Tính giá trị của biểu thức sau: A = 4 3 + 5 − 8 1 + 5 + 15 5
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\dfrac{3}{2}\times\dfrac{4}{5}\times\dfrac{2}{3}\times\dfrac{15}{4}\\ =\dfrac{3\times4\times2\times15}{2\times5\times3\times4}\\ =\dfrac{3\times4\times2\times5\times3}{2\times5\times3\times4}\\ =3\)
\(b,\dfrac{6}{7}\times\dfrac{5}{8}\times\dfrac{7}{3}\times\dfrac{7}{6}\times\dfrac{8}{5}\\ =\dfrac{6\times5\times7\times7\times8}{7\times8\times3\times6\times5}\\ =\dfrac{7}{3}\)
a) \(\dfrac{3}{2}\times\dfrac{4}{5}\times\dfrac{2}{3}\times\dfrac{15}{4}\)
\(=\left(\dfrac{3}{2}\times\dfrac{2}{3}\right)\times\left(\dfrac{4}{5}\times\dfrac{15}{4}\right)\)
\(=1\times3\)
\(=3\)
b) \(\dfrac{6}{7}\times\dfrac{5}{8}\times\dfrac{7}{3}\times\dfrac{7}{6}\times\dfrac{8}{5}\)
\(=\left(\dfrac{6}{7}\times\dfrac{7}{6}\right)\times\left(\dfrac{5}{8}\times\dfrac{8}{5}\right)\times\dfrac{7}{3}\)
\(=1\times1\times\dfrac{7}{3}\)
\(=\dfrac{7}{3}\)
Chúc bạn học tốt
\(A=\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}\) \(\left(a\ne0\right)\)
Tại a = 12 biểu thức \(A=\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{12}\right):\dfrac{4}{5}=\dfrac{5}{4}-\dfrac{11}{24}:\dfrac{4}{5}=\dfrac{5}{4}-\dfrac{11}{24}.\dfrac{5}{4}=\dfrac{5}{4}-\dfrac{55}{96}=\dfrac{65}{96}\)
Để \(A=\dfrac{15}{23}< =>\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}=\dfrac{15}{23}\)
\(\Leftrightarrow\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}=\dfrac{55}{92}< =>\dfrac{3}{8}+\dfrac{1}{a}=\dfrac{11}{23}< =>\dfrac{1}{a}=\dfrac{19}{184}< =>a=\dfrac{184}{19}\)
Thay \(a=12\) vào A ta có:
\(A=\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{12}\right):\dfrac{4}{5}=\dfrac{65}{96}\)
Vậy:
____________________
Ta có:
\(A=\dfrac{15}{23}\) khi \(\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}=\dfrac{15}{23}\)
\(\Rightarrow\left(\dfrac{3}{8}+\dfrac{1}{a}\right)\cdot\dfrac{5}{4}=\dfrac{5}{4}-\dfrac{15}{23}\)
\(\Rightarrow\dfrac{3}{8}+\dfrac{1}{a}=\dfrac{55}{92}:\dfrac{5}{4}\)
\(\Rightarrow\dfrac{3}{8}+\dfrac{1}{a}=\dfrac{11}{23}\)
\(\Rightarrow\dfrac{1}{a}=\dfrac{11}{184}\)
\(\Rightarrow a=\dfrac{1\cdot184}{11}=\dfrac{184}{11}\)
`5`
`a, -7/21 +(1+1/3)`
`=-7/21 + ( 3/3 + 1/3)`
`=-7/21+ 4/3`
`=-7/21+ 28/21`
`= 21/21`
`=1`
`b, 2/15 + ( 5/9 + (-6)/9)`
`= 2/15 + (-1/9)`
`= 1/45`
`c, (9-1/5+3/12) +(-3/4)`
`= ( 45/5-1/5 + 3/12)+(-3/4)`
`= ( 44/5 + 3/12)+(-3/4)`
`= 9,05 +(-0,75)`
`=8,3`
`6`
`x+7/8 =13/12`
`=>x= 13/12 -7/8`
`=>x=5/24`
`-------`
`-(-6)/12 -x=9/48`
`=> 6/12 -x=9/48`
`=>x= 6/12-9/48`
`=>x=5/16`
`---------`
`x+4/6 =5/25 -(-7)/15`
`=>x+4/6 =1/5 + 7/15`
`=> x+ 4/6=10/15`
`=>x=10/15 -4/6`
`=>x=0`
`----------`
`x+4/5 = 6/20 -(-7)/3`
`=>x+4/5 = 6/20 +7/3`
`=>x+4/5 = 79/30`
`=>x=79/30 -4/5`
`=>x= 79/30-24/30`
`=>x= 55/30`
`=>x= 11/6`
\(5)\)
\(A=\dfrac{-7}{21}+\left(1+\dfrac{1}{3}\right)\)
