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Ta có : \(A=8\frac{2}{7}-\left(3\frac{4}{9}+4\frac{2}{7}\right)\)
\(\Rightarrow A=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\)
\(\Rightarrow A=\frac{58}{7}-\frac{487}{63}=\frac{5}{9}\)
P/s:Câu B tương tự nhé
Tiếp B của @Phạm Tuấn Đạt
\(B=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)
\(\Rightarrow B=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)
\(B=\left(\frac{92}{9}-\frac{56}{9}\right)+\frac{13}{5}\)
\(B=\frac{36}{9}+\frac{13}{5}\)
\(B=4+\frac{13}{5}\)
\(B=\frac{20}{5}+\frac{13}{5}=\frac{33}{5}\)
ính giá trị của các biểu thức sau:
A=827−(349+427)A=827−(349+427)
B=(1029+235)−629B=(1029+235)−629
Giải:
A=827−(349+427)A=827−(349+427)
=587−(319+307)=58−307−319=4−319=587−(319+307)=58−307−319=4−319
= 36−319=5936−319=59
B=(1029+235)−629B=(1029+235)−629
=1029−629+235=4+235=635
ính giá trị của các biểu thức sau:
A
=
8
2
7
−
(
3
4
9
+
4
2
7
)
A=827−(349+427)
B
=
(
10
2
9
+
2
3
5
)
−
6
2
9
B=(1029+235)−629
Giải:
A
=
8
2
7
−
(
3
4
9
+
4
2
7
)
A=827−(349+427)
=
58
7
−
(
31
9
+
30
7
)
=
58
−
30
7
−
31
9
=
4
−
31
9
=587−(319+307)=58−307−319=4−319
=
36
−
31
9
=
5
9
36−319=59
B
=
(
10
2
9
+
2
3
5
)
−
6
2
9
B=(1029+235)−629
=
10
2
9
−
6
2
9
+
2
3
5
=
4
+
2
3
5
=
6
3
5
Xem thêm tại: http://loigiaihay.com/bai-100-trang-47-sgk-toan-6-tap-2-c41a24737.html#ixzz4eUGN0ooE
=58/7-(40/9+30/7)
=58/7-40/9-30/7
=(58/7-30/7)-40/9
==...................
k nha bn
a)\(A=\frac{5.2^{13}.2^{22}-2^{36}}{\left(3.2^{17}\right)^2}\)
\(A=\frac{5.2^{35}-2^{36}}{3^2.2^{34}}\)
\(A=\frac{2^{35}\left(5-2\right)}{3^2.2^{34}}\)
\(A=\frac{2.3}{3^2}=\frac{2}{3}\)
b) \(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(B=\frac{2^{18}.3^9\left(2+5\right)}{2^{19}.3^9\left(2+3\right)}\)
\(B=\frac{7}{2.5}=\frac{7}{10}\)
\(B=\frac{7.6^{10}.2^{20}.3^6-2^{19}.6^{15}}{9.6^{19}.2^9-4.3^{17}.2^{26}}\)
\(B=\frac{7.\left(2.3\right)^{10}.2^{20}.3^6-2^{19}.\left(2.3\right)^{15}}{3^2.\left(2.3\right)^{19}.2^9-2^2.3^{17}.2^{26}}\)
\(B=\frac{7.2^{30}.3^{16}-2^{34}.3^{15}}{3^{21}.2^{11}-2^{28}.3^{17}}\)
\(B=\frac{2^{30}.3^{15}.\left(7.3-2^4\right)}{3^{17}.2^{11}.\left(3^4-2^{17}\right)}\)
\(B=\frac{2^{19}.\left(21-2^4\right)}{3^2.\left(3^4-2^{17}\right)}\)
@_@ tự tính!
\(B=\frac{7.\left(2.3\right)^{10}.2^{20}.3^6-2^{19}.\left(2.3\right)^{15}}{3^2.\left(2.3\right)^{19}.2^9-4.3^{17}.2^{26}}\\ B=\frac{7.2^{10}.3^{10}.2^{20}.3^6-2^{19}.2^{15}.3^{15}}{3^2.2^{19}.3^{19}.2^9-2^2.3^{17}.2^{26}}\\ B=\frac{7.2^{30}.3^{16}-2^{34}.3^{15}}{3^{21}.2^{28}-2^{28}.3^{17}}\)
\(B=\frac{7.2^{30}.3.3^{15}-2^{34}.3^{15}}{3^{15}.3^6.2^{28}-2^{28}.3^{15}.3^2}\\ B=\frac{3^{15}.\left(7.2^{30}.3-3^{34}\right)}{3^{15}.\cdot\left(3^6-3^2\right)}\\ B=\frac{7.2^{30}.3-3^{34}}{3^6-3^2}\)
a) \(2^3.3^2=8.9=72\)
b) \(5^{10}:5^7=5^2=25\)
c) \(2^6:2=2^5=32\)
d) \(7^4:7^4=7^0=1\)
e) \(9^5:9^5=9^0=1\)
a) 23.32=8.9=7223.32=8.9=72
b) 510:57=52=25510:57=52=25
c) 5
32
d) 74:74=70=11
e) 95:95=90=11
\(a,a^3\cdot a^9=a^{12}\)
\(b,\left(a^5\right)^7=a^{35}\)
\(c,\left(a^6\right)^4\cdot a^{12}=a^{24}\cdot a^{12}=a^{36}\)
\(d,4\cdot5^2-2\cdot3^2=2^2\cdot5-2\cdot3^2=2\cdot\left(2\cdot5+3^2\right)=2\cdot19=38\)
\(e,5^6:5^3+3^3\cdot3^2=5^3+3^5\)
a,a3.a9=a3+9=a12
b,(a5)7=a5.7=a35
Mấy câu tiếp theo bn lám tương tự!
