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\(1,0,75-\dfrac{2}{3}-0,5=\dfrac{3}{4}-\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{9}{12}-\dfrac{8}{12}-\dfrac{1}{2}=\dfrac{1}{12}-\dfrac{1}{2}\)
\(=\dfrac{2}{24}-\dfrac{12}{24}=\dfrac{-10}{24}=\dfrac{-5}{12}\)
\(2,\dfrac{1}{5}-0,125-\dfrac{5}{4}=\dfrac{1}{5}-\dfrac{1}{8}-\dfrac{5}{4}=\dfrac{8}{40}-\dfrac{5}{40}-\dfrac{5}{4}=\dfrac{3}{40}-\dfrac{5}{4}\)
\(=\dfrac{3}{40}-\dfrac{50}{40}=\dfrac{-47}{40}\)
\(3,1,25-\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{5}{4}-\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{2}{4}+\dfrac{4}{3}=\dfrac{6}{12}+\dfrac{16}{12}=\dfrac{22}{12}=\dfrac{11}{6}\)
\(4,0,15-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{3}{20}-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{3}{20}-\dfrac{5}{20}+\dfrac{2}{5}=\dfrac{-2}{20}+\dfrac{2}{5}\)
\(=\dfrac{-2}{20}+\dfrac{8}{20}=\dfrac{6}{20}=\dfrac{3}{10}\)
\(5,5-3,4+\dfrac{1}{5}=\dfrac{5}{1}-\dfrac{17}{5}+\dfrac{1}{5}=\dfrac{25}{5}-\dfrac{17}{5}+\dfrac{1}{5}=\dfrac{25-17+1}{5}=\dfrac{9}{5}\)
\(6,\dfrac{1}{4}-0,3+\dfrac{4}{3}=\dfrac{1}{4}-\dfrac{3}{10}+\dfrac{4}{3}=\dfrac{10}{40}-\dfrac{12}{40}+\dfrac{4}{3}=\dfrac{-2}{40}+\dfrac{4}{3}\)
\(=\dfrac{-1}{20}+\dfrac{4}{3}=\dfrac{-3}{60}+\dfrac{80}{60}=\dfrac{77}{60}\)
\(7,0,2-3,25+4,7=\dfrac{1}{5}-\dfrac{13}{4}+\dfrac{47}{10}=\dfrac{4}{20}-\dfrac{65}{20}+\dfrac{47}{10}=\dfrac{-61}{20}+\dfrac{47}{10}\)
\(=\dfrac{-61}{20}+\dfrac{94}{20}=\dfrac{33}{20}=1,65\)
\(8,5,4+\dfrac{-7}{3}-\dfrac{-5}{7}=\dfrac{27}{5}+\dfrac{-7}{3}-\dfrac{-5}{7}=\dfrac{81}{15}+\dfrac{-35}{15}-\dfrac{-5}{7}\)
\(=\dfrac{46}{15}-\dfrac{-5}{7}=\dfrac{322}{105}-\dfrac{-75}{105}=\dfrac{397}{105}\)
\(9,\dfrac{-4}{2}+\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{-12}{6}+\dfrac{2}{6}-\dfrac{1}{4}=\dfrac{-10}{6}-\dfrac{1}{4}=\dfrac{-5}{3}-\dfrac{1}{4}\)
\(=\dfrac{-20}{12}-\dfrac{3}{12}=\text{ }\dfrac{-23}{12}\)
\(10,5,4-1,5-\left(7,2-1\right)=3,9-6,2=-2,3\)
\(11,4,9-\left(1,5-7,7+3\right)=4,9-\left(-3,2\right)=8,1\)
\(12,7,8-4,7+\left(5,3-1,4\right)=3,1+3,9=7\)
\(14,\dfrac{1}{2}-0,4+\dfrac{1}{5}\text{=}0,5-0,4+0,2=0,3\)
\(15,4,2-\dfrac{4}{5}+\dfrac{1}{2}=4,2-0,8+0,5=3,9\)
a: \(A=4-4.2\left(15.187+4.813\right)+1.16\)
\(=4-4.2\cdot20+1.16\)
\(=5.16-84=-78.84\)
b: \(B=13.14-4.59=8.55\)
c: \(C=3.5x^2-0.4xy+y^2\)
Trường hợp 1: x=0,5 và y=1,5
\(C=3.5\cdot0.5^2-0.4\cdot0.5\cdot1.5+1.5^2=2.825\)
Trường hợp 2: x=-0,5 và y=1,5
\(C=3.5\cdot0.5^2+0.4\cdot0.5\cdot1.5+1.5^2=3.425\)
\(a;C1:A=3+21-4=20\)
\(C2:A=3,43+20,51-4,2=19,74=20\)
\(\)
a, C1 : 3,43 + 20,51 - 4,2
\(\approx\)3 + 21 - 4
= 24 - 4
= 20
C2: 3,43 + 20,51 - 4,2
= 23,94 -4,2
= 19,74
= 20
Vì 20=20=>C1=C2
b, C1: \(\frac{72,8-4,75:0,8}{3,2}\)
= \(\frac{73-5:1}{3}\)
= \(\frac{68}{3}\)
C2: \(\frac{72,8-4,75:0,8}{3,2}\)
= \(\frac{72,8-5,9375}{3,2}\)
= \(\frac{66,8625}{3,2}\)
= \(\frac{67}{3}\)
Vì \(\frac{68}{3}>\frac{67}{3}\) => C1 > C2
\(\frac{\left(-3\right)^2.3^2.32}{3^4.\left(-2\right)^6}=\frac{3^4.2^5}{3^4.2^6}=\frac{1}{2}\)
a) 0,(1) + 0,(13) - 0,(123)
=0,(24)-0,(123)
=0,(119301)
b) 4,(14) + 2,(133)
\(\approx6,2745\)