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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.
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\(A=15.\left(\dfrac{3}{5}-\dfrac{2}{3}\right)+1\\ A=15.\left(\dfrac{9}{15}-\dfrac{10}{15}\right)+1\\ A=15.\dfrac{-1}{15}+1\\ A=-1+1\\ A=0\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\\ C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{9}.\dfrac{9}{11}+\dfrac{12}{7}\\ C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\\ C=\dfrac{-5}{7}.1+\dfrac{12}{7}\\ C=\dfrac{-5}{7}+\dfrac{12}{7}\\ C=1\)
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
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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!!!
\(M=\dfrac{8}{3}\cdot\dfrac{2}{5}\cdot\dfrac{3}{8}\cdot10\cdot\dfrac{19}{92}\\ =\dfrac{8\cdot2\cdot3\cdot10\cdot19}{3\cdot5\cdot8\cdot92}\\ =\dfrac{8\cdot2\cdot3\cdot2\cdot5\cdot19}{3\cdot5\cdot8\cdot2\cdot2\cdot23}\\ =\dfrac{19}{23}\)
\(N=\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\\ =\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)\\ =-\dfrac{5}{11}\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot0\\ =0\)
a/ 7x - 3x = 3,2 ; b/ \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
x ( 7 - 3 ) = 3,2 ; x ( \(\dfrac{2}{3}-\dfrac{1}{2}\) ) = \(\dfrac{5}{12}\)
x. 4 = 3,2 ; x ( \(\dfrac{4}{6}-\dfrac{3}{6}\) ) = \(\dfrac{5}{12}\)
x = 3,2 : 4 ; x \(\dfrac{1}{6}=\dfrac{5}{12}\)
x = 0,8 ; x = \(\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{12}.6\)
x = \(\dfrac{5}{2}\)
c/\(2\dfrac{1}{4}.\left(x-7\dfrac{1}{3}\right)=1,5\)
\(\dfrac{9}{4}\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}\)
\(x-\dfrac{22}{3}=\dfrac{3}{2}:\dfrac{9}{4}=\dfrac{3}{2}.\dfrac{4}{9}\)
\(x-\dfrac{22}{3}=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}+\dfrac{22}{3}\)
\(x=\dfrac{24}{3}=8\)
d/\(\left(1-\dfrac{3}{10}-x\right):\left(\dfrac{19}{10}-1-\dfrac{2}{5}\right)+\dfrac{4}{5}=1\)
\(\left(\dfrac{10}{10}-\dfrac{3}{10}-x\right):\left(\dfrac{19}{10}-\dfrac{10}{10}-\dfrac{4}{10}\right)+\dfrac{4}{5}=1\)
\(\left(\dfrac{7}{10}-x\right):\dfrac{5}{10}+\dfrac{4}{5}=1\)
\(\left(\dfrac{7}{10}-x\right):\dfrac{1}{2}=1-\dfrac{4}{5}\)
\(\left(\dfrac{7}{10}-x\right).2=\dfrac{1}{5}\)
\(\dfrac{7}{10}-x=\dfrac{1}{5}:2=\dfrac{1}{5}.\dfrac{1}{2}=\dfrac{1}{10}\)
\(x=\dfrac{7}{10}-\dfrac{1}{10}\)
\(x=\dfrac{6}{10}=\dfrac{3}{5}\)
Chúc bạn học tốt!!!
a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
\(A=\dfrac{-10}{3}+\dfrac{19}{6}\cdot\dfrac{7}{5}-\dfrac{19}{3}\cdot\dfrac{1}{10}+\dfrac{19}{10}\cdot\dfrac{4}{3}\)
\(=\dfrac{-10}{3}+\dfrac{19}{3}\cdot\dfrac{7}{10}-\dfrac{19}{3}\cdot\dfrac{1}{10}+\dfrac{19}{3}\cdot\dfrac{4}{10}\)
\(=\dfrac{-10}{3}+\dfrac{19}{3}\cdot\left(\dfrac{7}{10}-\dfrac{1}{10}+\dfrac{4}{10}\right)\)
\(=\dfrac{-10}{3}+\dfrac{19}{3}\cdot\dfrac{10}{10}=\dfrac{-10}{3}+\dfrac{19}{3}\)
\(=\dfrac{9}{3}=3\)