ta có

\(\frac{1}{x}-\frac{1}{2y}=\frac{1}{3}\Leftrightarrow6y-3x=2xy\Leftrightarrow2xy-6y+3x-9=-9\)

\(\Leftrightarrow\left(2y+3\right)\left(x-3\right)=-9\)\(\Rightarrow x-3\in\left\{-9,-1,1,9\right\}\Rightarrow x\in\left\{-6,2,4,12\right\}\)

tương ứng ta có cặp x,y là \(\left(-6,-1\right),\left(2,3\right),\left(4,-6\right),\left(12,-2\right)\)

Ta có:

 \(xy-3y=x^2+2\)\(\Rightarrow xy-3y-x^2=2\)

\(\rightarrow y\left(x-3\right)-x^2+9=11\)

\(\rightarrow y\left(x-3\right)-\left(x^2-3^2\right)=11\)

\(\rightarrow y\left(x-3\right)-\left(x-3\right)\left(x+3\right)=11\)

\(\rightarrow\left(y-x-3\right)\left(x-3\right)=11\)

Xét ΔEAC và ΔBAD có :

AD = AC ( gt )

ˆCAE=ˆDAB( hai góc đối đỉnh )

AE = AB ( gt )

nên ΔEAC=ΔBAD(c.g.c)

=> BD = CE ( hai cạnh tương ứng )

\(A=\frac{x^2-2x+1}{x+1}=\frac{x^2-2x-3+4}{x+1}=\frac{\left(x+1\right)\left(x-3\right)+4}{x+1}=x-3+\frac{4}{x+1}\inℤ\)

mà \(x\inℤ\)nên \(\frac{4}{x+1}\inℤ\)do đó \(x+1\inƯ\left(4\right)=\left\{-4,-2,-1,1,2,4\right\}\)

\(\Leftrightarrow x\in\left\{-5,-3,-2,0,1,3\right\}\).

\(\left(\frac{1}{9}\right)^{2017}.9^{2017}-96^2:24^2=\left(3x-2\right)^3+12\)

\(\left(\frac{1}{9}.9\right)^{2017}-\left(96:24\right)^2=\left(3x-2\right)^3+12\)

\(1^{2017}-4^2=\left(3x-2\right)^3+12\)

\(-15=\left(3x-2\right)^3+12\)

\(\Rightarrow\left(3x-2\right)^3=-27\rightarrow3x-2=3\)

\(\Rightarrow x=\frac{5}{3}\)

Bi xanh : 10 viên

Bi vàng : 14 viên

Bi đỏ : 18 viên

( cần lời giải chi tiết thì bảo sau )

~HT~

Tên các góc vuông là :

\(\widehat{xAm};\widehat{mAy};\widehat{nAy};\widehat{nAx}\)

~~Học tốt~~

thank...

có : \(y\inℕ\)

\(\Rightarrow m-3⋮m+2\)

\(\Rightarrow m+2-5⋮m+2\)

có :\(m+2⋮m+2\)       \(\Rightarrow-5⋮m+2\)

\(\Rightarrow m+2\inƯ_{\left(5\right)}\)

    Vậy để \(y\inℕ\) thì  \(m=\left\{-7;-3;-1;3\right\}\)