Tìm x :
a. x x 45 + x x 55 = 1000
b. 6 x + 1 2 = 2
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Mình dùng dấu . thay cho dấu nhân nha
a) \(x.45+x.55=100\)
\(\Rightarrow x\left(45+55\right)=100\)
\(\Rightarrow x.100=100\Rightarrow x=1\)
b) \(\frac{6}{x}+\frac{1}{2}=2\)
\(\Rightarrow\frac{6}{x}=2-\frac{1}{2}\)
\(\Rightarrow\frac{6}{x}=\frac{5}{2}\)
\(\Rightarrow5.x=2.6=12\)
\(\Rightarrow x=\frac{12}{5}\)
\(a,x\times45+x\times55=100\)
\(x\times\left(45+55\right)=100\)
\(x\times100=100\)
\(x=100\div100\)
\(x=1\)
\(b,\frac{6}{x}+\frac{1}{2}=2\)
\(\frac{6}{x}=2-\frac{1}{2}\)
\(\frac{6}{x}=\frac{3}{2}\)
\(\frac{6}{x}=\frac{6}{4}\)
\(x=4\)
x . 45 + x . 55 = 1000
x . ( 45 + 55 ) = 1000
x . 100 = 1000
x = 1000 : 100
x = 10
Vậy x = 10
6 / x + 1 / 2 = 2
6 / x = 2 - 1 / 2
6 / x = 3 / 2
6 / x = 6 / 4
=> x = 4
Vậy x = 4
a, \(\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\ne0\right)=0\Leftrightarrow x=100\)
b, \(\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)
\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}=\dfrac{x+2005}{2002}+\dfrac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\ne0\right)=0\Leftrightarrow x=-2005\)
a. lấy mỗi phân số e cộng vs 2 là bt làm ra liền
b, - 1 hoặc + 1 vs mỗi phân số nha
a) \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\Leftrightarrow\left(\frac{x-45}{55}-1\right)+\left(\frac{x-47}{53}-1\right)=\left(\frac{x-55}{45}-1\right)+\left(\frac{x-53}{47}-1\right)\)
\(\Leftrightarrow\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
\(\Leftrightarrow\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
Vì \(\hept{\begin{cases}\frac{1}{55}< \frac{1}{45}\\\frac{1}{53}< \frac{1}{47}\end{cases}}\Rightarrow\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}< 0\)
\(\Rightarrow x-100=0\Rightarrow x=100\)
Vậy x = 100
a. x x 45 + x x 55 = 1000
x x (45 + 55) = 1000
x x 100 = 1000
x = 1000 : 100
x = 10
b.
6 x + 1 2 = 2 6 : x + 1 2 = 2 6 : x = 2 - 1 2 6 : x = 3 2 x = 6 : 3 2 x = 4