22-x=32
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\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
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2^2.(x+3^2)-5=55
4.(x+9)=55+5=60
(x+9)=60:4=15
x=15-9=6
vậy x=6
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\(22.\left(x+32\right)-5=55.179^0\\ 22.\left(x+32\right)-5=55.1\\ 22.\left(x+32\right)-5=55\\ 22.\left(x+32\right)=55+5\\ 22.\left(x+32\right)=60\\ x+32=60:22\\ x+32=\dfrac{30}{11}\\ x=\dfrac{30}{11}-32\\ x=\dfrac{-322}{11}\)
\(22.\left(x+32\right)-5=55.179^0\)
=>22.(x+32)-5=55.1
=>22.(x+32)-5=55
=>22.(x+32)=55+5
=>22.(x+32)=60
=>x+32=60:22
=>x+32=\(\dfrac{30}{11}\)
=>x=\(\dfrac{30}{11}-32\)
\(=>x=-\dfrac{322}{11}\)
Vậy............
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Phương pháp giải:
Muốn tìm một số hạng ta lấy tổng trừ đi số hạng kia.
Lời giải chi tiết:
a) x + 9 = 22
x = 22 – 9
x = 13
b) 6 + x = 32
x = 32 – 6
x = 26
22-x = 32
22-x = 25
2 - x = 5
x = 2 - 5
x = - 3
Vậy x = - 3
x= -5 nha