tính giá trị biểu thức:
a) A = a.\(\frac{1}{3}\) + a.\(\frac{1}{4}\) - a.\(\frac{1}{6}\) vs a =\(-\frac{3}{5}\)
b) B = b.\(\frac{5}{6}\) + b.\(\frac{3}{4}\) - b.\(\frac{1}{2}\) vs b = \(\frac{12}{13}\)
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PT
15 tháng 8 2018
\(1)A=a\frac{1}{3}+a\frac{1}{4}-a\frac{1}{6}=a\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)=a\frac{5}{12}\)
Thay \(a=-\frac{3}{5}\) vào A,ta đc:
\(A=-\frac{3}{5}.\frac{5}{12}=-\frac{1}{4}\)
\(2)B=b\frac{5}{6}+b\frac{3}{4}-b\frac{1}{2}=b\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)=b\frac{13}{12}\)
Thay \(b=\frac{12}{13}\) vào B, ta đc: \(B=b\frac{13}{12}=\frac{12}{13}.\frac{13}{12}=1\)
18 tháng 4 2021
A = -4/5x(1/2+1/3+1/4)= -4/5x1 = -4/5
B = 6/19 x ( 3/4+4/3+-1/2)= 6/19x 19 = 6
C = 2002/2003x(3/4+5/6-19/12)=2003/2002x0=0
25 tháng 1 2017
Bài 1:
\(B=\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
\(=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-\left(0,625-0,5+\frac{5}{11}+\frac{5}{12}\right)}+\frac{3\left(0,5+\frac{1}{3}-0,25\right)}{5\left(0,5+\frac{1}{3}-0,25\right)}\)
\(=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-\left[5\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)\right]}+\frac{3}{5}\)
\(=\frac{-3}{5}+\frac{3}{5}\)
\(=0\)
Bài 2:
b) Giải:
Ta có: \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^6}{b^6}=\frac{c^6}{d^6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a^6}{b^6}=\frac{c^6}{d^6}=\frac{3a^6}{3b^6}=\frac{c^6}{d^6}=\frac{3a^6+c^6}{3b^6+d^6}\) (1)
\(\frac{a}{b}=\frac{c}{d}=\frac{a+b}{b+d}\)
\(\Rightarrow\left(\frac{a}{b}\right)^6=\left(\frac{a+c}{b+d}\right)^6=\frac{a^6}{b^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{3a^6+c^6}{3b^6+d^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\left(đpcm\right)\)
HQ
Hà Quang Minh
Giáo viên
18 tháng 9 2023
\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)
5 tháng 2 2022
a: \(A=\dfrac{1}{9}:\dfrac{1}{9}:\left(\dfrac{10+7}{15}:\dfrac{12-5}{30}\right)\)
\(=1:\left(\dfrac{17}{15}\cdot\dfrac{30}{7}\right)=1:\dfrac{34}{7}=\dfrac{7}{34}\)
b: \(=\left(5.6+0.64\right)\cdot1.25\cdot\dfrac{19}{3}+31.64\)
\(=\dfrac{39}{5}\cdot\dfrac{19}{3}+\dfrac{791}{25}=\dfrac{2026}{25}\)
3 tháng 6 2015
\(\frac{\left(\frac{53}{4}-\frac{59}{27}-\frac{65}{6}\right).\frac{5751}{25}+\frac{187}{4}}{\left(\frac{10}{7}+\frac{10}{3}\right):\left(\frac{37}{3}-\frac{100}{7}\right)}=\frac{\left(\frac{4293}{324}-\frac{708}{324}-\frac{3510}{324}\right).\frac{5751}{25}+\frac{187}{4}}{\left(\frac{30}{21}+\frac{70}{21}\right):\left(\frac{259}{21}-\frac{300}{21}\right)}=\frac{\frac{25}{108}.\frac{5751}{25}+\frac{187}{4}}{\frac{100}{21}:\left(-\frac{41}{21}\right)}\)=\(\frac{\frac{213}{4}+\frac{187}{4}}{-\frac{100}{41}}=100:\left(-\frac{100}{4}\right)=-4\)
3 tháng 6 2015
\(30+\frac{14}{5}:\left(\frac{24}{150}-\frac{270}{150}-\frac{25}{150}\right)=30+\frac{14}{5}:\left(-\frac{271}{150}\right)=30+\left(-\frac{420}{271}\right)=\frac{7710}{271}\)
a ,A = \(a.\frac{1}{3}+a.\frac{1}{4}-a.\frac{1}{6}\)
\(=a.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)\)
\(=\frac{-3}{5}.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)\\ =\frac{-3}{5}.\frac{5}{12}\)
\(=\frac{-1}{4}\)
b, B = \(b.\frac{5}{6}+b.\frac{3}{4}-b.\frac{1}{2}\)
\(=b.\left(\frac{5}{6}+\frac{1}{4}-\frac{1}{2}\right)\)
\(=\frac{12}{13}.\left(\frac{5}{6}+\frac{1}{4}-\frac{1}{2}\right)\)
\(=\frac{12}{13}.\frac{7}{12}\)
\(=\frac{7}{13}\)
a) Thay \(a=\frac{-3}{5}\)vào biểu thức A ta có :
\(A=\frac{-3}{5}.\frac{1}{3}+\frac{-3}{5}.\frac{1}{4}-\frac{-3}{5}.\frac{1}{6}\)
\(A=\frac{-3}{5}.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)\)
\(A=\frac{-3}{5}.\frac{5}{12}\)
\(A=\frac{-1}{4}\)
Vậy giá trị của biểu thức A tại \(a=\frac{-3}{5}\)là \(\frac{-1}{4}\)
b) Thay \(b=\frac{12}{13}\)vào biểu thức B ta có :
\(B=\frac{12}{13}.\frac{5}{6}+\frac{12}{13}.\frac{3}{4}-\frac{12}{13}.\frac{1}{2}\)
\(B=\frac{12}{13}.\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(B=\frac{12}{13}.\frac{13}{12}\)
\(B=1\)
Vậy giá trị của biểu thức B tại \(b=\frac{12}{13}\)là 1
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