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a, ĐKXĐ:\(x\ne-3\)
\(x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}\\ \Leftrightarrow x+1=\dfrac{x+5}{x+3}-\dfrac{2}{x+3}\\ \Leftrightarrow x+1=\dfrac{x+3}{x+3}\\ \Leftrightarrow x+1=1\\ \Leftrightarrow x=0\left(tm\right)\)
b, ĐKXĐ:\(x>2\)
\(\dfrac{x^2-4x-2}{\sqrt{x-2}}=\sqrt{x-2}\\ \Leftrightarrow x^2-4x-2=x-2\\ \Leftrightarrow x^2-5x=0\\ \Leftrightarrow x\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
mình nghĩ đề nó như thế này
\(\sqrt{a^2+b^2}-\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2-\left(b+d^{ }\right)^2}\)
hai zế BĐT ko âm nên bình phương 2 zế ta có
\(a^2+b^2+c^2+d^2+2\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge a^2+2ac+c^2+b^2+2bd+d^2\)
\(\Leftrightarrow\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge ac+bd\left(1\right)\)
Nếu \(ac+bd< 0\)thì BĐT đc c/m
Nêu \(ac+bd\ge0\left(1\right)\Leftrightarrow\left(a^2+b^2\right)\left(c^2+d^2\right)\ge a^2c^2+b^2d^2+2acbd\)
\(\Leftrightarrow a^2c^2+a^2d^2+b^2c^2+b^2d^2\ge a^2c^2+b^2d^2+2acbd\)
\(\Leftrightarrow a^2d^2+b^2c^2-2acbd\ge0\Leftrightarrow\left(ad-bc\right)^2\ge0\)( luôn đúng )
dấu = xảy ra khi \(ad=bc\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)
\(\left|\vec{AD}+\vec{AB}\right|=\left|\vec{AC}\right|=AC=a\sqrt{2}\)
\(\dfrac{a}{a+2\sqrt{\left(a+bc\right)}}=\dfrac{a}{a+2\sqrt{a\left(a+b+c\right)+bc}}=\dfrac{a}{a+2\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(=\dfrac{a}{a+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\)
\(\le\dfrac{a}{5^2}\left(\dfrac{1}{a}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\right)\)
\(=\dfrac{a}{25}\left(\dfrac{1}{a}+\dfrac{8}{\sqrt{\left(a+b\right)\left(a+c\right)}}\right)=\dfrac{1}{25}+\dfrac{8}{25}.\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
Tương tự:
\(\dfrac{b}{b+2\sqrt{b+ac}}\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\)
\(\dfrac{c}{c+2\sqrt{c+ab}}\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)
Cộng vế:
\(P\le\dfrac{3}{25}+\dfrac{4}{25}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{15}{25}=\dfrac{3}{5}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
\(A\cap B\\ \Leftrightarrow\left\{{}\begin{matrix}m+2< 0\\m+1< 1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< -2\\m< 0\end{matrix}\right.\\ hay.m\in\left[-\infty;-1\right]\cap\left[1;+\infty\right]\)
\(A=4\sqrt{2}sinx+1-2sin^2x+2=-2sin^2x+4\sqrt{2}sinx+3\)
Đặt \(sinx=t\Rightarrow t\in\left[-1;1\right]\)
\(A=f\left(t\right)=-2t^2+4\sqrt{2}t+3\)
Xét hàm \(f\left(t\right)\) trên \(\left[-1;1\right]\)
\(-\dfrac{b}{2a}=-\sqrt{2}\notin\left[-1;1\right]\)
\(f\left(-1\right)=1-4\sqrt{2}\) ; \(f\left(1\right)=1+4\sqrt{2}\)
\(\Rightarrow A_{max}=f\left(1\right)=1+4\sqrt{2}\)
\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=4\\c=2\end{matrix}\right.\)
Ủa đề bài sai, \(c>a\) chứ sao \(c\le a\) được?
//Em xem lại câu hỏi hồi nãy nhé, lúc nhấn gửi đáp án mới làm được 1 nửa nên chưa đúng đâu
Lời giải:
Áp dụng bất đẳng thức AM-GM:
\(a^2+2=(a^2+1)+1\geq 2\sqrt{a^2+1}\)
Do đó mà \(\frac{a^2+2}{\sqrt{a^2+1}}\geq \frac{2\sqrt{a^2+1}}{\sqrt{a^2+1}}=2\) (đpcm)
Dấu bằng xảy ra khi \(a^2+1=1\Leftrightarrow a=0\)