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a, (X-35)-120=0
X-35 =120+0
X-35 =120
X =120+35
X =155
b,124+(118-X)=217
118-X =217-124
118-X=93
X=118-93
X=25
c,156-(X+61)=82
X+61=156-82
X+61=74
X =74-61
X =13
k cho mình đi rồi mình k lại cho
a,(x-35)-120=0
x-35=120
x=120+35
x=155
b,124+(118-x)
118-x=217-124
118-x=93
x=118-93
x=25
c,156-(x+61)=82
x+61=156-82
x+61=74
x=74-61
x=13
a) \(\frac{1}{2}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{4}+\frac{1}{4}\times\frac{1}{5}+\frac{1}{5}\times\frac{1}{6}=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{2}-\frac{1}{6}=\frac{3}{6}-\frac{1}{6}=\frac{2}{6}=\frac{1}{3}\)
b) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}=\frac{1}{4}\)
\(a)\frac{1}{2}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{4}+\frac{1}{4}\times\frac{1}{5}+\frac{1}{5}\times\frac{1}{6}\)
\(\Rightarrow\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{6}\)
\(\Rightarrow\frac{1}{3}\)
\(b)\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\)
\(=\frac{1\times2\times3}{2\times3\times4}\)
\(=\frac{1}{4}\)
\(b)\left(1-\frac{1}{2}\right)\div\left(1-\frac{1}{3}\right)\div\left(1-\frac{1}{4}\right)\)
\(=\frac{1}{2}\div\frac{2}{3}\div\frac{3}{4}\)
\(=\frac{1}{2}\times\frac{3}{2}\times\frac{4}{3}\)
\(=\frac{1\times3\times2\times2}{2\times2\times3}\)
\(=1\)
a) X : ( 1800 + 600 ) : 30 = 560 : ( 315 - 35 )
X:2400:30=560:280
X:80=2
X=2x80
X=160
b) 134 - 2 x 156 - 6 x 54 - 2 x ( 9 + 6 )x X = 86
134-312-324-2x15xX=86
-502-30xX=86
-532xX=86
X=86:(-532)
X=-43/266
Chỉ làm bài khó thôi nhé:::::::::::::::
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2016}{2018}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x.\left(x+1\right)}=\frac{2016}{2018}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2016}{2018}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{1013}{2018}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1013}{2018}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1013}{2018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2018}\Rightarrow x+1=2018\Rightarrow x=2017\)
1> a) \(\frac{5}{7}x4:\frac{5}{9}=\frac{5}{7}:\frac{5}{9}x4=\frac{5}{7}x\frac{9}{5}x4=\frac{9}{7}x4=\frac{9x4}{7}=\frac{36}{7}\)
\(b,8x\frac{2}{3}:\frac{1}{2}=8x\frac{2}{3}x\frac{2}{1}=8x2x\frac{2}{3}=16x\frac{2}{3}=\frac{32}{3}\)
\(c,6:\frac{3}{5}-\frac{7}{6}x\frac{6}{7}=6x\frac{5}{3}-1=10-1=9\)
\(\frac{21}{5}x\frac{10}{11}+\frac{57}{11}=\frac{42}{11}+\frac{57}{11}=\frac{99}{11}=9\)
2) a) \(\frac{35}{9}:x=\frac{35}{6}\)
\(x=\frac{35}{9}:\frac{35}{6}\)
\(x=\frac{35}{9}x\frac{6}{35}\)
\(x=\frac{2}{3}\)
b) \(\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}\right)x10-X=0\)
\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{5}-\frac{1}{6}\right)x10-X=0\)
\(\left(\frac{1}{1}-\frac{1}{6}\right)x10-X=10\)
\(\frac{5}{6}x10-X=0\)
\(X=\frac{5}{6}x10=\frac{25}{3}\)
Đúng nha !!!!
1/a/\(\frac{5}{7}\cdot4:\frac{5}{9}=\frac{20}{7}:\frac{5}{9}=\frac{20}{7}\cdot\frac{9}{5}=\frac{36}{7}\)
b/\(8\cdot\frac{2}{3}:\frac{1}{2}=\frac{16}{3}:\frac{1}{2}=\frac{16}{3}\cdot\frac{2}{1}=\frac{32}{3}\)
c/\(6:\frac{3}{5}-\frac{7}{6}\cdot\frac{6}{7}=6\cdot\frac{5}{3}-1=10-1=9\)
2/a/\(\frac{35}{9}:x=\frac{35}{6}\)
\(x=\frac{35}{9}:\frac{35}{6}=\frac{35}{9}\cdot\frac{6}{35}\)
\(x=\frac{2}{3}\)
b/\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\right)\cdot10-x=0\)
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\cdot10-x=0\)
\(\left(\frac{30}{60}+\frac{10}{60}+\frac{5}{60}+\frac{2}{30}\right)\cdot10-x=0\)
\(\frac{47}{60}\cdot10-x=0\)
\(\frac{47}{6}-x=0\)
\(x=\frac{47}{6}-0\)
\(x=\frac{47}{6}\)
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+10\right)=65\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(1+2+3+...+10\right)=65\) (10 số hạng x)
\(\Leftrightarrow10x+55=65\Leftrightarrow10x=10\Leftrightarrow x=1\)
ko cần CTV, đấng Ed nhờ ta tới giúp con :))
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+10\right)=65\)
\(\Leftrightarrow\)\(\left(x+x+x+...+x\right)+\left(1+2+3+...+10\right)=65\)
\(\Leftrightarrow\)\(10x+\frac{10\left(10+1\right)}{2}=65\)
\(\Leftrightarrow\)\(10x+55=65\)
\(\Leftrightarrow\)\(10x=10\)
\(\Leftrightarrow\)\(x=1\)
Vậy \(x=1\)
ta phắn đey..
a, 134: [x-3] = 35 + 160: 5
134 : [x-3] = 35 + 32
134 : [x-3] = 67
x - 3 = 134: 67
x-3 = 2
x = 2 + 3
x = 5
Bài b bạn ghi lại đề nha
a,134:(x-3)=35+160:5
=> 134:(x-3)=35+32
=> 134:(x-3)= 67
=> x - 3 = 134 : 67
=> x-3 = 2
=> x = 2 + 3 = 5
b,[(10-x).2]:3-2=3
=> [(10-x).2]:3 = 3+2
=> [(10-x).2]:3 = 5
=> (10 -x) . 2 = 5.3
=> (10 - x).2 = 15
=> 10-x = 15: 2
=> 10 - x =7,5
=> x = 10 - 7,5 = 2,5