4+32+3=
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a) 32 : 4 : 2 = 8 : 2 = 4
32 : 4 : 2 = 32 : 2 = 16
b) 18 : 2 x 3 = 18 : 6 = 3
18 : 2 x 3 = 9 x 3 = 18
bạn cứ đọc kkĩ lại đi, tại sao chỗ D=2*4*+ lại có cả dấu + lẫn dấu nhân
2*4+2*4*8+4*8*16+8*16*32/3*4+2*6*8+4*12*16+8*24*32
bai toan tinh nhanh
2/3 x 4/5 + 4/5 x 8/3 = 4/5 x (2/3 + 8/3) = 4/5 x 10/3 = 8/3
27/32 x 16/9 -27/32 x7/9 + 27/32
= 27/32 x (16/9 - 7/9 + 1 )
=27/32 x 2
=27/16
\(\frac{2}{3}\) x \(\frac{4}{5}\) + \(\frac{4}{5}\) x \(\frac{8}{3}\)
=\(\frac{4}{5}\) x ( \(\frac{2}{3}\) + \(\frac{8}{3}\) )
= \(\frac{4}{5}\) x \(\frac{10}{3}\)
= \(\frac{40}{15}\) = \(\frac{8}{5}\)
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
a: \(=\sqrt{8+2\cdot2\sqrt{2}\cdot\sqrt{5}+5}+\sqrt{8-2\cdot2\sqrt{2}\cdot\sqrt{5}+5}\)
\(=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)
b: \(=2\cdot\sqrt{17-3\sqrt{32}}\)
\(=2\cdot\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}\)
\(=2\left(3-2\sqrt{2}\right)=6-4\sqrt{2}\)