(-60) + 22. {[(−35) + 52. 3]: (−8) − 15}
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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) \(\sqrt{4-\sqrt{15}}=\dfrac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)
= ( -60) + 4.{ [ ( - 35)+ 125 . 3] : (-8)-15}
= ( - 60 ) +4. { [(-35) +375] : (-8) -15}
= (-60) + 4. { 340 : ( - 8) - 15}
= ( - 60) + 4. { 42,5 - 15}
= ( -60 ) + 4 . 27,5
= ( -60) + 110
= 50
(-60)+22. { [ (-35)+53.3]: (-8)-15}
=(−60)+4.{[(−35)+25.3]:(−8)−15}
=(-60)+4.{[(-35)+75]:(-8)-15}
=(−60)+4.[40:(−8)−15]
=(−60)+4.[(−5)−15]
=(−60)+4.(−20)
=(−60)+(−80)
=(−140)
a: \(2^3-5^3:5^2+12\cdot2^2\)
\(=8-5+48\)
\(=51\)
b: \(5\cdot\left[\left(85-35:7\right):8+90\right]-5\)
\(=5\cdot\left[10+90\right]-5\)
=495
a: 23−53:52+12⋅2223−53:52+12⋅22
=8−5+48=8−5+48
=51=51
b: 5⋅[(85−35:7):8+90]−55⋅[(85−35:7):8+90]−5
=5⋅[10+90]−5=5⋅[10+90]−5
=495
a) -( -35 + 7 - 13 ) + ( 2 - 35 ) - ( 13 - 7 )
= 35 - 7 + 13 + 2 - 35 - 13 + 7
= ( 35 - 35 ) + ( -7 + 7 ) + ( 13 - 13 ) + 2
= 0 + 0 + 0 + 2
= 2
b) ( 70 + 8 - 35 ) - ( 8 - 37 - 10 ) - 70
= 70 + 8 - 35 - 8 + 37 + 10 - 70
= ( 70 - 70 ) + ( 8 - 8 ) + ( -35 + 37 ) + 10
= 0 + 0 + 2 + 10
= 12
c) 5 - ( 14 + 4 - 52 ) - ( 52 - 14 ) + 4
= 5 - 14 - 4 + 52 - 52 + 14 + 4
= 5 - ( 14 + 14 ) + ( 52 - 52 ) + ( -14 + 14 )
= 5 - 14 - 14 + 0 + 0
= 5 - ( 14 - 14 )
= 5 - 0
= 5
Làm luôn nhé
\(A=\frac{3}{8}+\frac{1}{5}+\frac{5}{6}>\frac{1}{6}+\frac{5}{6}=1\)
\(A=\frac{3}{8}+\frac{1}{5}+\frac{5}{6}< \frac{3}{8}+\frac{1}{4}+\frac{5}{4}=\frac{3}{8}+\frac{2}{8}+\frac{10}{8}=\frac{15}{8}< \frac{16}{8}=2\)
Vậy 1<A<2
\(B=\frac{5}{11}+\frac{5}{12}+\frac{5}{13}+\frac{5}{14}>\frac{5}{14}.4=\frac{10}{7}>1\)
\(B=\frac{5}{11}+\frac{5}{12}+\frac{5}{13}+\frac{5}{14}< \frac{5}{10}.4=2\)
Vậy 1<B<2