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\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)
\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)
=1/64
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a)
\(\frac{{28}}{9} \cdot 0,7 + \frac{{28}}{9} \cdot 0,5 = \frac{{28}}{9}.\left( {0,7 + 0,5} \right)\)
b)
\(\begin{array}{l}\frac{{36}}{{13}}:4 + \frac{{36}}{{13}}:9\\ = \frac{{36}}{{13}}.\frac{1}{4} + \frac{{36}}{{13}}.\frac{1}{9}\\ = \frac{{36}}{{13}}.\left( {\frac{1}{4} + \frac{1}{9}} \right)\\ = \frac{{36}}{{13}}.\frac{{13}}{{36}} = 1\end{array}\)
\(\begin{array}{l}\frac{{36}}{{13}}:(4 + 9)\\ = \frac{{36}}{{13}}:13\\ = \frac{{36}}{{13}}.\frac{1}{{13}}\\ = \frac{{36}}{{169}}\end{array}\)
Suy ra \(\frac{{36}}{{13}}:4 + \frac{{36}}{{13}}:9\) \( \ne \) \(\frac{{36}}{{13}}:(4 + 9)\).
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a)Đặt \(A=8^9+7^9+6^9+5^9+4^9+3^9+2^9+1^9\)
\(A< 8^9+8^9+8^9+8^9+8^9+8^9+8^9+8^9\)
\(A< 8\cdot8^9\)
\(A< 8^{10}< 9^{10}\)
\(\Rightarrow9^{10}>8^9+7^9+6^9+5^9+4^9+3^9+2^9+1^9\)
a) \(8^9+7^9+6^9+5^9+4^9+3^9+2^9+1^9\)
(8+7+6+5+4+3+2+1)9
369
Vậy369>99
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\(=-\dfrac{6}{5}:\left(\dfrac{6}{5}\cdot\dfrac{5}{4}\right):\dfrac{16}{25}-\dfrac{1}{25}+\left(1.04\right)\cdot\dfrac{4}{7}:\left(5:\dfrac{5}{9}-\dfrac{9}{4}\right)\cdot\dfrac{36}{17}+\dfrac{3}{10}\cdot\dfrac{5}{2}\)
\(=\dfrac{-6}{5}\cdot\dfrac{4}{6}\cdot\dfrac{25}{16}-\dfrac{1}{25}+\dfrac{1664}{8925}+\dfrac{3}{4}\)
\(=\dfrac{-27}{50}+\dfrac{1664}{8925}=-\dfrac{6311}{17850}\)
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1+2 =3
3+9=12
còn lại tự làm nha đây kiến thức lớp 7 á lớp 3 còn làm được
1+2=3
3+9=12
9-5=4
125-(-124)=249
35-9+3x158=500
36+4-3-39+10=8
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5(x - 4)2 - 9 = 36
5(x - 4)2 = 45
(x - 4)2 = 9
\(\Rightarrow\) x - 4 = 3 hoặc x - 4 = - 3
\(\Rightarrow\) x = 7 hoặc x = 1
\(5\left(x-4\right)^2-9=36\)
\(5\left(x-4\right)^2=36+9=45\)
\(\left(x-4\right)^2=45:5=9\)
\(\left(x-4\right)^2=3^3=\left(-3\right)^2\)
Trường hợp 1: \(x-4=3\Leftrightarrow x=3+4=7\)
Trường hợp 2: \(x-4=-3\Leftrightarrow x=-3+4=1\)
Vậy x = 7 hoặc x = 1
giup voiiiiiiiiiiiiiiiiiiiii
4+9+36 = 49 nhé!
Học tốt~