300

300

= 300 

đây đâu phải bài lớp 2 bn

=55 nhé

50-51+52-53+54-55+56-57+58-59+60

=(-1)+(-1)+(-1)+(-1)+(-1)+60

=(-5)+60

=55

a)  >

b) > 

c) >   

d) < 

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 !

1+2+3-4-5-6+7+8+9+...+55+56+57-58-59-60

=(1+2+3-4-5-6)+(7+8+9-10-11-12)+.......+(55+56+57-58-59-60)

=(-9)+(-9)+......+(-9)

còn lại bạn làm theo cách tính tổng của dãy là đc

Ta có:

1+2+3-4-5-6+7+8+9-..........+55+56+57-58-59-60

=(1+2+3-4-5-6)+(7+8+9-10-11-12)..........+(55+56+57-58-59-60)

=  -3+ (-3)+...+(-3)

---12 số------------

=(-3).12

=-36

57.\(\sqrt{8-\sqrt{55}}=\sqrt{\dfrac{16-2.\sqrt{5}.\sqrt{11}}{2}}=\sqrt{\dfrac{\sqrt{11}^2-2.\sqrt{5}.\sqrt{11}+\left(\sqrt{5}\right)^2}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{11}-\sqrt{5}\right)^2}{2}}=\dfrac{\left|\sqrt{11}-\sqrt{5}\right|}{\sqrt{2}}=\dfrac{\sqrt{11}-\sqrt{5}}{\sqrt{2}}\)

58. \(\sqrt{7+\sqrt{33}}=\sqrt{\dfrac{14+2\sqrt{3}.\sqrt{11}}{2}}=\sqrt{\dfrac{\left(\sqrt{11}\right)^2+2\sqrt{3}.\sqrt{11}+\left(\sqrt{3}\right)^2}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{11}+\sqrt{3}\right)^2}{2}}=\dfrac{\left|\sqrt{11}+\sqrt{3}\right|}{\sqrt{2}}=\dfrac{\sqrt{11}+\sqrt{3}}{\sqrt{2}}\)

mấy câu dưới bạn cũng làm tương tự thôi

60) \(\sqrt{7-3\sqrt{5}}=\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{2}}=\dfrac{3-\sqrt{5}}{\sqrt{2}}=\dfrac{3\sqrt{2}-\sqrt{10}}{2}\)

61) \(\sqrt{23+3\sqrt{5}}=\dfrac{\sqrt{46+6\sqrt{5}}}{\sqrt{2}}=\dfrac{3\sqrt{5}+1}{\sqrt{2}}=\dfrac{3\sqrt{10}+\sqrt{2}}{2}\)

62) \(\sqrt{7-\sqrt{33}}=\dfrac{\sqrt{14-2\sqrt{33}}}{\sqrt{2}}=\dfrac{\sqrt{11}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{22}-\sqrt{6}}{2}\)

63) \(\sqrt{8+\sqrt{55}}=\dfrac{\sqrt{16+2\sqrt{55}}}{\sqrt{2}}=\dfrac{\sqrt{11}+\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{22}+\sqrt{10}}{2}\)