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bn ơi???

\(\Leftrightarrow9x\left(x+2\right)+9y\left(y-\dfrac{2}{3}\right)=10\\ \Leftrightarrow9x^2+18x+9y^2-6y-10=0\\ \Leftrightarrow\left(9x^2+18x+9\right)+\left(9y^2-6y+1\right)=0\\ \Leftrightarrow9\left(x+1\right)^2+\left(3y-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)

Do x, y nguyên => \(\left\{{}\begin{matrix}\left(x+2\right)^2nguyên\ge0\\y-1nguyên\end{matrix}\right.\)

(x+2)² . (y-1) = -9

Ta có bảng:

\(-\dfrac{2}{3}=\dfrac{x}{-6}\Rightarrow x=\left(-\dfrac{2}{3}\right)\left(-6\right)=4\)

\(-\dfrac{2}{3}=\dfrac{10}{-y}\Rightarrow y=\left(-10\right):\left(-\dfrac{2}{3}\right)=15\)

\(-\dfrac{2}{3}=\dfrac{z}{9}\Rightarrow z=\left(-\dfrac{2}{3}\right).9=-6\)

\(\dfrac{-2}{3}=\dfrac{x}{-6}=\dfrac{10}{-y}=\dfrac{z}{9}\)

\(x=\left(-6.-2\right):3=4;y=\left(-6.10\right):-4=15;z=\left(10.9\right):-15=-6\)

\(\dfrac{-4}{x}=\dfrac{x}{-49}\\ \Rightarrow x^2=\left(-4\right)\left(-49\right)\\ \Rightarrow x^2=196\\ \Rightarrow x=\pm14\)

\(\dfrac{3.6}{x-3}=\dfrac{5}{3}\\ \Rightarrow5\left(x-3\right)=3.3.6\\ \Rightarrow5\left(x-3\right)=54\\ \Rightarrow x-3=\dfrac{54}{5}\\ \Rightarrow x=\dfrac{54}{5}+3\\ \Rightarrow x=\dfrac{69}{15}\)

\(\left(2x+1\right):2=12:3\\ \left(2x+1\right):2=4\\2x+1=2\\ 2x=1\\ x=\dfrac{1}{2} \)

\(\left(2x-14\right):3=12:9\\ \left(2x-14\right):3=\dfrac{4}{3}\\ 2x-14=4\\ 2x=16\\ x=8\)

-8+14=6-5=1

-20+29=14-5=9

[-8]+14=[6]-5=1

[-15]+29=[14]-5=9

(P/s: Mấy bài này sai đề hay sao mà mình tính ra số âm mất rồi? Lớp 1 đã bao giờ học đến số âm đâu?! Hay là bạn ghi nhầm lớp?)

_{Hoặc có thể là mình tính sai kkk =))))))))))}

Chúc bạn hoc tốt!! ^^

-8+14=6-5=1

-20+29=14-5=9

Chọn B

A

A

\(840=2^3\times3\times5\times7\)

\(150=2\times3\times5^2\)

\(990=2\times3^2\times5\times11\)

ƯCLN(840;150;990)=\(2\times3\times5=30\)