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\(A=\left(1+3+3^2\right)+...+\left(3^{99}+3^{100}+3^{101}\right)\\ A=\left(1+3+3^2\right)+...+3^{99}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(1+...+3^{99}\right)=13\left(1+...+3^{99}\right)⋮13\)
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a: \(B=3+3^2+3^3+...+3^{120}\)
\(=3\left(1+3+3^2+...+3^{119}\right)⋮3\)
b: \(B=3+3^2+3^3+3^4+...+3^{2020}\)
\(=3\left(1+3\right)+...+3^{2019}\left(1+3\right)\)
\(=4\cdot\left(3+...+3^{2019}\right)⋮4\)
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\(50\%\dfrac{-3}{4}x^2=\dfrac{-5}{2}\)
\(\Leftrightarrow\dfrac{-3}{8}x^2=\dfrac{5}{2}\)
\(\Leftrightarrow x^2=\dfrac{5}{2}-\dfrac{-3}{8}\)
\(\Leftrightarrow x^2=\dfrac{23}{8}\)
\(\Leftrightarrow x=\sqrt{\dfrac{23}{8}}\approx1,696\)
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a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
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a: =-5/6-3/7=-35/42-18/42=-53/42
b: =18/45-20/45=-2/45
c: =-24/35
d: =2/3x-5/4=-10/12=-5/6
\(\dfrac{-5}{6}+\dfrac{3}{7}=\dfrac{-5\times7+3\times6}{6\times7}=\dfrac{-17}{42}.\)
\(\dfrac{2}{5}-\dfrac{4}{9}=\dfrac{2\times9-4\times5}{5\times9}=\dfrac{-2}{45}.\)
\(\dfrac{-6}{5}\times\dfrac{4}{7}=\dfrac{-6\times4}{5\times7}=\dfrac{-24}{35}.\)
\(\dfrac{2}{3}:\dfrac{-4}{5}=\dfrac{2}{3}\times\dfrac{5}{-4}=\dfrac{-5}{6}.\)
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a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)
\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{-2}{3}\)
b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{-11}{20}\)
c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)
\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{-4}{5}\)
d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)
\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{9}\)
e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)
\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)
\(\Rightarrow x=\dfrac{-7}{2}\)
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\(\Leftrightarrow x:3+367\cdot\left(-2\right)=-60\)
=>x:3=674
hay x=2022
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Bài 7:
a, \(x\) = \(\dfrac{1}{5}\) + \(\dfrac{2}{11}\)
\(x\) = \(\dfrac{11}{55}\) + \(\dfrac{10}{55}\)
\(x=\dfrac{21}{55}\)
b, \(\dfrac{x}{15}\) = \(\dfrac{3}{5}\) - \(\dfrac{2}{3}\)
\(\dfrac{x}{15}\) = \(\dfrac{9}{15}\) - \(\dfrac{10}{15}\)
\(\dfrac{x}{15}\) = \(\dfrac{1}{15}\)
\(x\) = 1
c, \(\dfrac{11}{8}\) + \(\dfrac{13}{6}\)= \(\dfrac{85}{x}\)
\(\dfrac{33}{24}\) + \(\dfrac{52}{24}\) = \(\dfrac{85}{x}\)
\(\dfrac{85}{24}\) = \(\dfrac{85}{x}\)
24 = \(x\)
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Giải :
Để 2x37y chia hết cho 5 thì y = 0 hoặc 5 .
Vậy y = 2 hoặc 7 thì chia 5 sẽ dư 2 .
Muốn 2x37y chia hết cho 2 thì y = { 0 ; 2 ; 4 ; 6 ; 8 }
Suy ra 2x37y chia hết 2 , chia 5 dư 2 thì y = 2
Nếu y = 2 , ta có : 2x37y = 2x372
Muốn 2x372 chia hết cho 9 thì ( 2 + x + 3 + 7 + 2 ) chia hết cho 9 .
x + 14 chia hết cho 9 .
Mà x \(\in\)N , 0 \(\le\)x \(\le\)9 => x = 4 .
Vậy số cần tìm là : 24372 .
Đáp số : 24372 .
Để 2x37y chi hết 2 và chia 5 dư 2 thì y = 2
Để 2x372 chia hết cho 9 thì :( 2+x+3+7+2) chia hết 9
=> x = 4
Vậy số cần tìm là: 24372
\(\left(x+5\right)^2-3=13\)
\(\left(x+5\right)^2\) \(=13+3\)
\(\left(x+5\right)^2\) \(=16\)
\(\Leftrightarrow x+5\) \(=\sqrt{16}\)
\(x+5\) \(=4\)
\(x\) \(=4-5\)
\(x\) \(=-1\)
\(\left(x+5\right)^2-3=13\\ \left(x+5\right)^2=13+3\\ \left(x+5\right)^2=16\\ TH1:\\ \left(x+5\right)^2=4^2\\ x+5=4\\ x=4-5\\ x=-1\\ TH2:\left(x+5\right)^2=\left(-4\right)^2\\ \\ x+5=-4\\ x=-4-5\\ x=-9\)