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1) \(\dfrac{1}{4}x-\dfrac{1}{3}=\dfrac{-5}{9}\)
\(\Rightarrow\dfrac{1}{4}x=-\dfrac{2}{9}\Rightarrow x=-\dfrac{8}{9}\)
2) \(2^{x-3}-3.2^x=-92\)
\(\Rightarrow2^x\left(2^{-3}-3\right)=-92\)
\(\Rightarrow2^x.\dfrac{-23}{9}=-92\)
\(\Rightarrow2^x=32\Rightarrow x=5\)
...\(a=\left[\left(\left(50-1\right):1+1\right):2\right]\left(50+1\right)=25.51=1275\)
\(...a1=\left[\left(\left(98-35\right):3+1\right):2\right]\left(35+98\right)=11.133=1463\)
1.
\(\frac{x}{2}=\frac{y}{3}=>\frac{x}{10}=\frac{y}{15}\)
\(\frac{y}{5}=\frac{z}{7}=>\frac{y}{15}=\frac{z}{21}\)
=>\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{92}{46}=2\)
=> x=2x10=20
y=2x15=30
z=2x21=42
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
M=100
Xét tử N
92-(1/9)-(2/10)-(3/11)- ... -(90/98)-(91/99)-(92/100)
=(1+1+1+...+1)-(1/9)-(2/10)-(3/11)- ... -(90/98)-(91/99)-(92/100)
=1-(1/9)+1-(2/10)+1-(3/11)+......+1-(90/98)+1-(91/99)+1-(92/100)
=(8/9)+(8/10)+(8/11)+ ...+ (8/98)+(8/99)+(8/100)
=8.[(1/9)+(1/10)+(1/11)+...+(1/98)+(1/99)+(1/100)]
=40[(1/45)+(1/50)+(1/55)+...+(1/495)+(1/500)]
=>N=40
=>M/N=5/2
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