cho bài tìm
x nhân 1.2-3.45=4.68
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Yx,2-3,45=4,68
Yx1,2=4,68+3,45
Yx1,2=8,13
Y=8,13:1,2
Y=6,775
x.1,2-3,45=4,68
x.1,2=4,68+3,45
x.1,2=8,13
x=8,13:1,2
x=6,775
\(x-3,45=\frac{90,5}{15}=\frac{181}{30}\)
=> \(x=\frac{181}{30}+3,45=\frac{569}{60}\)
x - 3 . 45 = 90 . 5 : 15
x - 135 = 450 : 15
x - 135 = 30
x = 30 + 135
x = 165
Vậy x = 165
đề k rõ j hết phải là zầy k p :
X * 1,2 - 3.45=4.68
==> x=6,775
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(100-10\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=1.2=2\)
\(\Rightarrow\left(x+\dfrac{206}{100}\right)=\dfrac{5}{2}:2=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{5}{4}-\dfrac{206}{100}=\dfrac{125}{100}-\dfrac{206}{100}\)
\(\Rightarrow x=-\dfrac{81}{100}\)
Xx1,2-3,45=4,68
=>Xx1,2=4,68+3,45=8,13
=>x=8,13:1,2=6,775