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\(\text{a)}25\%-\frac{3}{2}+0,5.\frac{12}{5}\)
\(=\frac{1}{4}-\frac{3}{2}+\frac{1}{2}.\frac{12}{5}\)
\(=-\frac{5}{8}+\frac{6}{5}\)
\(=-\frac{25}{40}+\frac{48}{40}\)
\(=\frac{23}{40}\)
\(b)0,25.\frac{7}{3}.30.\text{ }0,5.\frac{8}{45}\)
\(=\frac{1}{4}.\frac{7}{3}.30.\frac{1}{2}.\frac{8}{45}\)
\(=\frac{7}{12}.(30.\frac{8}{90})\)
\(=\frac{7}{12}.\frac{8}{3}\)
\(=\frac{14}{9}\)
\(c)\frac{9}{23}.\frac{5}{8}+\frac{9}{23}.\frac{3}{8}-\frac{9}{23}\)
\(=\frac{9}{23}.\left(\frac{5}{8}+\frac{3}{8}-1\right)\)
\(=\frac{9}{23}.0\)
\(=0\)
A = \(\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\) . \(\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . \(\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . \(\left(\dfrac{4}{12}-\dfrac{3}{12}-\dfrac{1}{12}\right)\)
A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . 0
A = 0*
*Vì số nào nhân với 0 cũng bằng 0 nên không cần tính kết quả của phép tính\(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\)
a: \(24,8:0,5+24,8\cdot3+24,8:0,2\)
\(=24,8\cdot2+24,8\cdot3+24,8\cdot5\)
\(=24,8\left(2+3+5\right)\)
\(=24,8\cdot10=248\)
b: \(19,5+19,5\cdot3+19,5:0,5+19,5:0,25\)
\(=19,5+19,5\cdot3+19,5\cdot2+19,5\cdot4\)
\(=19,5\left(1+3+2+4\right)\)
\(=19,5\cdot10=195\)
c: \(121,07\cdot4+121,07:0,25+121,07:0,5\)
\(=121,07\cdot4+121,07\cdot4+121,07\cdot2\)
\(=121,07\left(4+4+2\right)\)
\(=121,07\cdot10=1210,7\)
1: 0,35*12,4=0,35*2,4+0,35*10=3,5+0,84=4,34
2: =0,1-2,34=-2,24
3: =5-2,9=2,1
4: \(=2,5\left(10,124-0,124\right)=10\cdot2,5=25\)
5: =-3/7+1/13
=-39/91+7/91
=-32/91
6: =-1/3+1/3=0
a: \(\Leftrightarrow2x+\dfrac{7}{2}=\dfrac{16}{3}:\dfrac{11}{3}=\dfrac{16}{11}\)
=>2x=-45/22
hay x=-45/44
b: =>x/7=-1/28:1/4=-1/7
=>x=-1
a)(7/2+2x).11/3=16/3
7/2+2x=16/3:11/3
7/2+2x=16/3.3/11
7/2+2x=16/11
2x=16/11-7/2
2x= -45/22
x= -45/22:2
x= -45/44
Vậy x= -45/44
b)x/7 +1/4= -1/28
x/7= -1/28-1/4
x/7= -2/7
=>x= -2
1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
2\(x\)( 0,5 + \(\dfrac{2}{3}\)) : \(\dfrac{7}{3}\) = 0,25
2\(x\)( \(\dfrac{1}{2}\) + \(\dfrac{2}{3}\)) = 0,25 \(\times\) \(\dfrac{7}{3}\)
2\(x\) \(\times\) \(\dfrac{7}{6}\) = \(\dfrac{7}{12}\)
2\(x\) = \(\dfrac{7}{12}\) : \(\dfrac{7}{6}\)
2\(x\) = \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{1}{2}\):2
\(x\) = \(\dfrac{1}{4}\)