1: 0,35*12,4=0,35*2,4+0,35*10=3,5+0,84=4,34

2: =0,1-2,34=-2,24

3: =5-2,9=2,1

4: \(=2,5\left(10,124-0,124\right)=10\cdot2,5=25\)

5: =-3/7+1/13

=-39/91+7/91

=-32/91

6: =-1/3+1/3=0

1−2−3+4+5−6−7+8+...+21−22−23+24+25

= (1 - 2 - 3 + 4) + (5 - 6 - 7 + 8) + ... + (21 - 22 - 23 + 24) + 25=(1−2−3+4)+(5−6−7+8)+...+(21−22−23+24)+25

= 0 + 0 + ... + 0 + 25=0+0+...+0+25

= 25

1) \(35.18-5.7.28\)

\(=35.18-35.28\)

\(=35.\left(18-28\right)\)

\(=35.\left(-10\right)\)

\(=-350\)

2) \(45-5.\left(12+9\right)\)

\(=45-60-45\)

\(=\left(45-45\right)-60\)

\(=0-60\)

\(=-60\)

3) \(24.\left(16-5\right)-16.\left(24-5\right)\)

\(=\left(384-120\right)-\left(384-120\right)\)

\(=264-264\)

\(=0\)

4) \(13.\left(23+22\right)-3.\left(17+28\right)\)

\(=13.45-3.45\)

\(=\left(13-3\right).45\)

\(=10.45\)

\(=450\)

tk ủng hộ nha!!!!!!!!1

1/ 35 . 18 - 5 . 7 . 28

= 35 . 18 - 35 . 28

= 35 . (18 - 28)

= 35 . (-10)

= -350

2/ 45 - 5 . (12 + 9)

= 5 . 9 - 5 . 12 + 5 . 9

= 5 . (9 - 12 + 9)

= 5 . 6

= 30

3/ 24 . (16 - 5) - 16 . (24 - 5)

= 24 . 16 - 24 . 5 - 16 . 24 - 16 . 5

= (24 . 16 - 16 . 24) - (24 . 5 - 16 . 5)

= 0 - [5 . (24 - 16)]

= 0 - [5 . 8]

= 0 - 40

= -40

4/ 13 . (23 + 22) - 3 . (17 + 28)

= 13 . 45 - 3 . 45

= 45 . (13 - 3)

= 45 . 10

= 450

Xin lỗi, cho mk sửa lại ý 2, các ý còn lại đúng rùi đó!

2/ 45 - 5 . (12 + 9)

= 5 . 9 - 5 . 12 - 5 . 9

= -(5 . 12)

= -60

\(=\frac{2}{5}\)

a)=3/4+6/13+7/13+8/45+0,25

=(6/13+7/13)+(3/4+0,25)+8/45

=1+1+8/45

=98/45

a)=3/4+6/13+7/13+8/45+1/4

=(3/4+1/4)+(6/13+7/13)+8/45

=1+1+8/45=2+8/45=90/45+8/45=98/45

1. TÍNH 

`5/7 xx 4 : 5/9 = 5/7 xx 4 xx 9/5 = 20/7 xx 9/5 = 36/7`

`4/9 : 2 xx 5/7 = 4/9 xx 1/2 xx 5/7 = 2/9 xx 10/63 `

`8 xx 2/3 : 1/2= 8xx 2/3 xx 2/1 = 8 xx 2/3 xx 2 = 16/3 xx 2=32/3`

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 !

a: \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{3}+1+\sqrt{3}-1\right)=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

b: \(\dfrac{1}{\sqrt{7-\sqrt{24}}+1}-\dfrac{1}{\sqrt{7+\sqrt{24}}+1}\)

\(=\dfrac{1}{\sqrt{6}-1+1}-\dfrac{1}{\sqrt{6}+1+1}\)

\(=\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{6}+2}=\dfrac{\sqrt{6}+2-\sqrt{6}}{\sqrt{6}\left(\sqrt{6}+2\right)}\)

\(=\dfrac{2}{\sqrt{6}\left(\sqrt{6}+2\right)}=\dfrac{2}{6+2\sqrt{6}}=\dfrac{1}{3+\sqrt{6}}=\dfrac{3-\sqrt{6}}{3}\)