\(A=\dfrac{-7}{21}+\dfrac{4}{3}\)
\(A=\dfrac{-7}{21}+\dfrac{28}{21}\)
\(A=1\)
\(--------------\)
\(B=\dfrac{2}{15}+\left(\dfrac{5}{9}+\dfrac{-6}{9}\right)\)
\(B=\dfrac{2}{15}+\dfrac{-1}{9}\)
\(B=\dfrac{18}{135}+\dfrac{-15}{135}\)
\(B=\dfrac{1}{45}\)
\(------------\)
\(C=9-\dfrac{1}{5}+\dfrac{3}{12}+\dfrac{-3}{4}\)
\(C=\dfrac{44}{5}+\dfrac{3}{12}+\dfrac{-3}{4}\)
\(C=\dfrac{528}{60}+\dfrac{15}{60}+\dfrac{-3}{4}\)
\(C=\dfrac{181}{20}+\dfrac{-3}{4}\)
\(C=\dfrac{181}{20}+\dfrac{-15}{20}\)
\(C=\dfrac{83}{10}\)
\(6)\)
\(a)\) \(x+\dfrac{7}{8}=\dfrac{13}{12}\)
\(x=\dfrac{13}{12}-\dfrac{7}{8}\)
\(x=\dfrac{104}{96}-\dfrac{84}{96}\)
\(x=\dfrac{5}{24}\)
\(b)\) \(\dfrac{-6}{12}-x=\dfrac{9}{48}\)
\(\dfrac{-1}{2}-x=\dfrac{3}{16}\)
\(x=\dfrac{-1}{2}-\dfrac{3}{16}\)
\(x=\dfrac{-8}{16}-\dfrac{3}{16}\)
\(x=\dfrac{-11}{16}\)
\(c)\) \(x+\dfrac{4}{6}=\dfrac{5}{25}-\left(-\dfrac{7}{15}\right)\)
\(x+\dfrac{4}{6}=\dfrac{5}{25}+\dfrac{7}{15}\)
\(x+\dfrac{4}{6}=\dfrac{75}{375}+\dfrac{105}{375}\)
\(x+\dfrac{4}{6}=\dfrac{12}{25}\)
\(x=\dfrac{12}{25}-\dfrac{4}{6}\)
\(x=\dfrac{72}{150}-\dfrac{100}{150}\)
\(x=\dfrac{-14}{75}\)
\(d)\) \(x+\dfrac{4}{5}=\dfrac{6}{20}-\left(-\dfrac{7}{3}\right)\)
\(x+\dfrac{4}{5}=\dfrac{6}{20}+\dfrac{7}{3}\)
\(x+\dfrac{4}{5}=\dfrac{18}{60}+\dfrac{140}{60}\)
\(x+\dfrac{4}{5}=\dfrac{79}{30}\)
\(x=\dfrac{79}{30}-\dfrac{4}{5}\)
\(x=\dfrac{79}{30}-\dfrac{24}{30}\)
\(x=\dfrac{11}{6}\)
Lời giải:
a.
$\frac{5}{15}-\frac{1}{6}\times \frac{2}{5}=\frac{5}{15}-\frac{1}{15}=\frac{4}{15}$
b.
$\frac{8}{24}+\frac{3}{4}:\frac{1}{8}=\frac{1}{3}+6=\frac{19}{3}$
c.
$\frac{1}{7}: \frac{2}{8}-\frac{1}{7}=\frac{1}{7}\times 4-\frac{1}{7}$
$=\frac{1}{7}\times (4-1)=\frac{1}{7}\times 3=\frac{3}{7}$
\(\dfrac{15}{16}.\dfrac{8}{5}.\dfrac{3}{4}\)= \(\dfrac{5.3.4.2.3}{4.2.2.5.4}=\dfrac{9}{8}\)
a) Cách 1:
\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = 8 + \frac{7}{3} - \frac{3}{5} - 5 - \frac{2}{5} - \frac{{10}}{3} + 2\\ = (8 - 5 + 2) + (\frac{7}{3} - \frac{{10}}{3}) - (\frac{3}{5} + \frac{2}{5})\\ = 5 + \frac{{ - 3}}{3} - \frac{5}{5}\\ = 5 + ( - 1) - 1\\ = 3\end{array}\)
Cách 2:
\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = (\frac{{120}}{{15}} + \frac{{35}}{{15}} - \frac{9}{{15}}) - (\frac{{25}}{5} + \frac{2}{5}) - (\frac{{10}}{3} - \frac{6}{3})\\ = \frac{{146}}{{15}} - \frac{{27}}{5} - \frac{4}{3}\\ = \frac{{146}}{{15}} - \frac{{81}}{{15}} - \frac{{20}}{{15}}\\ = \frac{{45}}{{15}}\\ = 3\end{array}\)
b)
\(\begin{array}{l}(7 - \frac{1}{2} - \frac{3}{4}):(5 - \frac{1}{4} - \frac{5}{8})\\ = (\frac{{28}}{4} - \frac{2}{4} - \frac{3}{4}):(\frac{{40}}{8} - \frac{2}{8} - \frac{5}{8})\\ = \frac{{23}}{4}:\frac{{33}}{8}\\ = \frac{{23}}{4}.\frac{8}{{33}}\\ = \frac{{46}}{{33}}\end{array}\)
A = 4 3 + 5 − 8 1 + 5 + 15 5 = 4 ( 3 − 5 ) 4 − 8 ( 1 − 5 ) − 4 + 15 5 5 = 3 − 5 + 2 − 2 5 + 3 5 = 5