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`a)(4,5-2x)*1 4/7=11/14`
`=>(4,5-2x)*11/7=11/14`
`=>4,5-2x=1/2`
`=>2x=4,5-0,5=4`
`=>x=2`
Vậy `x=2`
`b)(2,8x-32):2/3=-90`
`=>2,8x-32=-90*2/3=-60`
`=>2,8x=-28`
`=>x=-10`
Vậy `x=-10`
A/\(\left(2,8x-32\right):\frac{2}{3}=-90\)
\(\left(\frac{28}{10}x-32\right)=\frac{-90}{1}.\frac{2}{3}\)
\(\left(\frac{14}{5}x-32\right)=\frac{-30}{1}.\frac{2}{1}\)
\(\left(\frac{14}{5}x-32\right)=-60\)
\(\frac{14}{5}x=-60+32\)
\(\frac{14}{5}x=-28\)
\(x=\frac{-28}{1}:\frac{14}{5}\)
\(x=\frac{-28}{1}.\frac{5}{14}\)
\(x=\frac{-2}{1}.\frac{5}{1}=-10\)
B/\(\left(4,5-2x\right).1\frac{4}{7}=\frac{11}{14}\)
\(\left(\frac{45}{10}-2x\right).\frac{11}{7}=\frac{11}{14}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}:\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}.\frac{7}{11}\)
\(\left(\frac{9}{2}-2x\right)=\frac{1}{2}.\frac{1}{1}=\frac{1}{2}\)
\(2x=\frac{9}{2}-\frac{1}{2}\)
\(2x=\frac{8}{2}\)
\(x=\frac{8}{2}:\frac{2}{1}=\frac{8}{2}.\frac{1}{2}\)
\(x=\frac{4}{2}.\frac{1}{1}=\frac{4}{2}=2\)
a, \(\left(2,8x-32\right):\dfrac{2}{3}=-90\)
\(\Rightarrow2,8x-32=-60\)
\(\Rightarrow2,8x=-28\)
\(\Rightarrow x=-10\)
Vậy x = -10
b, \(\left(4,5-2x\right):1\dfrac{4}{7}=\dfrac{11}{14}\)
\(\Rightarrow\left(4,5-2x\right):\dfrac{11}{7}=\dfrac{11}{14}\)
\(\Rightarrow4,5-2x=\dfrac{121}{98}\)
\(\Rightarrow2x=\dfrac{160}{49}\)
\(\Rightarrow x=\dfrac{80}{49}\)
Vậy \(x=\dfrac{80}{49}\)
\(a,\left(2,8x-32\right):\dfrac{2}{3}=-90\)
\(2,8x-32=-90.\dfrac{2}{3}\)
\(2,8x-32=-60\)
\(2,8x=-60+32\)
\(2,8x=-28\)
\(x=-28:2,8\)
\(x=-10\)
Vậy \(x=-10\)
\(b,\left(4,5-2x\right):1\dfrac{4}{7}=\dfrac{11}{14}\)
\(\left(4,5-2x\right):\dfrac{11}{7}=\dfrac{11}{14}\)
\(4,5-2x=\dfrac{11}{14}.\dfrac{11}{7}\)
\(4,5-2x=\dfrac{121}{98}\)
\(2x=4,5-\dfrac{121}{98}\)
\(2x=\dfrac{160}{49}\)
\(x=\dfrac{160}{49}:2\)
\(x=\dfrac{80}{49}\)
Vậy \(x=\dfrac{80}{49}\)
\(\left(5+\frac{4}{7}\right):x=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\left[5+\frac{4}{7}\right]:x=13\)
\(\Rightarrow\left[\frac{5\cdot7+4}{7}\right]:x=13\)
\(\Rightarrow\frac{39}{7}:x=13\)
\(\Rightarrow x=\frac{39}{7}:13=\frac{39}{7}\cdot\frac{1}{13}=\frac{3}{7}\cdot\frac{1}{1}=\frac{3}{7}\)
\(\left[2,8\cdot x+32\right]:\frac{2}{3}=90\)
\(\Rightarrow\left[2,8\cdot x+32\right]=90\cdot\frac{2}{3}\)
\(\Rightarrow\left[2,8\cdot x+32\right]=60\)
\(\Rightarrow2,8\cdot x+32=60\)
\(\Rightarrow2,8\cdot x=60-32\)
\(\Rightarrow2,8\cdot x=28\)
Đến đây dễ rồi :v
\(\left(2,8x-32\right)\div\frac{2}{3}=-90\)
\(\Rightarrow2,8x-32=-60\)
\(\Rightarrow2,8x=-28\)
\(\Rightarrow x=-10\)
\(\left(4,5-2x\right)\times1\frac{4}{7}=\frac{11}{14}\)
\(\Rightarrow4,5-2x=\frac{1}{2}\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
\(a.\left(2,8x-32\right):\frac{2}{3}=-90\)
\(2,8x-32=-90\cdot\frac{2}{3}=-30\cdot\frac{2}{1}\)
\(2,8x-32=-60\)
\(2,8x=-60+32\)
\(2,8x=28\)
\(x=\frac{28}{2,8}=10\). Vậy x=10
\(b.\left(4,5-2x\right)\cdot1\frac{4}{7}=\frac{11}{14}\)
\(4,5-2x=\frac{11}{14}:\frac{11}{7}=\frac{11}{14}\cdot\frac{7}{11}\)
\(4,5-2x=\frac{1}{2}\)
\(2x=\frac{45}{10}-\frac{1}{2}=\frac{9}{2}-\frac{1}{2}=\frac{8}{2}\)
\(2x=4\)
\(x=\frac{4}{2}=2\) . Vậy x=2
Chúc bạn học tốt!^_^
a)(2,8x-32):\(\frac{2}{3}\)=-90
(2,8x-32) =(-90).\(\frac{2}{3}\)
(2,8x-32) =(-60)
2,8x =(-60)+32
2,8x =(-28)
x =(-28):2,8
x =(-10)
b)(4,5-2x).1\(\frac{4}{7}\)=\(\frac{11}{14}\)
(4,5-2x).\(\frac{11}{7}\)=\(\frac{11}{14}\)
(4,5-2x) =\(\frac{11}{14}\):\(\frac{11}{7}\)
(4,5-2x) =\(\frac{1}{2}\)
2x =4,5-\(\frac{1}{2}\)
2x =4
x =4:2
x =2
a)(2,8x-32):\(\frac{2}{3}\)=-90
2,8x-32=-90.\(\frac{2}{3}\)
2,8x =-60+32
x =-28:2,8
x=-10
b)(4,5-2x).1\(\frac{4}{7}\)=\(\frac{11}{14}\)
4,5-2x =\(\frac{11}{14}:\frac{11}{7}\)
-2x =\(\frac{1}{2}\)
x=-1/4