Ta có: \(\frac{3}{7}x\frac{21}{15}:\frac{3}{5}=\frac{3}{7}x\frac{21}{15}x\frac{5}{3}=\frac{3x21x5}{7x15x3}\)\(=\frac{3x3x7x5}{7x3x5x3}=1\)

Mọi người tk cho mình nha. Mình cảm ơn nhiều ^-^

3/7x3x7/3x5:3/5

=3/7x3x7/3x5;3/5

3/5:3/5=1

\(\frac{3}{7}x\frac{21}{15}:\frac{3}{5}\)

= \(\frac{3}{5}:\frac{3}{5}\)

= 1

                          Nếu các bn thấy đúng thì **** mk heg..........♥♥♥♥♥

                             Ai tk mk mk tk lại ~   ♥♥♥♥

                                   ! chúc ai tk mk thi tốt ~ ♥♥♥♥

a) 15/21 + 7/21 + 5/21

= (15/21 + 5/21) + 7/21

= 20/21 + 7/21

= 27/21

b) 4/7 + 8/15 + 22/15

= 4/7 + 2

= 18/7

nha bn

a,\(\frac{15}{21}+\frac{7}{21}+\frac{5}{21}\)

= \(\left(\frac{15}{21}+\frac{5}{21}\right)+\frac{7}{21}\)

=  \(\frac{20}{21}+\frac{7}{21}\)

=\(\frac{27}{21}\)

b, \(\frac{4}{7}+\frac{8}{15}+\frac{22}{15}\)

=\(\frac{4}{7}+\left(\frac{8}{15}+\frac{22}{15}\right)\)

= \(\frac{4}{7}+2\)

= \(\frac{18}{7}\)

6.42857142857

a,\(\frac{1}{5}\times\frac{2}{9}\div\frac{1}{15}\)

\(=\frac{1}{5}\times\frac{2}{9}\times15\)

\(=\left(\frac{1}{5}\times15\right)\times\frac{2}{9}\)

\(=3\times\frac{2}{9}\)

\(=\frac{2}{3}\)

b\(\frac{4}{15}\times\frac{7}{15}\times\frac{5}{4}\)

\(=\left(\frac{4}{15}\times\frac{5}{4}\right)\times\frac{7}{15}\)

\(=\frac{1}{3}\times\frac{7}{15}\)

\(=\frac{7}{45}\)

c,\(\frac{21}{23}\times\frac{5}{11}\times\frac{44}{?}\)

d,\(26\times\frac{13}{42}\div13\)

\(=\left(26\div13\right)\times\frac{13}{42}\)

\(=2\times\frac{13}{42}\)

\(=\frac{13}{21}\)

ờ ờ .......................... hong bt

21/23 x 5/11x 44/35

Bài 1: 

1) Ta có: \(\left(-12\right)+6\cdot\left(-3\right)\)

\(=-12-18\)

=-30

2) Ta có: \(\left(36-2020\right)+\left(2019-136\right)-27\)

\(=36-2020+2019-136-27\)

\(=1-100-27\)

\(=-126\)

3) Ta có: \(\left(144-97\right)-\left(244-197\right)\)

\(=144-97-244+197\)

\(=-100+100=0\)

4) Ta có: \(\left(-24\right)\cdot13-24\cdot\left(-3\right)\)

\(=-24\cdot13+24\cdot3\)

\(=24\cdot\left(-13+3\right)\)

\(=24\cdot\left(-10\right)=-240\)

5) Ta có: \(54+55+56+57+58-\left(64+65+66+67+68\right)\)

\(=54+55+56+57+58-64-65-66-67-68\)

\(=\left(54-64\right)+\left(55-65\right)+\left(56-66\right)+\left(57-67\right)+\left(58-68\right)\)

\(=\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)\)

=-50

6) Ta có: \(24\cdot\left(16-5\right)-16\cdot\left(24-5\right)\)

\(=24\cdot16-24\cdot5-16\cdot24+16\cdot5\)

\(=-24\cdot5+16\cdot5\)

\(=5\cdot\left(-24+16\right)\)

\(=-5\cdot8=-40\)

7) Ta có: \(47\cdot\left(23+50\right)-23\cdot\left(47+50\right)\)

\(=47\cdot23+47\cdot50-23\cdot47-23\cdot50\)

\(=47\cdot50-23\cdot50\)

\(=50\cdot\left(47-23\right)\)

\(=50\cdot24=1200\)

8) Ta có: \(\left(-31\right)\cdot47+\left(-31\right)\cdot52+\left(-31\right)\)

\(=-31\cdot\left(47+52+1\right)\)

\(=-31\cdot100=-3100\)

Bài 2: 

1) Ta có: \(-17-\left(2x-5\right)=-6\)

\(\Leftrightarrow-17-2x+5+6=0\)

\(\Leftrightarrow-2x-6=0\)

\(\Leftrightarrow-2x=6\)

hay x=-3

Vậy: x=-3

2) Ta có: \(10-2\left(4-3x\right)=-4\)

\(\Leftrightarrow10-8+6x+4=0\)

\(\Leftrightarrow6x+6=0\)

\(\Leftrightarrow6x=-6\)

hay x=-1

Vậy: x=-1

3) Ta có: \(-12+3\left(-x+7\right)=-18\)

\(\Leftrightarrow-12-3x+21+18=0\)

\(\Leftrightarrow-3x+27=0\)

\(\Leftrightarrow-3x=-27\)

hay x=9

Vậy: x=9

4) Ta có: \(-45:\left[5\cdot\left(-3-2x\right)\right]=3\)

\(\Leftrightarrow5\cdot\left(-3-2x\right)=-15\)

\(\Leftrightarrow-2x-3=-3\)

\(\Leftrightarrow-2x=0\)

hay x=0

Vậy: x=0

5) Ta có: x(x+3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-3\right\}\)

6) Ta có: (x-2)(x+4)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{2;-4\right\}\)

7) Ta có: \(x\left(x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=3\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-1;3\right\}\)

Bài 1: 

1) Ta có: (−12)+6⋅(−3)(−12)+6⋅(−3)

=−12−18=−12−18

=-30

2) Ta có: (36−2020)+(2019−136)−27(36−2020)+(2019−136)−27

=36−2020+2019−136−27=36−2020+2019−136−27

=1−100−27=1−100−27

=−126

Tớ chcs cậu học thật giỏi nha !

Lời giải:

a. $=-(37+63)+[25+(-25)]+(-9)=-100+0+(-9)=-(100+9)=-109$

b. $=[1+(-3)]+[5+(-7)]+....+[21+(-23)]$

$=\underbrace{(-2)+(-2)+....+(-2)}_{6}=(-2).6=-12$

c. $=-(280+20)+[-(79+21)]=-300+(-100)=-(300+100)=-400$

d. $=[-(27+43)]+[-(208+102)]=-70+(-310)=-(70+310)=-380$

e. $=(38+120)-(12+46)=158-58=100$

f. $=9+15+11+24=(9+11)+(15+24)=20+39=59$