\([(10-x)\times2+5]\div3-2=3\)
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\(\left[\left(10-x\right).2+5\right]\div3-x=3\)
\(\Rightarrow\left[20-2x+5\right]\div3-x=3\)
\(\Rightarrow25-2x=\left(3+x\right).3\)
\(\Rightarrow25-2x=9+3x\)
\(\Rightarrow25=9+5x\)
\(\Rightarrow5x=16\)
\(\Rightarrow x=3,2\)
Ta có: \(\left(-\dfrac{5}{8}\right)-x:3\dfrac{5}{6}+7\dfrac{3}{4}=-2\)
\(\Leftrightarrow\dfrac{-5}{8}-x:\dfrac{23}{6}+\dfrac{31}{4}+2=0\)
\(\Leftrightarrow x:\dfrac{-23}{6}+\dfrac{73}{8}=0\)
\(\Leftrightarrow x\cdot\dfrac{-6}{23}=-\dfrac{73}{8}\)
\(\Leftrightarrow x=-\dfrac{73}{8}:\dfrac{-6}{23}=\dfrac{-73}{8}\cdot\dfrac{23}{-6}\)
hay \(x=\dfrac{1679}{48}\)
Vậy: \(x=\dfrac{1679}{48}\)
a) \(3\times\dfrac{4}{11}=\dfrac{3\times4}{11}=\dfrac{12}{11}\)
b) \(1\times\dfrac{5}{4}=\dfrac{1\times5}{4}=\dfrac{5}{4}\)
c) \(0\times\dfrac{2}{5}=\dfrac{0\times2}{5}=\dfrac{0}{5}=0\)
a: \(=\dfrac{3\cdot4}{11}=\dfrac{12}{11}\)
b: \(=\dfrac{1\cdot5}{4}=\dfrac{5}{4}\)
c: \(=\dfrac{0\cdot2}{5}=0\)
a: Số số hạng là \(\dfrac{2018-2}{2}+1=1009\left(số\right)\)
Tổng là: \(\dfrac{2018+2}{2}\cdot1009=1009\cdot1010=1019090\)
b: \(10S=10^2+10^3+...+10^{101}\)
\(\Rightarrow9S=10^{101}-10\)
hay \(S=\dfrac{10^{101}-10}{9}\)
c: \(5S=1+\dfrac{1}{5}+...+\dfrac{1}{5^{99}}\)
\(\Leftrightarrow4S=1-\dfrac{1}{5^{100}}\)
hay \(S=\dfrac{1}{4}\left(1-\dfrac{1}{5^{100}}\right)\)
a) \(32< 2^x< 128\)
=> \(2^5< 2^x< 2^7\)
=> x = 6
b) \(2^{x-1}+4\cdot2^x=9\cdot2^5\)
=> \(2^{x-1}+2^2\cdot2^x=9\cdot2^5\)
=> \(2^{x-1}+2^{2+x}=9\cdot2^5\)
=> 9.2x-1 = 9.25
=> 2x-1 = \(\frac{9\cdot2^5}{9}=2^5\)
=> x - 1 = 5 => x = 6
c) \(9\cdot27\le3^x\le243\)
=> \(243\le3^x\le243\)
=> x = 5
d) Giống câu b)
e) \(3^{x-1}+5\cdot3^{x-2}=216\)
=> 8.3x-2 = 216
=> 3x-2 = 27
=> 3x-2 = 33
=> x - 2 = 3 => x = 5
f) 27x-3 = 9x+3
=> 27x-3 = 9x+3
=> (33)x-3 = (32)x+3
=> 33x-9 = 32x + 6
=> không thỏa mãn x vì x là phân số mà theo đề bài là số nguyên
g) x2019 = x => x2019 - x = 0 => x(x2018 - 1) = 0 => x = 0 hoặc x = 1
a)
\(2^5< 2^x< 2^7\)
\(5< x< 7\)
\(x=6\)
b)
\(2^{x-1}+2^2\cdot2^x=9\cdot2^5\)
\(2^{x-1}+2^{2+x}=9\cdot2^5\)
\(2^{x-1}\left(1+2^3\right)=9\cdot2^5\)
\(2^{x-1}\cdot9=9\cdot2^5\)
\(2^{x-1}=2^5\)
\(x-1=5\)
\(x=6\)
\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{9}-\frac{1}{10}\)
\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{10}=\frac{9}{10}\)
\(\frac{2}{5}x=\frac{9}{10}-\frac{3}{10}=\frac{3}{5}\)
\(x=\frac{\frac{3}{5}}{\frac{2}{5}}=\frac{3}{2}\)
Ta có: \(\frac{1}{1x2}\)+ \(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ \(\frac{1}{4x5}\)+ .....+ \(\frac{1}{9x10}\)
= \(1-\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
= 1 - \(\frac{1}{10}\)
= \(\frac{9}{10}\)
[(10-x).2+5]:3-2=3
[(10-x).2+5]:3=3+2
[(10-x).2+5]:3=5
[(10-x).2+5]=5.3
(10-x).2+5=15
(10-x).2=15-5
(10-x).2=10
10-x=10:2
10-x=5
x=10-5
x=5
\(\left[\left(10-x\right).2+5\right]:3-2=3\)
\(\Rightarrow\left(10-x\right).2+5=\left(3+2\right).3\)
\(\Rightarrow\left(10-x\right).2+5=15\)
\(\Rightarrow10-x=\left(15-5\right):2\)
\(\Rightarrow10-x=5\)
\(\Rightarrow x=10-5\)
\(\Rightarrow x=5\)