Y: 0,24= 0,75 - 1/5
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Bài giải:
Câu 1: a, \(\left(-2\right).4.5.38.\left(-25\right)\)
\(=\left[\left(-2\right).5\right].\left[4.\left(-25\right)\right].38\)
\(=\left(-10\right).\left(-100\right).38\)
\(=1000.38=38000\)
b,\(\frac{1}{3}+\frac{3}{8}-\frac{7}{12}\)
\(=\left(\frac{1}{3}+\frac{3}{8}\right)-\frac{7}{12}\)
\(=\frac{17}{24}-\frac{7}{12}=\frac{1}{8}\)
c, \(\frac{-5}{8}.\frac{5}{12}+\frac{-5}{8}.\frac{7}{12}+2\frac{1}{8}\)
\(=\frac{-5}{8}.\left(\frac{5}{12}+\frac{7}{12}\right)+\frac{17}{8}\)
\(=\frac{-5}{8}.1+\frac{17}{8}\)
\(=\frac{3}{2}\)
Câu 2: a, \(x-\frac{2}{5}=0,24\)
\(x-0,4=0,24\)
\(x=0,24+0,4\)
\(\Rightarrow x=0,64\left(\frac{16}{25}\right)\)
b,\(\frac{2}{3}.x+\frac{1}{12}=\frac{1}{10}\)
\(\frac{2}{3}.x=\frac{1}{10}-\frac{1}{12}\)
\(\frac{2}{3}.x=\frac{1}{60}\)
\(x=\frac{1}{60}:\frac{2}{3}\)
\(\Rightarrow x=\frac{1}{40}\)
c, \(\left(3\frac{1}{2}-2x\right).1\frac{1}{3}=7\frac{1}{3}\)
\(\frac{7}{2}-2x=\frac{22}{3}:\frac{4}{3}\)
\(\frac{7}{2}-2x=\frac{11}{2}\)
\(2x=\frac{7}{2}-\frac{11}{2}\)
\(2x=-2\)
\(\Rightarrow x=-2:2\)
\(x=-1\)
* \(\dfrac{15}{12}:x=\left(\dfrac{5}{6}-\dfrac{1}{2}\right).\dfrac{5}{4}\)
\(\dfrac{15}{12}:x=\dfrac{1}{3}.\dfrac{5}{4}\)
\(\dfrac{15}{12}:x=\dfrac{5}{12}\)
\(x=\dfrac{15}{12}:\dfrac{5}{12}\)
\(x=3\)
* \(y:0,15=\left(2,8+0,38\right):0,75\)
\(y:0,15=3,18:0,75\)
\(y:15=4,24\)
\(y=4,24.15\)
\(y=63,6\)
1) 2,75 - 5/6 × 2/5 = 2,75 - (5/6) × (2/5) = 2,75 - 1/3 = 2,75 - 0,33 = 2,42
2) 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 3/4) - 3/5 = 1,25 - (5/6 - 9/12) - 3/5 = 1,25 - (10/12 - 9/12) - 3/5 = 1,25 - 1/12 - 3/5 = 1,25 - 0,08 - 0,6 = 1,25 - 0,68 = 0,57
3) 4/9 × 0,75 + 8/5 + 3,125 = (4/9) × 0,75 + 8/5 + 3,125 = 0,44 + 8/5 + 3,125 = 0,44 + 1,6 + 3,125 = 0,44 + 4,725 = 5,165
4) 1,125 - 4/7 - 0,12 = 1,125 - (4/7) - 0,12 = 1,125 - 0,57 - 0,12 = 0,435 - 0,12 = 0,315
5) (1/3 + 0,4) × 3,5 + (1/6 + 0,75) × 6/5
y:0,15 = ( 2,8 + 0,38 ) : 0,75
y:0,15= 3,18 : 0,75
y:0,15= 4,24
y = 4,24 *0,15
y= 0,636
\(\dfrac{y}{9}\) = \(\dfrac{z}{5}\) = \(\dfrac{t}{2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{y}{9}\) = \(\dfrac{z}{5}\) = \(\dfrac{y-z}{9-5}\) = \(\dfrac{0,24}{4}\) = 0,06
\(y\) = 0,06 \(\times\) 9 = 0,54
\(z\) = 0,06 \(\times\) 5 = 0,3
\(t\) = 0,06 \(\times\) 2 = 0,12
Vậy \(y\) = 0,54; \(z\) = 0,3; \(t\) = 0,12
Ta có: \(\dfrac{y}{9}\) = \(\dfrac{z}{5}\) = \(\dfrac{t}{2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{y}{9}\) = \(\dfrac{z}{5}\) = \(\dfrac{y-z}{9-5}\) = \(\dfrac{0,24}{4}\) = 0,06
\(z\) = 0,06 \(\times\) 5 = 0,3
y = 0,06 \(\times\) 9 = 0,54
\(t\) = 0,06 \(\times\) 2 = 0,12
Vậy: \(y\) = 0,54; \(z\) = 0,3; \(t\) = 0,12
a) \(0,75:4,5=\dfrac{1}{5}:\left(2y\right)\)
\(\Rightarrow\dfrac{3}{4}.\dfrac{1}{4,5}=\dfrac{1}{5}:\left(2y\right)\)
\(\Rightarrow\dfrac{1}{6}=\dfrac{1}{5}:\left(2y\right)\)
\(\Rightarrow2y=\dfrac{1}{5}:\dfrac{1}{6}\)
\(\Rightarrow2y=\dfrac{1}{5}.\dfrac{6}{1}\)
\(\Rightarrow2y=\dfrac{6}{5}\)
\(\Rightarrow y=\dfrac{6}{5}:2\)
\(\Rightarrow y=\dfrac{6}{5}.\dfrac{1}{2}\)
\(\Rightarrow y=\dfrac{6}{10}\)
\(\Rightarrow y=\dfrac{3}{5}\)
Vậy \(y=\dfrac{3}{5}\)
b) \(\left|y-4\right|-12=2y\)
\(\Leftrightarrow\left|y-4\right|=2y+12\)
Xét trường hợp 1: \(y-4=2y+12\)
\(\Rightarrow y-4-\left(2y+12\right)=0\)
\(\Rightarrow y-4-2y-12=0\)
\(\Rightarrow-16-y=0\)
\(\Rightarrow y=-16\)
Xét trường hợp 2: \(y-4=-\left(2y+12\right)\)
\(\Rightarrow y-4-\left[-\left(2y+12\right)\right]=0\)
\(\Rightarrow y-4+2y+12=0\)
\(\Rightarrow8+3y=0\)
\(\Rightarrow3y=-8\)
\(\Rightarrow y=\dfrac{-8}{3}\)
Vậy \(y=-16\) hoặc \(y=\dfrac{-8}{3}\)
`y:0,24=0,75-1/5`
`y:0,24=0,75-0,2`
`y:0,24=0,55`
`y=0,55xx0,24`
`y=0,132`
y:0,24=0,75−1/5
y:0,24=0,75−0,2y:0,24=0,75-0,2
y:0,24=0,55y:0,24=0,55
y=0,55×0,24y=0,55×0,24
y=0,132