\(a,\dfrac{-4}{5}+\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{9}{7}=\left(\dfrac{-4}{5}-\dfrac{1}{5}\right)+\left(\dfrac{2}{7}-\dfrac{9}{7}\right)=-1+\left(-1\right)=-2\)

\(b,\dfrac{1}{12}-\left(\dfrac{1}{2}+\dfrac{7}{12}\right)=\dfrac{1}{12}-\left(\dfrac{6}{12}+\dfrac{7}{12}\right)=\dfrac{1}{12}-\dfrac{13}{12}=\dfrac{-12}{12}=-1\)

\(c,\dfrac{-12}{5}-\left(\dfrac{3}{5}-\dfrac{1}{8}\right)=\dfrac{-12}{5}-\left(\dfrac{24}{40}-\dfrac{5}{40}\right)=\dfrac{-12}{5}-\dfrac{19}{40}=\dfrac{-96}{40}-\dfrac{16}{40}=\dfrac{-100}{40}=\dfrac{-5}{2}\)

\(d,\dfrac{-5}{7}-\left(\dfrac{-7}{6}+\dfrac{2}{7}\right)+\dfrac{7}{-6}=\dfrac{-5}{7}+\dfrac{7}{6}-\dfrac{2}{7}+\dfrac{-7}{6}=\left(\dfrac{-5}{7}-\dfrac{2}{7}\right)+\left(\dfrac{7}{6}+\dfrac{-7}{6}\right)=-1\)

\(e,\dfrac{-6}{5}-\dfrac{2}{7}+\dfrac{1}{5}-\dfrac{5}{7}=\left(\dfrac{-6}{5}+\dfrac{1}{5}\right)+\left(\dfrac{-2}{7}-\dfrac{5}{7}\right)=-1+\left(-1\right)=-2\)

a: \(=\dfrac{3}{13}\left(\dfrac{6}{11}+\dfrac{9}{11}-\dfrac{4}{11}\right)+1+\dfrac{10}{13}=\dfrac{3}{13}+\dfrac{10}{13}+1=2\)

b: \(=4\left(2.86+3.14\right)-25\cdot6.01+9\cdot0.75\)

=4*6-25*6,01+9*0,75

=-119,5

c: \(=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}+\dfrac{15}{4}=\dfrac{7}{8}+\dfrac{15}{4}=\dfrac{7+30}{8}=\dfrac{37}{8}\)

d: \(=\dfrac{-3}{31}-\dfrac{28}{31}-\dfrac{6}{17}-\dfrac{11}{17}+\dfrac{1}{25}-\dfrac{5}{25}=\dfrac{-4}{25}-2=-\dfrac{54}{25}\)

\(1.678\simeq1.68;1.734\simeq1.73\)
\(2.014\simeq2.01;2.536\simeq2.54\)

\(2.834\simeq2.83;2.927\simeq2.93\)

28 = 7 x 2²

48 = 2⁴ x 3

40 = 2³ x 5

=> ƯCLN(28,48,40) = 2² = 4

ƯCLN (28,48,40) = ước chung lớn nhất của 28,48,40

Ư(28) = { 1 ; 2 ; 4 ; 7 ; 8 ; 14 ; 28 } 

Ư(48)=(1,2,3,4,6,7,12,14,24,48)

Ư(40) = (1, 2, 4, 5, 8, 10, 20, 40)

Vậy ƯCLN (28,48,40)= 4

2:

a: AC=BC=6/2=3cm

b: Vì BD=BC

nên B là trung điểm của DC

c: AD=6+3=9cm

bài 3.

1.\(\dfrac{-51}{136}=\dfrac{-3}{8};\dfrac{-60}{108}=\dfrac{-5}{9};\dfrac{26}{-156}=\dfrac{-1}{6}\)

\(\dfrac{-3}{8}=\dfrac{-24}{72};\dfrac{-5}{9}=\dfrac{-40}{72};\dfrac{-1}{6}=\dfrac{-12}{72}\)

2.\(\dfrac{-165}{270}=\dfrac{-11}{18};\dfrac{-91}{-156}=\dfrac{7}{12};\dfrac{-210}{1134}=\dfrac{-5}{27}\)

\(\dfrac{-11}{18}=\dfrac{-66}{108};\dfrac{7}{12}=\dfrac{63}{108};\dfrac{-5}{27}=\dfrac{-20}{108}\)

a) \(\dfrac{x}{12}=\dfrac{6}{9}\)

\(\Rightarrow x=\dfrac{12\cdot6}{9}=8\)

b) \(\dfrac{9}{15}=x+\dfrac{11}{35}\)

\(\Rightarrow x=\dfrac{9}{15}-\dfrac{11}{35}\)

\(\Rightarrow x=\dfrac{2}{7}\)

c) \(x+\dfrac{6}{5}=\dfrac{x}{2}\)

\(\Rightarrow2\left(x+\dfrac{6}{5}\right)=x\)

\(\Rightarrow2x+\dfrac{12}{5}=x\)

\(\Rightarrow2x-x=-\dfrac{12}{5}\)

\(\Rightarrow x=-\dfrac{12}{5}\)

\(a,\dfrac{x}{12}=\dfrac{6}{9}\\ \Leftrightarrow x=12\cdot\dfrac{6}{9}\\ \Leftrightarrow x=8\\ b,\dfrac{9}{15}=x+\dfrac{11}{35}\\ \Leftrightarrow x=\dfrac{9}{15}-\dfrac{11}{35}\\ \Leftrightarrow x=\dfrac{2}{7}\\ c,x+\dfrac{6}{5}=\dfrac{x}{2}\\ \Leftrightarrow\dfrac{x}{2}=-\dfrac{6}{5}\\ \Leftrightarrow x=-\dfrac{12}{5}\)

C

C.75 min