5 và 3/5+7 và 21/48 ;10/7-5 và 21/48:10/7
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\(\dfrac{7}{15}và\dfrac{5}{13}=\dfrac{7.13}{15.13}=\dfrac{91}{195};\dfrac{5}{13}=\dfrac{5.15}{13.15}=\dfrac{75}{195}\)
\(\dfrac{11}{12}và\dfrac{7}{48}=\dfrac{11.4}{12.4}=\dfrac{44}{48};\dfrac{7}{48}\)
\(\dfrac{24}{35}và\dfrac{21}{7}=\dfrac{24}{35};\dfrac{21}{7}=\dfrac{21.5}{7.5}=\dfrac{105}{35}\)
\(\dfrac{15}{16}và\dfrac{7}{8}=\dfrac{15}{16};\dfrac{7}{8}=\dfrac{7.2}{8.2}=\dfrac{14}{16}\)
\(a,MSC:195\\ \dfrac{7}{15}=\dfrac{7.13}{15.13}=\dfrac{91}{195}\\ \dfrac{5}{13}=\dfrac{5.15}{13.15}=\dfrac{75}{195}\\ b,MSC:48\\ \dfrac{11}{12}=\dfrac{11.4}{12.4}=\dfrac{44}{48}\)
Giữ nguyên \(\dfrac{7}{48}\)
\(c,MSC:30\\ \dfrac{3}{2}=\dfrac{3.15}{2.15}=\dfrac{45}{30}\\ \dfrac{2}{3}=\dfrac{2.10}{3.10}=\dfrac{20}{30}\\ \dfrac{7}{5}=\dfrac{7.6}{5.6}=\dfrac{42}{30}\\ d,MSC:75\\ \dfrac{5}{3}=\dfrac{25.5}{3.25}=\dfrac{125}{75}\\ \dfrac{15}{25}=\dfrac{15.3}{25.3}=\dfrac{45}{75}\)
\(\dfrac{7}{15}=\dfrac{7.13}{15.13}=\dfrac{91}{195};\dfrac{5}{13}=\dfrac{5.15}{13.15}=\dfrac{75}{195}\)
\(\dfrac{11}{12}=\dfrac{11.4}{12.4}=\dfrac{44}{48}\) ; giữ nguyên phân số còn lại
\(\dfrac{3}{2}=\dfrac{3.15}{2.15}=\dfrac{45}{30};\dfrac{2}{3}=\dfrac{2.10}{3.10}=\dfrac{20}{30};\dfrac{7}{5}=\dfrac{7.6}{5.6}=\dfrac{42}{30}\)
\(\dfrac{5}{3}=\dfrac{5.25}{3.25}=\dfrac{125}{75};\dfrac{15}{25}=\dfrac{15.3}{25.3}=\dfrac{45}{75}\)
\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
a)
\(\dfrac{48}{92}=\dfrac{48:4}{92:4}=\dfrac{12}{23}\)
\(\dfrac{36}{69}=\dfrac{36:3}{69:3}=\dfrac{12}{23}\)
Ta có:
Mẫu số chung 2 phân số: 23
Vì \(12=12\) nên \(\dfrac{12}{23}=\dfrac{12}{23}\)
Vậy \(\dfrac{48}{92}=\dfrac{36}{69}\)
b)
\(\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{7}{8}=\dfrac{573}{280}\)
Ta có:
Mẫu số chung 2 phân số: 280
\(\dfrac{3}{1}=\dfrac{3*280}{1*280}=\dfrac{840}{280}\)
Vì \(573< 840\) nên \(\dfrac{573}{280}< \dfrac{840}{280}\)
Vậy \(\dfrac{573}{280}< \dfrac{3}{1}\)
c)
Ta có:
Mẫu số chung 2 phân số: 10
\(\dfrac{2}{5}=\dfrac{2*2}{5*2}=\dfrac{4}{10}\)
Vì \(1< 4\) nên\(\dfrac{1}{10}< \dfrac{4}{10}\)
Vậy \(\dfrac{1}{10}< \dfrac{2}{5}\)
d)
\(\dfrac{4}{10}=\dfrac{4:2}{10:2}=\dfrac{2}{5}\)
Ta có:
Mẫu số chung 2 phân số: 5
Vì \(2=2\) nên \(\dfrac{2}{5}=\dfrac{2}{5}\)
Vậy \(\dfrac{4}{10}=\dfrac{2}{5}\)
k)(-37)+14+26+37
=(-37)+37+14+26
=0+14+26
=40
l)15+23+(-25)+(-23)
=15+(-25)+23+(-23)
=-10+0
=-10
n)25+37-48-25-37
=25-25+37-37-48
=0+0-48
=-48
o)24.(16-5)-16.(24-5)
=24.16-24.5-16.24-16.5
=0-24.5-16.5
=0-(24-16).5
=0-8.5
=0-40
=-40
p)31.(-18)+31.(-81)-31
=31.(-18)+31.(-81)+31.(-1)
=31.[-18+(-81)+(-1)]
=31.(-100)
=-3100
q)-48+48.(-78)+48.(-21)
=48.(-1)+48.(-78)+48.(-21)
=48.[(-1)+(-78)+(-21)]
=48.(-100)
=-4800
r)(7.3-3):(-6)
=(7.3-1.3):(-6)
=(6.3):(-6)
=-3
s)-3.7-4.(-5)+1
=-21-(-20)+1
=-20-(-20)
=0
t(6.8-10:5)+3.(-7)
=(48-2)+(-21)
=46+(-21)
=25
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