^2+1}\)
Áp dụng BĐT AM-GM ta có: \(x^2+1\geq 2\sqrt{x^2}\Leftrightarrow x^2+1\geq 2|x|\)
\(\Rightarrow (x^2+1)^2\geq 4x^2\)
\(\Rightarrow \left(\frac{2x}{x^2+1}\right)^2\leq 1\Leftrightarrow -1\leq \frac{2x}{x^2+1}\leq 1\)
Từ đây suy ra \(\left\{\begin{matrix} y=2-\frac{2x}{x^2+1}\geq 1\Leftrightarrow x=1\\ y=2-\frac{2x}{x^2+1}\leq 3\Leftrightarrow x=-1\end{matrix}\right.\)
Vậy \(y_{\min}=1;y_{\max}=3\)
Biểu thức 2:
ĐKXĐ: $x,y$ không đồng thời bằng 0
\(Q=\frac{2x^2+4xy+5y^2}{x^2+y^2}=\frac{(x^2+y^2)+(x+2y)^2}{x^2+y^2}\)
\(\Leftrightarrow Q=1+\frac{(x+2y)^2}{x^2+y^2}\)
Ta thấy \((x+2y)^2\geq 0\forall x,y\in\mathbb{R}; x^2+y^2>0\) (nằm trong khoảng xác định)
\(\Rightarrow \frac{(x+2y)^2}{x^2+y^2}\geq 0\Rightarrow Q\geq 1\)
Vậy \(Q_{\min}=1\Leftrightarrow x=-2y\) và \(x,y \neq 0\)
Mặt khác theo BĐT Bunhiacopxky:
\((x+2y)^2\leq (x^2+y^2)(1+2^2)=5(x^2+y^2)\); \(x^2+y^2>0\) trong khoảng xác định
\(\Rightarrow \frac{(x+2y)^2}{x^2+y^2}\leq \frac{5(x^2+y^2)}{x^2+y^2}=5\)
\(\Rightarrow Q\leq 1+5\Leftrightarrow Q\leq 6\Leftrightarrow Q_{\max}=6\)
Dấu bằng xảy ra khi \(\frac{x}{1}=\frac{y}{2}\Leftrightarrow 2x=y\) và \(x,y\neq 0\)
a. A=\(\dfrac{-2}{x^{2^{ }}-2x+5}\)= \(\dfrac{-2}{\left(x-1\right)^{2^{ }}+4}\)
Ta có: (x-1) 2 ≥ 0 với mọi x
⇔ (x- 1)2 +4 ≥4
⇔ \(\dfrac{-2}{\left(x-1\right)^{2^{ }}+4}\)≤ \(\dfrac{-2}{4}\) = \(\dfrac{-1}{2}\)
Dấu''='' xảy ra ⇔ x-1=0
⇔x=1
Vậy maxA= -0,5 ⇔ x=1
b. B=\(\dfrac{3}{x^{2^{ }}-2x+1}\)=\(\dfrac{3}{\left(x-1\right)^2}\)
Ta có: (x-1)2 ≥ 0 với mọi x
⇔ \(\dfrac{3}{\left(x-1\right)^2}\)≤0
Đề bài sai: Khi \(x=4\) thì \(A=\dfrac{1}{2};B=\dfrac{28}{9};\dfrac{A}{B}=\dfrac{9}{56};\dfrac{x-2}{4\sqrt{x}}=\dfrac{1}{4}\Rightarrow\dfrac{A}{B}\ne\dfrac{x-2}{4\sqrt{x}}\)
\(\left(\dfrac{x-4}{2x-4}+\dfrac{2}{x^2-2x}\right):\dfrac{x-2}{x+1}\)
\(=\left(\dfrac{x-4}{2\left(x-2\right)}+\dfrac{2}{x\left(x-2\right)}\right).\dfrac{x+1}{x-2}\)
\(=\dfrac{x\left(x-4\right)+4}{2x\left(x-2\right)}.\dfrac{x+1}{x-2}\)
\(=\dfrac{x^2-4x+4}{2x\left(x-2\right)}.\dfrac{x+1}{x-2}\)
\(=\dfrac{\left(x-2\right)^2\left(x+1\right)}{2x\left(x-2\right)\left(x-2\right)}\)
\(=\dfrac{x+1}{2x}\)
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: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)
\(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\)
\(A=\dfrac{x^2-2x+2}{x^2+2x+2}\)
\(\Leftrightarrow Ax^2+2Ax+2A=x^2-2x+2\)
\(\Leftrightarrow\left(A-1\right)x^2+\left(2A+2\right)x+\left(2A-2\right)=0\) (*)
Để (*) có nghiệm thì
\(\Delta'\ge0\Leftrightarrow\left(A+1\right)^2-2\left(A-1\right)^2\ge0\Leftrightarrow-A^2+6A-1\ge0\)
\(\Leftrightarrow3-2\sqrt{2}\le A\le3+2\sqrt{2}\)
Vậy GTNN của A là \(3-2\sqrt{2}\); GTLN của A là \(3+2\sqrt{2}\)
\(B=\dfrac{x^2+2x+2}{x^2+1}\)
Làm tương tự câu a ta được \(\dfrac{3-\sqrt{5}}{2}\le B\le\dfrac{3+\sqrt{5}}{2}\)
A=\(\dfrac{x^2-2x+2}{x^2+2x+2}\)