\(\frac{31}{47}.\frac{14}{29}+\frac{31}{47}.\frac{15}{29}+\frac{16}{47}=\frac{31}{47}(\frac{14}{29}+\frac{15}{29})+\frac{16}{47}=\frac{31}{47}+\frac{16}{47}\)

=1

a) \(\dfrac{43}{47}=\dfrac{47-4}{47}=\dfrac{47}{47}-\dfrac{4}{47}=1-\dfrac{4}{47}\)

\(\dfrac{53}{57}=\dfrac{57-4}{57}=\dfrac{57}{57}-\dfrac{4}{57}=1-\dfrac{4}{57}\)

Vì \(\dfrac{4}{47}>\dfrac{4}{57}\)=>\(1-\dfrac{4}{47}< 1-\dfrac{4}{57}\)

=>\(\dfrac{43}{47}< \dfrac{53}{57}\)

b) Ta chọn phân số trung gian là: \(\dfrac{64}{81}\)

Vì \(\dfrac{64}{81}>\dfrac{64}{85}\)

Mà \(\dfrac{64}{81}< \dfrac{73}{81}\)

=> \(\dfrac{64}{85}< \dfrac{64}{81}< \dfrac{73}{81}\)

Vậy \(\dfrac{64}{85}< \dfrac{73}{81}\)

c) Ta chọn phân số trung gian là: \(\dfrac{18}{37}\)

Vì \(\dfrac{18}{37}>\dfrac{15}{37}\)

Mà \(\dfrac{18}{37}< \dfrac{18}{31}\)

=> \(\dfrac{15}{37}< \dfrac{18}{37}< \dfrac{18}{31}\)

Vậy \(\dfrac{18}{31}>\dfrac{15}{37}\)

Phần d) mình không biết lời giải

Mình chỉ biết \(\dfrac{12}{47}>\dfrac{19}{77}\) thôi !

b) Thay x=-28

\(\left(-2\right)+17+\left(-28\right)=-13\)

c) Thay x=-4

\(\left(-4\right)+25+\left(-47\right)=-26\)

d) Thay x=8

\(25+8+\left(-13\right)=20\)

a) \(27^{64}:81^{20}=3^{192}:3^{80}=3^{112}\)

b) \(\left(\dfrac{1}{8}\right)^{20}:\left(\dfrac{1}{16}\right)^9=\left(\dfrac{1}{2}\right)^{60}:\left(\dfrac{1}{2}\right)^{36}=\left(\dfrac{1}{2}\right)^{24}\)

c) \(\dfrac{1}{3}:\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{5}{3}-\dfrac{1}{6}=\dfrac{10}{6}-\dfrac{1}{6}=\dfrac{9}{6}=\dfrac{3}{2}\)

a: =25x100-150=2500-150=2350

c: \(=520:\left\{515\cdot25\right\}\)

=104/2575

a: \(\left(36-16\right)\cdot\left(-15\right)+16\cdot\left(-14-6\right)\)

\(=-15\cdot20+16\cdot\left(-20\right)=-20\cdot\left(-15+16\right)=-20\)

b: \(\left(-47\right)\cdot\left(-69\right)-\left(-31\right)\cdot47\)

\(=47\cdot69+31\cdot47\)

\(=47\cdot100=4700\)

c: \(38\cdot\left(-41\right)+59\cdot\left(-38\right)\)

\(=38\left(-41-59\right)\)

\(=-100\cdot38=-3800\)

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 !

-23 -47 = -70

-32 - (-23) = -32 +23= -9

-37+15 = -22

\(-23-47=-70\)

\(-31-\left(-23\right)=-8\)

\(-37+15=-22\)

sửa câu 2

\(-31-\left(-23\right)=-9\)

Hc tốt