a, \(A=-x-2,5-\left|4x+4,8\right|\)

\(b,B=-\left|-4-2,5x\right|-\dfrac{6}{5}x-6,2\)

\(c,C=\left|4,2-1,4x\right|-\dfrac{4}{5}x-\dfrac{24}{5}\)

a: \(A=2x-3.5-3x+1-\left|4x+4.8\right|\)

\(=-x-2.5-\left|4x+4.8\right|\)

Trường hợp 1: x>=-1,2

=>A=-x-2,5-4x-4,8=-5x-7,3

Trường hợp 2: x<-1,2

=>A=-x-2,5+4x+4,8=3x+2,3

b: \(B=-\left|2.5x+4\right|-1.2x-6.2\)

Trường hợp 1: x>=-1,6

=>B=-2,5x-4-1,2x-6,2=-3,7x-10,2

Trường hợp 2: x<-1,6

B=2,5x+4-1,2x-6,2=1,3x-2,2

c: \(C=-0.8x-4.8+\left|1.4x-4.2\right|\)

Trường hợp 1: x>=3

C=-0,8x-4,8+1,4x-4,2=0,6x-9

Trường hợp 2: x<3

C=-0,8x-4.8-1.4x+4,2=-2,2x-0,6

Có.

Vì \(\dfrac{4,2}{12,6}=\dfrac{1+3,2}{3+9,6}\) và \(\dfrac{-2,2}{-6,6}=\dfrac{1-3,2}{3-9,6}\)

\(\Rightarrow\dfrac{1}{3}=\dfrac{3,2}{9,6}=\dfrac{4,2}{12,6}=\dfrac{-2,2}{-6,6}\)

-Có

vì \(\dfrac{4,2}{12,6}=\dfrac{1+3,2}{3+9,6}\)và \(\dfrac{-2,2}{-6,6}=\dfrac{1-3,2}{3-9,6}\)

=> \(\dfrac{1}{3}=\dfrac{3,2}{9,6}=\dfrac{4,2}{12,6}\dfrac{-2,2}{-6,6}\)

a) \(\dfrac{24}{32}=\dfrac{15}{20};\dfrac{32}{24}=\dfrac{20}{15};\dfrac{15}{24}=\dfrac{20}{32};\dfrac{32}{24}=\dfrac{24}{15}\)

b) \(\dfrac{4,8}{25,6}=\dfrac{0,6}{3,2};\dfrac{25,6}{4,8}=\dfrac{3,2}{0,6};\dfrac{25,6}{3,2}=\dfrac{4,8}{0,6};\dfrac{3,2}{25,6}=\dfrac{0,6}{4,8}\)

a: \(\dfrac{24}{32}=\dfrac{15}{20};\dfrac{24}{15}=\dfrac{32}{20};\dfrac{32}{24}=\dfrac{20}{15};\dfrac{15}{24}=\dfrac{20}{32}\)

b: \(\dfrac{4.8}{25.6}=\dfrac{0.6}{3.2};\dfrac{4.8}{0.6}=\dfrac{25.6}{3.2};\dfrac{25.6}{4.8}=\dfrac{3.2}{0.6};\dfrac{0.6}{4.8}=\dfrac{3.2}{25.6}\)

bo chiu

Câu 2:

a: \(0,25\cdot5,7\cdot0,4\)

\(=5,7\cdot0,1\)

=0,57

b: 4,83+6,59+0,17

=4,83+0,17+6,59

=6,59+5

=11,59

c: \(10-0,34-0,66=10-\left(0,34+0,66\right)\)

=10-1

=9

d: \(\left(8+0,8\right)\cdot1,25\)

\(=8\cdot1,25+0,8\cdot1,25\)

=10+1

=11

2,9+3,7+(-4,2)+(-2,9)+4,2

=(2,9-2,9)+(-4,2+4,2)+3,7

=3,7

\(\left(3,6-4,8\right)-\left(-4,4+3,6-4,8\right)\)
<=> 3,6 - 4,8 + 4,4 - 3,6 + 4,8
<=> 4,4

thank

= (-4,2) + 4,2 + 2,9 + (-2,9) + 3,7 = 0+0+3,7 = 3,7

=(2,9-2,9)+(-4,2+4,2)+3,7=0+0+3,7=3,7