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\(A=\dfrac{\left(a+b\right)\left(-x-y\right)-\left(a-y\right)\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{a\left(-x-y\right)+b\left(-x-y\right)-a\left(b-x\right)+y\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-ax-ay-bx-by-ab+ax+by-xy}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-ay-bx-ab-xy}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-xy+ay+ab+by}{abxy\left(xy+ay+ab+by\right)}=\dfrac{-1}{abxy}\)
Với \(a=\dfrac{1}{3};b=-2;x=\dfrac{3}{2};y=1\)
\(\Rightarrow A=\dfrac{-1}{\dfrac{1}{3}.\left(-2\right).\dfrac{3}{2}.1}=-1\)
Bài 1:
a, \(2y.\left(y-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2y=0\\y-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=0\\y=\dfrac{1}{7}\end{matrix}\right.\)
Vậy \(y\in\left\{0;\dfrac{1}{7}\right\}\)
b, \(\dfrac{-2}{5}+\dfrac{2}{3}y+\dfrac{1}{6}y=\dfrac{-4}{15}\)
\(\Rightarrow\dfrac{5}{6}y=\dfrac{-4}{15}+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{5}{6}y=\dfrac{2}{15}\)
\(\Rightarrow y=\dfrac{4}{25}\)
Vậy \(y=\dfrac{4}{25}\)
Chúc bạn học tốt!!!
Bài 1:
a, \(2y\left(y-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2y=0\\y-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{1}{7}\end{matrix}\right.\)
Vậy...
b, \(\dfrac{-2}{5}+\dfrac{2}{3}y+\dfrac{1}{6}y=\dfrac{-4}{15}\)
\(\Rightarrow\dfrac{5}{6}y=\dfrac{2}{15}\)
\(\Rightarrow y=\dfrac{4}{25}\)
Vậy...
Bài 2:
a, \(x\left(x-\dfrac{4}{7}\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x-\dfrac{4}{7}>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x< 0\\x-\dfrac{4}{7}< 0\end{matrix}\right.\)
\(\Rightarrow x>\dfrac{4}{7}\left(x\ne0\right)\) hoặc \(x< \dfrac{4}{7}\left(x\ne0\right)\)
Vậy...
Các phần còn lại tương tự nhé
a) Theo bđt cauchy ta có:
\(a^3+b^3+b^3\ge3\sqrt[3]{a^3.b^6}=3ab^2\)
\(a^3+a^3+b^3\ge3a^2b\)
công vế theo vế ta có \(3\left(a^3+b^3\right)\ge3ab^2+3a^2b\)
\(\Leftrightarrow a^3+b^3+3\left(a^3+b^3\right)\ge a^3+3a^2b+3ab^2+b^3\)
\(\Leftrightarrow4\left(a^3+b^3\right)\ge\left(a+b\right)^3\)
suy ra đpcm
ta luôn có \(\left(a-b\right)^2\ge0\)
\(\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow a^2+b^2+a^2+b^2\ge a^2+2ab+b^2\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow\dfrac{2\left(a^2+b^2\right)}{4}\ge\dfrac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow\dfrac{\left(a^2+b^2\right)}{2}\ge\dfrac{\left(a+b\right)^2}{2^2}=\left(\dfrac{a+b}{2}\right)^2\)
suy ra đpcm
Ta có: \(\dfrac{a+b}{2}\ge\sqrt{ab}\)
\(\Leftrightarrow a+b\ge2\sqrt{ab}\)
\(\Leftrightarrow\left(\sqrt{a}\right)^2-2\sqrt{ab}+\left(\sqrt{b}\right)^2\ge0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) luôn đúng
Dấu \("="\) xảy ra khi a = b.
Cauchy-shwarz:
\(\dfrac{x^2}{a}+\dfrac{y^2}{b}\ge\dfrac{\left(x+y\right)^2}{a+b}\)
\(\Leftrightarrow bx^2\left(a+b\right)+ay^2\left(a+b\right)\ge\left(x+y\right)^2ab\)
\(\Leftrightarrow\left(abx^2-abx^2\right)+\left(aby^2-aby^2\right)+\left(bx\right)^2-2bxay+\left(ay\right)^2\ge0\)
\(\Leftrightarrow\left(bx-ay\right)^2\ge0\) luôn đúng
Dấu \("="\) xảy ra khi \(bx=ay\Leftrightarrow\dfrac{x}{a}=\dfrac{y}{b}\)
Hằng đẳng thức thứ 2 à