5x+x=39-3^101:3^99
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5x + 17 = x - 47
5x - x = -47 - 17
4x = -64
x = -64 : 4
x = -16
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\(a,=\dfrac{1}{3}\times\left(\dfrac{1}{2}+\dfrac{1}{3}\right)=\dfrac{1}{3}\times\dfrac{5}{6}=\dfrac{5}{18}\\ b,=\dfrac{4}{5}\times\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=\dfrac{4}{5}\times\dfrac{1}{6}=\dfrac{2}{15}\\ c,=456\times99-6\times99+456\\ =456\times\left(99+1\right)-594\\ =456\times100-594=45600-594=45006\\ d,=101\times\left(101-1\right)=101\times100=10100\)
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\(\frac{2}{x.\left(x+2\right)}+\frac{2}{3.5}+\frac{2}{5.7}.+.....+\frac{2}{99.101}=\frac{100}{101}\)
\(\Rightarrow\frac{1}{x\left(x+1\right)}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}=\frac{100}{101}\)
\(\Rightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}=\frac{100}{101}\)
\(\Rightarrow\frac{1}{x}-\frac{1}{101}=\frac{100}{101}\)
\(\Rightarrow\frac{1}{x}=\frac{101}{101}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{1}\)
\(\Rightarrow x=1\)
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(x+1)+(x+3)+(x+3^2)+...+(x+3^99)+(x+3^100)=3^101
=>x+1+x+3+x+3^2+...+x+3^99+x+3^100=3^101
=>(x+x+...+x) + (1+3+3^2+...+3^100) = 3^101
101 lần
=>101.x + (1+3+3^2+...+3^100) = 3^101.
Đặt A = 1 + 3 + 3^2 + ... + 3^100
=> 3A = 3.(1 + 3 + 3^2 + ... + 3^100)
=> 3A = 3 + 3^2 + 3^3 + ... + 3^101
=> 3A - A = 3 + 3^2 + 3^3 + ... + 3^101 - 1 - 3 - 3^2 - ... - 3^100
=> 2A = 3^101 - 1
=> A = (3^101 - 1)/2
Thay A = (3^101 - 1)/2 vào trên, ta có:
101.x + (3^101 - 1)/2 = 3^101
=>101x = 3^101 - (3^101-1)/2
=> x = [3^101 - (3^101 - 1)/2]/101
\(5x+x=39-3^{101}:3^{99}\)
\(\Rightarrow5x+x=39-9\)
\(6x=30\)
\(x=30:6\)
\(x=5\)
Vậy \(x=5\)
Chúc bạn học tốt !!!
3^101/3^99=3^2
nên 6x=39-9=30
x=30/6=5
(k) nha