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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.
hực hiện phép nhân hoặc phép chia hai hỗn số bằng cách viết hỗn số dưới dạng phân số:
a) 512.334512.334 b) 613:429613:429
Giải
a) 512.334=112.154=1658;512.334=112.154=1658;
b) 6{1 \over 3}:4{2 \over 9} = {{19} \over 3}:{{38} \over 9} = {{19} \over 3}.{9 \over {38}} = {3 \over 2}\)
Lưu ý: Khi cộng hai hỗn số ta có thể cộng phần nguyên với nhau, phần phân số với nhau. Nhưng nhân (hoặc chia) hai hỗn số ta không thể nhân (hoặc chia) phần nguyên với nhau và phần phân số với nhau.
\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}.\dfrac{5}{1}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{1}{3}.2+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{1}{3}.\left(\dfrac{2}{5}+\dfrac{3}{5}-2\right)\)=\(\dfrac{1}{3}.\left(-1\right)=\dfrac{-1}{3}\)
a) \(\dfrac{8}{5}-\dfrac{9}{5}=\dfrac{8-9}{5}=\dfrac{-1}{5}\)
b) \(\dfrac{5}{2}+\dfrac{2}{3}=\dfrac{15}{6}+\dfrac{4}{6}=\dfrac{15+4}{6}=\dfrac{19}{6}\)
c) \(\dfrac{-5}{9}\cdot\dfrac{2}{11}=\dfrac{-5\cdot2}{9\cdot11}=\dfrac{-10}{99}\)
d) \(\dfrac{-2}{9}:\dfrac{1}{3}=\dfrac{-2}{9}\cdot3=\dfrac{-2}{3}\)
e) \(\dfrac{3}{8}-\dfrac{1}{4}+\dfrac{5}{12}=\dfrac{9}{24}-\dfrac{6}{24}+\dfrac{10}{24}=\dfrac{9-6+10}{24}=\dfrac{13}{24}\)
f) \(\dfrac{-4}{3}\cdot\dfrac{5}{4}:\dfrac{7}{3}=\dfrac{-4}{3}\cdot\dfrac{5}{4}\cdot\dfrac{3}{7}=\dfrac{-4\cdot5\cdot3}{3\cdot4\cdot7}=\dfrac{-5}{7}\)
\(a.\dfrac{1}{3}+\dfrac{4}{5}-\left(\dfrac{4}{5}+\dfrac{5}{8}\right)\)
\(=\dfrac{1}{3}+\dfrac{4}{5}-\dfrac{4}{5}-\dfrac{5}{8}\)
\(=\dfrac{1}{3}+\left(\dfrac{4}{5}-\dfrac{4}{5}\right)-\dfrac{5}{8}\)
\(=\dfrac{1}{3}-\dfrac{5}{8}\)
=\(-\dfrac{7}{24}\)
\(b.8\dfrac{4}{9}-\left(4\dfrac{2}{7}+5\dfrac{4}{9}\right)\)
\(=8\dfrac{4}{9}-4\dfrac{2}{7}-5\dfrac{4}{9}\)
\(=\left(8\dfrac{4}{9}-5\dfrac{4}{9}\right)-4\dfrac{2}{7}\)
\(=\left(8+\dfrac{4}{9}-5-\dfrac{4}{9}\right)-4-\dfrac{2}{7}\)
\(=\left[8-5+\left(\dfrac{4}{9}-\dfrac{4}{9}\right)\right]-4-\dfrac{2}{7}\)
\(=3-4-\dfrac{2}{7}\)
\(=-1-\dfrac{2}{7}\)
\(=-\dfrac{9}{7}\)
\(c.\left(-\dfrac{5}{7}\right).\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\left(-\dfrac{5}{7}\right).\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1+\left(-\dfrac{5}{7}\right).\left(-1\right)\)
\(=\left(-\dfrac{5}{7}\right).\left[\dfrac{2}{11}+\dfrac{9}{11}+\left(-1\right)\right]+1\)
\(=\left(-\dfrac{5}{7}\right).0+1=1\)
Bài 1:
a) \(\dfrac{2}{5}\cdot x-\dfrac{1}{4}=\dfrac{1}{10}\)
\(\dfrac{2}{5}\cdot x=\dfrac{1}{10}+\dfrac{1}{4}\)
\(\dfrac{2}{5}\cdot x=\dfrac{7}{20}\)
\(x=\dfrac{7}{20}:\dfrac{2}{5}\)
\(x=\dfrac{7}{8}\)
Vậy \(x=\dfrac{7}{8}\).
b) \(\dfrac{3}{5}=\dfrac{24}{x}\)
\(x=\dfrac{5\cdot24}{3}\)
\(x=40\)
Vậy \(x=40\).
c) \(\left(2x-3\right)^2=16\)
\(\left(2x-3\right)^2=4^2\)
\(\circledast\)TH1: \(2x-3=4\\ 2x=4+3\\ 2x=7\\ x=\dfrac{7}{2}\)
\(\circledast\)TH2: \(2x-3=-4\\ 2x=-4+3\\ 2x=-1\\ x=\dfrac{-1}{2}\)
Vậy \(x\in\left\{\dfrac{7}{2};\dfrac{-1}{2}\right\}\).
