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Ta có:
Mẫu số chung 2 phân số: 84
\(\dfrac{3}{7}=\dfrac{3*12}{7*12}=\dfrac{36}{84}\)
\(\dfrac{5}{12}=\dfrac{5*7}{12*7}=\dfrac{35}{84}\)
Vì \(36>35\) nên\(\dfrac{36}{84}>\dfrac{35}{84}\)
Vậy \(\dfrac{3}{7}>\dfrac{5}{12}\)
Ta có:
\(\dfrac{9}{8}>1>\dfrac{2023}{2024}\) nên \(\dfrac{9}{8}>\dfrac{2023}{2024}\)
Ta có:
\(\dfrac{1+15}{16}=1\)
\(\dfrac{1+16}{15}=\dfrac{17}{15}>1\)
\(\Rightarrow\dfrac{1+15}{16}>\dfrac{1+16}{15}\)
a: =>7/9:x=1/18-2/9=1/18-4/18=-3/18=-1/6
=>x=-7/9:1/6=-7/9*6=-42/9=-14/3
b: =>x*7/5=2/15+2/5=8/15
=>x=8/15:7/5=8/21
c: =>x-1/2=3/14:4/7=3/8
=>x=3/8+4/8=7/8
d: =>0,4x+0,3x-0,2x=0,7
=>0,5x=0,7
=>x=1,4
a) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}\)
\(=\dfrac{5}{7}\times\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\)
\(=\dfrac{5}{7}\times1\)
\(=\dfrac{5}{7}\)
b) \(\dfrac{1}{10}+\dfrac{5}{9}+\dfrac{4}{9}+\dfrac{9}{10}-1\)
\(=\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{1}{10}+\dfrac{9}{10}-1\right)\)
\(=1+0\)
\(=1\)
c) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}+\dfrac{2}{7}\)
\(=\dfrac{5}{7}\times\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{2}{7}\)
\(=\dfrac{5}{7}+\dfrac{2}{7}\)
\(=1\)
d) \(\dfrac{2}{7}+\dfrac{2}{8}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{4}{7}\)
\(=\left(\dfrac{2}{8}+\dfrac{1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{4}{7}\right)\)
\(=\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+1\)
\(=\dfrac{1}{2}+1\)
\(=\dfrac{3}{2}\)
e) \(\dfrac{4}{5}+\dfrac{3}{10}+\dfrac{2}{10}+0,7\)
\(=\dfrac{4}{5}+\dfrac{5}{10}+\dfrac{7}{10}\)
\(=\dfrac{4}{5}+\dfrac{12}{10}\)
\(=\dfrac{4}{5}+\dfrac{6}{5}\)
\(=\dfrac{10}{5}\)
\(=2\)
g) \(362\times728+326\times272\)
\(=326\times\left(728+272\right)\)
\(=326\times1000\)
\(=326000\)
\(1\dfrac{1}{2}\times1\dfrac{1}{3}\times1\dfrac{1}{4}\times...\times1\dfrac{1}{2023}\times1\dfrac{1}{2024}\)
\(=\left(1+\dfrac{1}{2}\right)\times\left(1+\dfrac{1}{3}\right)\times\left(1+\dfrac{1}{4}\right)\times...\times\left(1+\dfrac{1}{2023}\right)\times\left(1+\dfrac{1}{2024}\right)\)
\(=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{6}{5}\times...\times\dfrac{2024}{2023}\times\dfrac{2025}{2024}\)
\(=\dfrac{3\times4\times5\times...\times2024\times2025}{2\times3\times4\times...\times2023\times2024}\)
\(=\dfrac{2025}{2}\)
\(=1012,5\)
1) (x - 35) - 120 = 0
x - 35 = 120
x = 120 + 35
x = 155
2) 310 - (118 - x) = 217
118 - x = 310 - 217
118 - x = 93
x = 118 - 93
x = 25
3) 156 - (x + 61) = 82
x + 61 = 156 - 82
x + 61 = 74
x = 74 - 61
x = 13
4) 814 - (x - 305) = 712
x - 305 = 814 - 712
x - 305 = 102
x = 102 + 305 = 407
5) 100 - 7 - (x - 5) = 58
x - 5 = 93 - 58
x - 5 = 35
x = 35 + 5 = 40
6) 12(x - 1) : 3 = 43 + 23
4(x - 1) = 72
x - 1 = 18
x = 18 + 1 = 19
7) 24 + 5x = 75 : 73
24 + 5x = 49
5x = 25
x = 25 : 5 = 5
8) 5(x - 1) : 3 = 43 + 23
\(\dfrac{5}{3}\left(x-1\right)=72\)
x - 1 = \(\dfrac{216}{5}\)
x = 221/5
9) 5(x - 4)2 - 7 = 13
5(x - 4)2 = 20
(x - 4)2 = 4
\(\Rightarrow\left[{}\begin{matrix}x-4=2\\x-4=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)
10) (x + 1) + (x + 2) + ... + (x + 30) = 795
=> (x + x + x + ... + x) + (1 + 2 + 3 +...+ 30) = 795 (1)
Đặt A = 1 + 2 + 3 +...+ 30
Số số hạng trong A là: (30 - 1) : 1 + 1 = 30 (số)
Tổng A bằng : (30 + 1).30 : 2 =465
Thay A = 465 vào (1) , ta được:
30x + 465 = 795
=> 30x =330
=> x =11
1: =>x-35=120
=>x=120+35=155
2: =>118-x=310-217=93
=>x=118-93=25
3: =>x+61=156-82=74
=>x=74-61=13
4: =>x-305=814-712=102
=>x=102+305=407
5: =>93-(x-5)=58
=>x-5=35
=>x=40
6: =>4(x-1)=64+8=72
=>x-1=18
=>x=19
7: =>5x+24=49
=>5x=25
=>x=5
8: =>5(x-1):3=4^3+2^3=64+8=72
=>5(x-1)=216
=>x-1=216/5
=>x=221/5
1/7 x 1/5 + 1/7 x 2/5 + 1/5 x 4/7
= 1/7 x 1/5 + 1/7 x 2/5 + 4/5 x 1/7
= 1/7 x (1/5 + 2/5 + 4/5)
= 1/7 x 7/5
= 1/5
\(x\times\dfrac{2}{7}+x\times\dfrac{5}{7}=2024\\ x\times\left(\dfrac{2}{7}+\dfrac{5}{7}\right)=2024\\ x\times1=2024\\ x=2024\)