5. (x-22) (x-32)=0
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\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
a) \(\left(x+22\right)\left(x-13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+22=0\\x-13=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-22\\x=13\end{cases}}}\)
Vậy \(x\in\left\{-22;13\right\}\)
b) \(\left(-x+25\right)\left(32-x\right)=0\)
\(\Leftrightarrow\left(25-x\right)\left(32-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}25-x=0\\32-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=25\\x=32\end{cases}}}\)
Vậy \(x\in\left\{25;32\right\}\)
a)\(\left(x+22\right)\left(x-13\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+22=0\\x-13=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-22\\x=13\end{cases}}}\)
b) \(\left(-x+25\right)\left(32-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}-x+25=0\\32-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=25\\x=32\end{cases}}}\)
hok tốt!!
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
\(5.\left[225-\left(x-10\right)-125\right]=0\)
\(225-\left(x-10\right)-125=0:5\)
\(225-\left(x-10\right)-125=0\)
\(\left(x-10\right)-125=225-0\)
\(\left(x-10\right)-125=225\)
\(x-10=225+125\)
\(x-10=350\)
Mấy bài này làm theo thứ tứ trái sang phải, nhân chia trc cộng trừ sau thoi mà :>>>?
\(22.\left(x+32\right)-5=55.179^0\\ 22.\left(x+32\right)-5=55.1\\ 22.\left(x+32\right)-5=55\\ 22.\left(x+32\right)=55+5\\ 22.\left(x+32\right)=60\\ x+32=60:22\\ x+32=\dfrac{30}{11}\\ x=\dfrac{30}{11}-32\\ x=\dfrac{-322}{11}\)
\(22.\left(x+32\right)-5=55.179^0\)
=>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............
Th 1 : x - 22 = 0
x = 0 + 22
x = 22
Th 2 : x - 32 = 0
x = 0 + 32
x = 32
5. (x-22)(x-32)=0
<=>\(\hept{\begin{cases}x-2^2=0\\x-3^2=0\end{cases}}\)
<=>\(\hept{\begin{cases}x=2^2=4\\x=3^2=9\end{cases}}\)
Vậy x ∈ {4;9}