Bài 2:
a) \(25\%-4\dfrac{2}{5}+0.3:\dfrac{6}{5}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}:\dfrac{6}{5}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}\cdot\dfrac{5}{6}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{1}{4}\)
\(=\dfrac{5}{20}-\dfrac{88}{20}+\dfrac{5}{20}\)
\(=\dfrac{5-88+5}{20}\)
\(=\dfrac{78}{20}=\dfrac{39}{10}\)
b) \(\left(\dfrac{1}{6}-\dfrac{1}{5^2}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{1}{6}-\dfrac{1}{25}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{1}{6}-\dfrac{1}{5}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{5}{30}-\dfrac{6}{30}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{5-6+1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=0\cdot\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=0\)
Bài 3:
a) \(\dfrac{4}{19}\cdot\dfrac{-3}{7}+\dfrac{-3}{7}\cdot\dfrac{15}{19}\)
\(=\dfrac{-3}{7}\left(\dfrac{4}{19}+\dfrac{15}{19}\right)\)
\(=\dfrac{-3}{7}\cdot1\)
\(=\dfrac{-3}{7}\)
b) \(7\dfrac{5}{9}-\left(2\dfrac{3}{4}+3\dfrac{5}{9}\right)\)
\(=\dfrac{68}{9}-\dfrac{11}{4}-\dfrac{32}{9}\)
\(=\dfrac{68}{9}-\dfrac{32}{9}-\dfrac{11}{4}\)
\(=4-\dfrac{11}{4}\)
\(=\dfrac{16}{4}-\dfrac{11}{4}\)
\(\dfrac{5}{4}\)
Bài 4:
\(\dfrac{4}{12\cdot14}+\dfrac{4}{14\cdot16}+\dfrac{4}{16\cdot18}+...+\dfrac{4}{58\cdot60}\)
\(=2\left(\dfrac{1}{12\cdot14}+\dfrac{1}{14\cdot16}+\dfrac{1}{16\cdot18}+...+\dfrac{1}{58\cdot60}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{18}+...+\dfrac{1}{58}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{5}{60}-\dfrac{1}{60}\right)\)
\(=2\cdot\dfrac{1}{15}\)
\(=\dfrac{2}{15}\)
d: =-2/7-3/11+2/7=-3/11
e: =2+3/7+1+4/7-17/7
=4-17/7=11/7
f: =-2/3*4/5+1/5*11/9
=-8/15+11/45
=-24/45+11/45=-13/45
h: =(-3,1+3,7):0,2=0,6:0,2=3
3/5
3/5
\(\dfrac{3}{5}\)
\(A=\dfrac{3}{5}.\left(\dfrac{5}{4}-\dfrac{1}{4}\right)\)
\(A=\dfrac{3}{5}\)
\(\dfrac{3}{5}\)x(\(\dfrac{5}{4}\)-\(\dfrac{1}{4}\))=3/5 x 1 =3/5
A=\(\dfrac{3}{5}\cdot\dfrac{5}{4}-\dfrac{3}{5}\cdot\dfrac{1}{4}\)
A= \(\dfrac{3}{5}\cdot\left(\dfrac{5}{4}-\dfrac{1}{4}\right)\)
A=\(\dfrac{3}{5}\cdot1\)
A= \(\dfrac{3}{5}\)
\(\dfrac{3}{5}\times\left(\dfrac{5}{4}-\dfrac{1}{4}\right)=\dfrac{3}{5}\times1=\dfrac{3}{5}\)
\(\dfrac{3}{5}\)
=\(\dfrac{3}{5}\)
a, A= 3/5.(5/4-1/4)
= 3/5.1
= 3/5
\(A=\dfrac{3}{5}\cdot\dfrac{5}{4}-\dfrac{3}{5}\cdot\dfrac{1}{4}=\dfrac{3}{5}\left(\dfrac{5}{4}-\dfrac{1}{4}\right)=\dfrac{3}{5}\cdot\dfrac{1}{1}=\dfrac{3}{5}\)
a. A= \(\dfrac{3}{5}.\dfrac{5}{4}-\dfrac{3}{5}.\dfrac{1}{4}\)
A=\(\dfrac{3}{5}.\left(\dfrac{5}{4}-\dfrac{1}{4}\right)\)
A=\(\dfrac{3}{5}.1=\dfrac{3}{5}\)
\(\dfrac{3}{5}\)