Tính hợp lí nếu có thể:
(72.48+72.33) : 49
49 . 55 + 46 . 49 - 49
=49(55+46)-49
=49.101-49
==> =49.100
=4900
49.55+46.49-49
=49.55-46.49-49.1
=49.(55+46-1)
=49.100
=4900
Tính bằng cách hợp lí:
a, 125+70+375+230
b, 7 2 . 33 + 7 2 . 67
c, 35.11+65.18+35.13+65.6
d, 490 – {[(128+22):3. 2 2 ] – 7}
a, 125+70+375+230
= (125+375)+(70+230)
= 500+300 = 800
b, 7 2 . 33 + 7 2 . 67
= 49.33+49.67 = 49.(33+67) = 49.100 = 4900
c, 35.11+65.18+35.13+65.6
= 35.(11+13)+65.(18+6) = 35.24+65.24 = 24.(35+65) = 24.100 = 2400
d, 490 – {[(128+22):3. 2 2 ] – 7}
= 490 – {[150:3.4] – 7}
= 490 – {[50.4] – 7}
= 490 – {200 – 7}
= 490 – 193 = 297
Thực hiện phép tính( hợp lí nếu có thể)
a) 49.(51 - 4) - 51.(49 + 4)
b) 71.64 + 32.(-7) - 13.32
c) 11 + (-13) + 15 + (-17) + ... + 59 + (-61)
a)
\( 49.(51 - 4) - 51.(49 + 4)\)
\(=49.51-49.4-51.49-51.4\)
\(=49.(51-4-51)-51.4\)
\(=49.(-4)-51.4\)
\(=4.(-49-51)\)
\(=4.(-100)\)
\(=-400\)
b)
\( 71.64 + 32.(-7) - 13.32\)
\(=71.32.2+32.(-7)-13.32\)
\(=142.32+32.(-7)-13.32\)
\(=32.(142-7-13)\)
\(=32.122\)
\(=3904\)
c)
\( 11 + (-13) + 15 + (-17) + ... + 59 + (-61)\)
\(=(-2)+(-2)+...+(-2)\)
\(=(-2).13\)
\(=-26\)
Tính hợp lí nếu có thể :3,5 x 4/49 -[2,(4) x 2/5/11] : (-8,4)
Thực hiện phép tính( hợp lí nếu có thể)
a) 49.(51 - 4) - 51.(49 + 4)
b) 71.64 + 32.(-7) - 13.32
c) 11 + (-13) + 15 + (-17) + ... + 59 + (-61)
a. 49.51- 49.4 -51.49 -51.4
= -49.4 - 51.4
=4.(-49-51)
=4.-100
=-400
b.71.64 + 32.(-7) -13.32
= 71.64 + 32.(-7-13)
=71.64 + 32.-20
=71.32.2 +32.-20
=32.(71x2-20)
=32. 122
= 3904
c. (11-13)+(15-17)+..+(59-61)
=-2 x 13
=-26
tính hợp lí (nếu có thể)
a, A=(5^30.7^49.4^5-49^24.125^10.2^8):(5^29.16^2.7^48)
Tính hợp lí ( nếu có thể)
\(P=\dfrac{32}{19}.\dfrac{-57}{64}+\dfrac{35}{-44}.\dfrac{-22}{21}\)
\(Q=\dfrac{75}{164}.\dfrac{82}{125}-\dfrac{49}{-105}.\dfrac{-35}{98}\)
\(P=\dfrac{32}{64}\cdot\dfrac{-57}{19}+\dfrac{35}{21}\cdot\dfrac{22}{44}=\dfrac{1}{2}\left(-3+\dfrac{5}{3}\right)=\dfrac{1}{2}\cdot\dfrac{-4}{3}=\dfrac{-2}{3}\)
\(Q=\dfrac{75}{125}\cdot\dfrac{82}{164}+\dfrac{49}{98}\cdot\dfrac{-35}{105}=\dfrac{1}{2}\left(\dfrac{3}{5}-\dfrac{1}{3}\right)=\dfrac{1}{2}\cdot\dfrac{4}{15}=\dfrac{2}{15}\)