\(A=\dfrac{6}{7}+\dfrac{1}{7}.\dfrac{2}{7}+\dfrac{1}{7}.\dfrac{5}{7}.\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}\left(\dfrac{2}{7}+\dfrac{5}{7}\right).\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}.1.\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}=1.\)
Vậy \(A=1.\)
\(B=\dfrac{40}{9}.\dfrac{13}{3}-\dfrac{4}{3}.\dfrac{40}{9}.\)
\(B=\dfrac{4}{9}.\dfrac{13}{3}-\dfrac{4}{9}.\dfrac{40}{3}.\)
\(B=\dfrac{4}{9}\left(\dfrac{13}{3}-\dfrac{40}{3}\right).\)
\(B=\dfrac{4}{9}.\left(-9\right).\)
\(B=-4.\)
Vậy \(B=-4.\)
a)
\(\begin{array}{l}\frac{2}{3} + \frac{{ - 2}}{5} + \frac{{ - 5}}{6} - \frac{{13}}{{10}}\\ = \frac{2}{3} + \frac{{ - 5}}{6} + \frac{{ - 2}}{5} - \frac{{13}}{{10}}\\ = \left( {\frac{2}{3} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 2}}{5} - \frac{{13}}{{10}}} \right)\\ = \left( {\frac{4}{6} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 4}}{{10}} - \frac{{13}}{{10}}} \right)\\ = \frac{{ - 1}}{6} + \frac{{ - 17}}{{10}}\\ = \frac{{ - 5}}{{30}} + \frac{{ - 51}}{{30}}\\ = \frac{{ - 56}}{{30}}\\ = \frac{{ - 28}}{{15}}\end{array}\)
b)
\(\begin{array}{l}\frac{{ - 3}}{7}.\frac{{ - 1}}{9} + \frac{7}{{ - 18}}.\frac{{ - 3}}{7} + \frac{5}{6}.\frac{{ - 3}}{7}\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 1}}{9} + \frac{7}{{ - 18}} + \frac{5}{6}} \right)\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 2}}{{18}} + \frac{{ - 7}}{{18}} + \frac{{15}}{{18}}} \right)\\ = \frac{{ - 3}}{7}.\frac{{ 6}}{{18}}\\ = \frac{-1}{7}\end{array}\).
Bài 35 :
\(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
\(A=\frac{2^{10}.\left(13+65\right)}{2^8.104}\)
\(A=\frac{2^8.2^2.98}{2^8.104}\)
\(A=\frac{2^8.4.98}{2^8.4.26}\)
\(A=\frac{49}{13}\)
Vậy \(A=\frac{49}{13}\)
\(B=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(B=\frac{11.3^{29}-9^{15}}{2^2.\left(3^{14}\right)^2}\)
\(B=\frac{11.3^{29}-9^{15}}{2^2.3^{28}}\)
\(B=\frac{11.3^{29}-\left(3^2\right)^{15}}{4.3^{28}}\)
\(B=\frac{11.3^{29}-3^{30}}{4.3^{28}}\)
\(B=\frac{11.3^{29}-3^{29}.3}{4.3^{28}}\)
\(B=\frac{3^{29}.\left(11-3\right)}{4.3^{28}}\)
\(B=\frac{3^{29}.8}{4.3^{28}}\)
\(B=\frac{3^{28}.3.4.2}{4.3^{28}}\)
\(B=3.2\)
\(B=6\)
Vậy B = 6
A = 2^10 . 13 + 2^10 . 65 / 2^8 . 104
= 2^10 ( 13 + 65 ) / 2^8 . 104 = 2^10 . 78 / 2^8 . 104 = 2^8 . 2^2 . 78 / 2^8 . 104 = 2^8 . 4 . 78 / 2^8 . 104 = 2^8 . 312 / 2^8 . 104
= 312/104
= 3
B = 11 . 3^22 . 3^7 - 9^15 / ( 2.3^14)^2
= 11 . 3^29 - (3^2)^15 / ( 3.2^14)^2
= 11 . 3^29 - 3^30 / ( 3. 2 )^28
= ( 8 + 3 ) . 3^29 - 3^30 / ( 3. 2)^28
= 8 . 3^29 + 3.3^29 - 3^30 / ( 3.2)^28
= 8 . 3^29 + 3^30 - 3^30 / ( 3 . 2)^28
= 8 . 3^29 / 3^28 . 2^28
= 2^3 . 3 / 2^28
= 3/ 2^25