50. √(98-16√3)
51. √(2-√3)
52. √(4+√15)
53. √(5-√21)
54. √(6-√35)
55. √(2+√3)
56. √(4-√15)
57. √(8-√55)
58. √(7+√33)
59. √(6+√35)
60. √(7-3√5)
K
Khách
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Những câu hỏi liên quan
3 tháng 2 2021
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: \(x\in\left\{0;-1;3\right\}\)
DT
25 tháng 1 2022
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 !
22 tháng 12 2016
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
MC
1
13 tháng 6 2015
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
AT
1 tháng 7 2021
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
1 tháng 7 2021
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) \(\sqrt{8+\sqrt{55}}=\dfrac{\sqrt{16+2\sqrt{55}}}{\sqrt{2}}=\dfrac{\sqrt{11}+\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{22}+\sqrt{10}}{2}\)
50) \(\sqrt{98-16\sqrt{3}}=4\sqrt{6}-\sqrt{2}\)
51) \(\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{\sqrt{6}-\sqrt{2}}{2}\)
52) \(\sqrt{4+\sqrt{15}}=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)
53) \(\sqrt{5-\sqrt{21}}=\dfrac{\sqrt{10-2\sqrt{21}}}{\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{6}}{2}\)
54) \(\sqrt{6-\sqrt{35}}=\dfrac{\sqrt{12-2\sqrt{35}}}{\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{10}}{2}\)
55) \(\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{6}+\sqrt{2}}{2}\)
56) \(\sqrt{4-\sqrt{15}}=\dfrac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)
Can bac 8