y - 3 x y + 7 x 7 = 30
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\(A=2^2+2^4+...+2^{20}\)
\(=2^2\left(1+2^2+...+2^{18}\right)=4\left(1+2^2+...+2^{18}\right)⋮4\)
\(A=2^2+2^4+...+2^{18}+2^{20}\)
\(=2^2\left(1+2^2\right)+...+2^{18}\left(1+2^2\right)\)
\(=5\left(2^2+2^6+...+2^{18}\right)⋮5\)
Sửa đề; DE//BC
Xét ΔABC có \(\dfrac{AD}{AB}=\dfrac{AE}{AC}\)
nên DE//BC
\(A=1+3+3^2+...+3^{11}\)
\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(3^8+3^9+3^{10}+3^{11}\right)\)
\(=40+3^4\left(1+3+3^2+3^3\right)+3^8\left(1+3+3^2+3^3\right)\)
\(=40\left(1+3^4+3^8\right)⋮40\)
Để ý thấy rằng \(1+3+3^2+3^3=40\)
\(A=1+3+3^2+3^3+...+3^{11}\)
\(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+3^8\left(1+3+3^2+3^3\right)\)
\(=40+3^4\times40+3^8\times40\)
\(=40\left(1+3^4+3^8\right)\)
Do đó A chia hết cho 40
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x+1}{3}=\dfrac{y-2}{4}=\dfrac{\left(x+1\right)-\left(y-2\right)}{3-4}=\dfrac{x+1-y+2}{-1}=\dfrac{x-y+3}{-1}=\dfrac{18}{-1}\)
`= -18`
Suy ra: \(\left\{{}\begin{matrix}\dfrac{x+1}{3}=-18\\\dfrac{y-2}{4}=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1=-54\\y-2=-72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-55\\y=-70\end{matrix}\right.\)
Vậy ....
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x+1}{3}=\dfrac{y-2}{4}=\dfrac{x-y+1+2}{3-4}=\dfrac{15+3}{-1}=-18\)
=>\(\left\{{}\begin{matrix}x+1=-18\cdot3=-54\\y-2=4\cdot\left(-18\right)=-72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-54-1=-55\\y=-72+2=-70\end{matrix}\right.\)
\(\left(-\dfrac{3}{5}\right)^2.\dfrac{5}{11}+\dfrac{9}{25}.\left(-\dfrac{16}{11}\right)\)
\(=\dfrac{9}{25}.\dfrac{5}{11}+\dfrac{9}{25}.\left(-\dfrac{16}{11}\right)\)
\(=\dfrac{9}{25}.\left[\dfrac{5}{11}+\left(-\dfrac{16}{11}\right)\right]\)
\(=\dfrac{9}{25}.\left(-1\right)\)
\(=-\dfrac{9}{25}\)
\(y-3y+7\cdot7=30\)
=>-2y=30-49=-19
=>\(y=\dfrac{19}{2}\)