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Ta có: \(A=\left[6.\left(-\frac{1}{3}\right)^2-3.\left(-\frac{1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(=\left(6.\frac{1}{9}-\left(-1\right)+1\right):\left(\frac{-4}{3}\right)\)
\(=\left(\frac{2}{3}+2\right).\left(\frac{-3}{4}\right)\)
\(=\frac{8}{3}.\left(-\frac{3}{4}\right)\)
\(=-2\)
\(B=\left(729-1^3\right)\left(729-3^3\right)...\left(729-125^3\right)\)
\(\Rightarrow B=\left(729-1^3\right)\left(729-3^3\right)...\left(729-9^3\right)...\left(729-125^3\right)\)
\(\Rightarrow B=\left(729-1^3\right)\left(729-3^3\right)...0...\left(729-125^3\right)\)
\(\Rightarrow B=0\)
Vì -2 < 0 nên A < B
Vậy A < B
a) \(\dfrac{1}{2}-\dfrac{3}{4}.\dfrac{-6}{5}\)
\(=\dfrac{1}{2}-\dfrac{3.\left(-6\right)}{4.5}\)
\(=\dfrac{1}{2}-\dfrac{-18}{20}\)
\(=\dfrac{1}{2}+\dfrac{9}{10}\)
\(=\dfrac{5}{10}+\dfrac{9}{10}\)
\(=\dfrac{5+9}{10}\)
\(=\dfrac{14}{10}\)
\(=\dfrac{7}{5}\)
b) \(\dfrac{\dfrac{1^0}{9}.3^2.9^3}{729}\)
\(=\dfrac{9^{-1}.3^2.9^3}{729}\)
\(=\dfrac{9^{-1}.9.9^3}{729}\)
\(=\dfrac{9^{-1+1+3}}{729}\)
\(=\dfrac{9^3}{729}\)
\(=\dfrac{729}{729}\)
\(=1\)
\(\frac{x}{7}=\frac{y}{3}=\frac{x-y}{7-3}=\frac{20}{4}=5\)
\(\frac{x}{7}=5\Rightarrow x=35\)
\(\frac{y}{3}=5\Rightarrow y=15\)
tíc mình nha
3 bài à
1/ ta có x/7 = y/3 và x-y=20
ADTCDTSBN
x/7=y/3 = x-y/ 7-3 = 20/4= 5
Suy ra
x/7=5 => x=7.5= 35
y/3=5=> y= 3.5 = 15
Vậy x = 35 và y=15