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\(=\dfrac{2^{12}\cdot3^{10}+2^9\cdot2^3\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot5}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot5}=\dfrac{2\cdot6}{3\cdot5}=\dfrac{12}{15}=\dfrac{4}{5}\)
a: \(=\left(\dfrac{-48}{12}+\dfrac{-8}{12}+\dfrac{21}{12}\right)\cdot\dfrac{-12}{13}\)
\(=\dfrac{-35}{12}\cdot\dfrac{-12}{13}=\dfrac{35}{13}\)
b: \(=\dfrac{-3}{6}+\dfrac{5}{6}-\dfrac{312}{100}+\dfrac{51}{10}\)
\(=\dfrac{1}{3}-\dfrac{312}{100}+\dfrac{51}{10}=\dfrac{347}{150}\)
c: \(=\left(\dfrac{48}{300}+\dfrac{175}{300}-\dfrac{135}{100}\right)\cdot\dfrac{5}{2}+\dfrac{1}{4}\)
\(=\dfrac{88}{300}\cdot\dfrac{5}{2}+\dfrac{1}{4}=\dfrac{59}{60}\)
Bài 2:
a: \(A=11+\dfrac{3}{13}-2-\dfrac{4}{7}-5-\dfrac{3}{13}\)
\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)
b: \(B=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}\)
\(=5+\dfrac{7}{11}=\dfrac{62}{11}\)
c: \(C=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{5}{7}=1\)
d: \(D=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)
\(=\dfrac{20}{10}\cdot7\cdot\dfrac{8}{3}\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}=2\cdot\dfrac{5}{4}=\dfrac{5}{2}\)
a: =31/9+31/6=155/18
b: =113/14-45/7=23/7
c: =7-3-6/7=4-6/7=24/7
1) \(A=1+2+2^2+2^3+......+2^{2015}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)
\(\Leftrightarrow A=2^{2016}-1\)
Vậy \(A=2^{2016}-1\)
6)Ta có: \(13+23+33+43+.......+103=3025\)
\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)
\(\Leftrightarrow26+46+66+86+.......+206=6050\)
\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)
\(\Leftrightarrow23+43+63+83+.......+203+=6020\)
Vậy S=6020
b, B có 19 thừa số
=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)
<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)
<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)
<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)
<=>\(B=\frac{-21}{40} \)
\(A=cos3a+2cos\left(\pi-3a\right)sin^2\left(\dfrac{\pi}{4}-1,5a\right)\)
\(=cos3a-2cos3a\dfrac{1-cos\left(\dfrac{\pi}{2}-3a\right)}{2}\)
\(=cos3a-cos3a\left(1-sin3a\right)\)
\(=cos3a-cos3a+cos3asin3a=\dfrac{1}{2}sin6a\)
\(=\dfrac{1}{2}sin\left(6\dfrac{5\pi}{6}\right)=\dfrac{1}{2}sin\left(4\pi+\pi\right)=\dfrac{1}{2}sin\pi=0\)
Vì a=\(\dfrac{5\pi}{6}\) nên: \(3a=\dfrac{5\pi}{2}\) => \(\cos3a=0\)
\(\pi-3a=\pi-\dfrac{5\pi}{2}=\dfrac{-3\pi}{2}\)
=> \(\cos\left(\pi-3a\right)=0\)
A = (1- 2) \(\times\) ( 4 - 3) \(\times\) (5 - 6) \(\times\) (8 - 7) \(\times\) (9 - 10) \(\times\) (12 - 11) \(\times\)(13 - 14)
A = (-1) \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1)
A = 1
a, Biến đổi ta được E = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b, Ta có E = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\) = \(1+\dfrac{4}{\sqrt{x}-3}\) .
. Nếu x không là số chính phương thì \(\sqrt{x}\) là số vô tỉ . Suy ra E là số vô tỉ ( loại )
. Nếu x là số chính phươn thì \(\sqrt{x}\) là số nguyên nên để E có giá trị nguyên thì \(4⋮\left(\sqrt{x}-3\right)\) .
Mà \(\sqrt{x}-3\ge-3\) nên \(\left(\sqrt{x}-3\right)\in\left\{-2;-1;1;2;4\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{1;2;4;5;7\right\}\Rightarrow x\in\left\{1;4;16;25;49\right\}\)
Kết hợp với ĐKXĐ ta được x = 1 ; 16 ; 25 ; 49
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-3^{11}\cdot2^{11}}=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot5}=\dfrac{2}{3}\cdot\dfrac{6}{5}=\dfrac{4}{5}\)