a, \(115^2+15^2-30.115=115^2-2.115.15+15^2=\left(115-15\right)^2\) b, \(\left(3x+1\right)^2+\left(3x-1\right)^2-2.\left(9x^2-1\right)=\left(3x+1\right)^2+\left(3x-1\right)^2-2.\left(3x+1\right).\left(3x-1\right)=\left(3x+1-3x+1\right)^2=2^2=4\)

b

115 2+152−30.115

9 số 0

tick cho tớ nhé

=\(\frac{10+9\sqrt{15}}{15}\)

Ta có: BAE +BAC +EAC = 360 mà BAE=100 , EAC=120 \(\Rightarrow\)BAC=140

Do BAC=ACD=140 và hai góc là hai góc ở vị trí so le trong => AB//CD

Vậy....chúc bạn học tốt

a, \(x-y=2\left(x+y\right)\Rightarrow x-y=2x+2y\Rightarrow x-2x=y+2y\Rightarrow-x=3y\Rightarrow x=-3y\)\(\left(1\right)\)

Thay (1)vào x - y = x : y \(\Rightarrow-3y-y=-3y:y\Rightarrow-4y=-3\Rightarrow y=\dfrac{3}{4}\)\(\Rightarrow x=\dfrac{3}{4}.\left(-3\right)=\dfrac{-9}{4}\)Vậy \(x=\dfrac{-9}{4};y=\dfrac{3}{4}\)

b,\(x+y=x.y\Rightarrow x=x.y-y\Rightarrow x=y.\left(x-1\right)\left(2\right)\)

Thay \(\left(2\right)vào\)\(x+y=x:y\Rightarrow x+y=\dfrac{y.\left(x-1\right)}{y}\Rightarrow x+y=x-1\Rightarrow y=-1\left(3\right)\)

Thay \(\left(3\right)\)vào x + y = x . y\(\Rightarrow x-1=x.\left(-1\right)\Rightarrow x-1=-x\Rightarrow x+x=1\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\)

Vậy y = -1 và x = 1/2

Ta có: \(C=\left|x-\dfrac{2}{5}\right|+\left|x-\dfrac{3}{5}\right|=\left|x-\dfrac{2}{5}\right|+\left|\dfrac{3}{5}-x\right|\ge\left|x-\dfrac{2}{5}+\dfrac{3}{5}-x\right|=\left|\dfrac{1}{5}\right|=\dfrac{1}{5}\)hay C \(\ge\dfrac{1}{5}\Rightarrow C=\dfrac{1}{5}khi\left\{{}\begin{matrix}x-\dfrac{2}{5}\ge0\\\dfrac{3}{5}-x\ge0\end{matrix}\right.\Rightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{2}{5}\\x\le\dfrac{3}{5}\end{matrix}\right.\)Vậy minC=\(\dfrac{1}{5}khi\dfrac{2}{5}\le x\le\dfrac{3}{5}\).....Chúc các bạn học tốt

Ta có: \(\dfrac{x}{4}-\dfrac{2}{y}\: =\dfrac{3}{2}\: \: \: \Leftrightarrow\dfrac{xy}{4y}\: -\dfrac{8}{4y}=\dfrac{6y}{4y}\Rightarrow xy-8=6y\)\(xy-6y=8\Rightarrow y.\left(x-6\right)=8\)

Vì x,y \(\in Z\Rightarrow x-6\in Z\)

\(\Rightarrow y.\left(x-6\right)=1.8=-1.\left(-8\right)=2.4=-2.\left(-4\right)\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-6=1\\y=8\end{matrix}\right.\\\left\{{}\begin{matrix}x-6=8\\y=1\end{matrix}\right.\end{matrix}\right.hoặc\left[{}\begin{matrix}\left\{{}\begin{matrix}x-6=-1\\y=-8\end{matrix}\right.\\\left\{{}\begin{matrix}x-6=-8\\y=-1\end{matrix}\right.\end{matrix}\right.hoặc\left[{}\begin{matrix}\left\{{}\begin{matrix}x-6=2\\y=4\end{matrix}\right.\\\left\{{}\begin{matrix}x-6=4\\y=2\end{matrix}\right.\end{matrix}\right.hoặc\left[{}\begin{matrix}\left\{{}\begin{matrix}x-6=-2\\y=-4\end{matrix}\right.\\\left\{{}\begin{matrix}x-6=-4\\y=-2\end{matrix}\right.\end{matrix}\right.\)

Rồi làm như tìm x,y bình thường nha bạn......chúc bạn học tốt