Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
Áp dụng tính chất phân phối, rồi tính giá trị biểu thức.
Chẳng hạn,
Với , thì
ĐS. ; C = 0.
Xem thêm tại: http://loigiaihay.com/bai-77-trang-39-phan-so-hoc-sgk-toan-6-tap-2-c41a5943.html#ixzz4eU1fQCGw
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
Bài 1. Tính giá trị của biểu thức:
a, \(A=\dfrac{3}{11}.\dfrac{-7}{19}+\dfrac{17}{11}.\dfrac{-3}{19}+\dfrac{3}{19}.\dfrac{25}{11}.\)
\(=\dfrac{3}{19}.\dfrac{-7}{11}+\dfrac{-17}{11}.\dfrac{3}{19}+\dfrac{3}{19}.\dfrac{24}{11}.\)
\(=\dfrac{3}{19}\left(\dfrac{-7}{11}+\dfrac{-17}{11}+\dfrac{24}{11}\right).\)
\(=\dfrac{3}{19}.0\)
\(=0.\)
Vậy A = 0.
b, \(B=\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}.....\dfrac{99^2}{99.100}.\)
\(=\dfrac{3.3.4.4.....99.99}{\left(3.4.5.....99\right)\left(4.5.6.....100\right)}.\)
\(=\dfrac{\left(3.4.5.....99\right)\left(3.4.5.....99\right)}{\left(3.4.5.....99\right)\left(4.5.6.....100\right)}.\)
\(=\dfrac{1.3}{1.100}.\)
\(=\dfrac{3}{100}.\)
Vậy \(B=\dfrac{3}{100}.\)
Bài 2. So sánh:
\(A=\dfrac{10^{2015}+7}{10^{2016}+7}\) và \(B=\dfrac{10^{2016}+7}{10^{2017}+7}.\)
Giải:
Ta có:
\(10A=\dfrac{\left(10^{2015}+7\right)10}{10^{2016}+7}.\)
\(=\dfrac{10^{2016}+70}{10^{2016}+7}.\)
\(=\dfrac{\left(10^{2016}+7\right)+63}{10^{2016}+7}.\)
\(=1+\dfrac{63}{10^{2016}+7}._{\left(1\right).}\)
\(10B=\dfrac{\left(10^{2016}+7\right)10}{10^{2017}+7}.\)
\(=\dfrac{10^{2017}+70}{10^{2017}+7}.\)
\(=\dfrac{\left(10^{2017}+7\right)+63}{10^{2017}+7}.\)
\(=1+\dfrac{63}{10^{2017}+7}._{\left(2\right).}\)
Mà \(\dfrac{63}{10^{2016}+7}>\dfrac{63}{10^{2017}+7}._{\left(3\right).}\)
Từ (1), (2) và (3) suy ra: \(10A>10B.\).
\(\Rightarrow A>B.\)
Vậy A > B.
CHÚC BN HỌC GIỎI!!! ^ - ^
Đừng quên bình luận nếu bài mik sai nhé!!! Và nếu bài mik đúng thì nhớ tick mik nha!!!
a, \(A=\dfrac{3}{11}.\dfrac{-7}{19}+\dfrac{17}{11}.\dfrac{-3}{19}+\dfrac{3}{19}.\dfrac{25}{11}.\)
\(=\dfrac{3}{19}.\dfrac{-7}{11}+\dfrac{-17}{11}.\dfrac{3}{19}+\dfrac{3}{19}.\dfrac{25}{11}.\)
\(=\dfrac{3}{19}\left(\dfrac{-7}{11}+\dfrac{-17}{11}+\dfrac{25}{11}\right).\)
\(=\dfrac{3}{19}.\dfrac{1}{11}.\)
\(=\dfrac{3}{209}.\)
Vậy \(A=\dfrac{3}{209}.\)
Do phần a có 1 chút nhầm lẫn của mik nên bài mik bị sai nhé, xin lỗi bn!!!
CHÚC BN HỌC GIỎI!!!
\(A=15.\left(\dfrac{3}{5}-\dfrac{2}{3}\right)+1\\ A=15.\left(\dfrac{9}{15}-\dfrac{10}{15}\right)+1\\ A=15.\dfrac{-1}{15}+1\\ A=-1+1\\ A=0\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\\ C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{9}.\dfrac{9}{11}+\dfrac{12}{7}\\ C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\\ C=\dfrac{-5}{7}.1+\dfrac{12}{7}\\ C=\dfrac{-5}{7}+\dfrac{12}{7}\\ C=1\)
a: \(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\)
\(=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{122}{8}\right)\)
\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=5\cdot\dfrac{3}{2}=\dfrac{15}{2}\)
b: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}\)
\(=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)
\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)
\(=\dfrac{35}{6}\cdot\dfrac{30}{-31}-\dfrac{11}{31}\)
\(=\dfrac{-35\cdot5-11}{31}=\dfrac{-186}{31}=-6\)
\(M=\dfrac{8}{3}\cdot\dfrac{2}{5}\cdot\dfrac{3}{8}\cdot10\cdot\dfrac{19}{92}\\ =\dfrac{8\cdot2\cdot3\cdot10\cdot19}{3\cdot5\cdot8\cdot92}\\ =\dfrac{8\cdot2\cdot3\cdot2\cdot5\cdot19}{3\cdot5\cdot8\cdot2\cdot2\cdot23}\\ =\dfrac{19}{23}\)
\(N=\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\\ =\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)\\ =-\dfrac{5}{11}\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot0\\ =0\)
\(A=\dfrac{-10}{3}+\dfrac{19}{6}\cdot\dfrac{7}{5}-\dfrac{19}{3}\cdot\dfrac{1}{10}+\dfrac{19}{10}\cdot\dfrac{4}{3}\)
\(=\dfrac{-10}{3}+\dfrac{19}{3}\cdot\dfrac{7}{10}-\dfrac{19}{3}\cdot\dfrac{1}{10}+\dfrac{19}{3}\cdot\dfrac{4}{10}\)
\(=\dfrac{-10}{3}+\dfrac{19}{3}\cdot\left(\dfrac{7}{10}-\dfrac{1}{10}+\dfrac{4}{10}\right)\)
\(=\dfrac{-10}{3}+\dfrac{19}{3}\cdot\dfrac{10}{10}=\dfrac{-10}{3}+\dfrac{19}{3}\)
\(=\dfrac{9}{3}=3\)