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Đặt: \(A=\frac{3^6.21^{12}}{175^9.7^3}=\frac{3^{18}.7^{12}}{7^{12}.25^9}=\frac{3^{18}}{5^{18}}=\left(\frac{3}{5}\right)^{18}\)
\(B=\frac{3^{10}.6^7.4}{10^9.5^8}=\frac{3^{10}.2^7.3^7.2^2}{2^9.5^9.5^8}=\frac{3^{17}.2^9}{2^9.5^{17}}=\left(\frac{3}{5}\right)^{17}\)
Vì: \(\left(\frac{3}{5}\right)^{18}< \left(\frac{3}{5}\right)^{17}\Rightarrow A< B\)
Ta có:
\(a< b,c< d,m< n\)
\(\Rightarrow a+c+m< b+d+n\Rightarrow2a+2c+2m< a+b+c+d+m+n\)
\(\Rightarrow a+c+m< \frac{1}{2}\left(a+b+c+d+m+n\right)\)
\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\) ( đpcm )
ta co: 6x-2y=x+y(nhan cheo)
\(\Rightarrow\)5x=3y
\(\Rightarrow\)x/y=3/5
M=(x+y)3-3x2+6xy+3y2-7x-7y+8
M=(x+y)3-3x2+3xy+3xy+3y2-(7x+7y)+8
M=(x+y)3-3x(x+y)+3y(x+y)-7(x+y)+8
Thay x+y=-2 vào ta đc:
M=(-2)3-3x.(-2)+3y.(-2)-7.(-2)+8
M=-8-[(-2).(3x+3y)]-(-14)+8
M=-8-[(-2).3.(x+y)]+14+8
M=-8-[(-2)-3.(-2)]+14+8
M=-8-[(-2)+6]+14+8=10
Vậy M=10
\(7x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{6}{-4}=-\dfrac{3}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{-3}{2}\\\dfrac{y}{7}=\dfrac{-3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=-3.3\\2y=-3.7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-9}{2}\\y=\dfrac{-21}{2}\end{matrix}\right.\)
\(Vậy\)\(x=\dfrac{-9}{2}và,y=\dfrac{-21}{2}\)