a: \(\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)

=>\(\frac{x+1}{65}+1+\frac{x+2}{64}+1=\frac{x+3}{63}+1+\frac{x+4}{62}+1\)

=>\(\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}\)

=>\(\left(x+66\right)\left(\frac{1}{62}+\frac{1}{63}-\frac{1}{64}-\frac{1}{65}\right)=0\)

=>x+66=0

=>x=-66

b: \(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-72}{17}\)

=>\(\frac{x-12}{77}-1+\frac{x-11}{78}-1=\frac{x-74}{15}-1+\frac{x-72}{17}-1\)

=>\(\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{17}\)

=>\(\left(x-89\right)\left(\frac{1}{15}+\frac{1}{17}-\frac{1}{77}-\frac{1}{78}\right)=0\)

=>x-89=0

=>x=89

a,\(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)

\(\Rightarrow\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}\)

\(\Rightarrow\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}=\dfrac{x+3+63}{63}+\dfrac{x+4+62}{62}\)

\(\Rightarrow\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)

\(\Rightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)

\(\Rightarrow x+66=0\) ( vì \(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}>0\) )

\(\Rightarrow x=-66\)

\(b,\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)

\(\Rightarrow\dfrac{x-12}{77}-1+\dfrac{x-11}{78}-1=\dfrac{x-74}{15}-1+\dfrac{x-73}{16}-1\)

\(\Rightarrow\dfrac{x-12-77}{77}+\dfrac{x-11-78}{78}=\dfrac{x-74-15}{15}+\dfrac{x-73-16}{16}\)

\(\Rightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\)

\(\Rightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)

\(\Rightarrow x-89=0\)

\(\Rightarrow x=89\)

b.

\(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\\ \Rightarrow\left(\dfrac{x-12}{77}-1\right)+\left(\dfrac{x-11}{78}-1\right)=\left(\dfrac{x-74}{15}-1\right)+\left(\dfrac{x-73}{16}-1\right)\\ \Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}=\dfrac{x-89}{15}+\dfrac{x-89}{16}\\ \Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\\ \\ \Leftrightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\\ \Leftrightarrow x-89=0\\ \Leftrightarrow x=89\)

Ý bạn là mỗi phép tính phải thêm dấu và ....để kết quả bằng 6 phải không ạ

(1+1+1)×2=6

2+2+2=6

3×3⁰+3=6

(4+4+4)÷2=6

(5⁰+5⁰+5⁰)×2=6

6×6⁰×6⁰=6

7-7⁰×7⁰=6

8-(8⁰+8⁰)=6

(9⁰+9⁰+9⁰)×2=6

(10⁰+10⁰+10⁰)×2=6

c: \(78\cdot31+78\cdot24+78\cdot17+72\cdot22\)

\(=78\cdot\left(31+24+17\right)+72\cdot22\)

\(=72\cdot78+72\cdot22=72\cdot100=7200\)

d: \(\left(456\cdot11+912\right)\cdot37:13:74\)

\(=\frac{456\left(11+2\right)}{13}\cdot\frac{37}{74}=456\cdot\frac12=\frac{456}{2}=228\)

e: [\(\left(315+372\right)\cdot3+\left(372+315\right)\cdot7\) ]:\(\left(26\cdot13+74\cdot14\right)\)

\(=687\cdot\left(3+7\right):\left\lbrack26\cdot13+74\cdot13+74\right\rbrack\)

=687*10:[1300+74]

=687*10:1374

=5

a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

=0

Bài 3:

a: Ta có: \(23\left(42-x\right)=23\)

\(\Leftrightarrow42-x=1\)

hay x=41

b: Ta có: 15(x-3)=30

nên x-3=2

hay x=5

Bài 1: 

a: 32+89+68=100+89=189

b: 64+112+236=300+112=412

c: \(1350+360+650+40=2000+400=2400\)

Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{9}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot...\cdot\dfrac{8}{9}\)

\(=\dfrac{1}{9}\)