Mình đang cần gấp

A = \(\dfrac{1+2}{1.2}-\dfrac{2+3}{2.3}+\dfrac{3+4}{3.4}-...-\dfrac{8+9}{8.9}\)

= \(\dfrac{1}{2}-\dfrac{1}{1}-\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{3}-...\dfrac{1}{9}-\dfrac{1}{8}\)

= \(1-\dfrac{1}{9}\)

=\(\dfrac{8}{9}\)

Bài 1:

\(82\times439+82\times562-82\)

\(=82\left(439+562-1\right)\)

\(=82\times1000=82000\)

Bài 2:

\(\dfrac{4}{5}+\dfrac{3}{10}\times\dfrac{1}{2}=\dfrac{4}{5}+\dfrac{3}{20}=\dfrac{16}{20}+\dfrac{3}{20}=\dfrac{19}{20}\)

Bài 3:

Chiều rộng hàng rào là:

18x2/3=12(m)

Số mét hàng rào cần làm là:

(18+12)x2=60(m)

Số tiền cần bỏ ra là:

60x40000=2400000(đồng)

a: \(\dfrac{CD}{CB}=\dfrac{2}{6}=\dfrac{1}{3}\)

=>\(S_{ADC}=\dfrac{1}{3}\cdot S_{ABC}\)

=>\(S_{ABC}=4\cdot3=12\left(cm^2\right)\)

b: ADCE là hình thang 

=>AD//CE

Xét ΔBEC có AD//CE

nên \(\dfrac{BA}{BE}=\dfrac{AD}{CE}=\dfrac{BD}{BC}=\dfrac{2}{3}\)

\(S_{ABD}=S_{ABC}-S_{ADC}=12-4=8\left(cm^2\right)\)

Vì BA/BE=2/3

nên \(S_{ABD}=\dfrac{2}{3}\cdot S_{BED}\)

=>\(S_{BED}=8:\dfrac{2}{3}=12\left(cm^2\right)\)

câu a: 12m2 vì ABC và ADC cùng chiều cao, CD/BC=1/3 => sADC = 1/3 sABC => sABC = 4x3=12m2.    
b, kết quả 16/3m2. Vì DE//AC => BE/BA = BD/BC = 2/3. Xét tam giác BAD và BED có chung chiều cao từ D nên => sBED/sBAD = 2/3 mà sBAD = 8cm2 => sBED = 2/3x8 = 16/3cm2

g: \(\left(3x^4-2x^3+x^2\right):\dfrac{1}{3}x^2\)

\(=\dfrac{3x^4}{\dfrac{1}{3}x^2}-\dfrac{2x^3}{\dfrac{1}{3}x^2}+\dfrac{x^2}{\dfrac{1}{3}x^2}\)

\(=9x^2-6x+3\)

h: \(\dfrac{x^3-3x^2+6x}{-3x}\)

\(=-\dfrac{x^3}{3x}+\dfrac{3x^2}{3x}-\dfrac{6x}{3x}\)

\(=-\dfrac{1}{3}x^2+x-2\)

i: \(\dfrac{2x^2-5x+3}{2x-3}\)

\(=\dfrac{2x^2-3x-2x+3}{2x-3}\)

\(=\dfrac{x\left(2x-3\right)-\left(2x-3\right)}{2x-3}=x-1\)

j: \(\dfrac{x^5+x+1}{x^3-x}\)

\(=\dfrac{x^5-x^3+x^3-x+2x+1}{x^3-x}\)

\(=\dfrac{x^2\left(x^3-x\right)+\left(x^3-x\right)+2x+1}{x^3-x}\)

\(=x^2+1+\dfrac{2x+1}{x^3-x}\)

làm giùm mình câu g h i j mình cần gấp

D. -1/3 và -13/39.

D

Bài 3:

a: \(x\times3,4=15,3\)

=>x=15,3:3,4=4,5

b: \(7,5:x=0,24\)

=>x=7,5:0,24=31,25

c: \(\dfrac{5}{6}:x=\dfrac{7}{8}\)

=>\(x=\dfrac{5}{6}:\dfrac{7}{8}=\dfrac{5}{6}\times\dfrac{8}{7}=\dfrac{40}{42}=\dfrac{20}{21}\)

d: \(\dfrac{2}{3}\times x=\dfrac{5}{9}\)

=>\(x=\dfrac{5}{9}:\dfrac{2}{3}=\dfrac{5}{9}\times\dfrac{3}{2}=\dfrac{15}{18}=\dfrac{5}{6}\)

Bài 2:

a: 2,5x1,23x0,4

=(2,5x0,4)x1,23

=1x1,23=1,23

b: 3,6x4,5+4,5x6,4

=4,5(3,6+6,4)

=4,5x10=45

a

mjx3oid

a: \(y-\dfrac{4}{5}=2-\dfrac{1}{3}\)

=>\(y-\dfrac{4}{5}=\dfrac{5}{3}\)

=>\(y=\dfrac{5}{3}+\dfrac{4}{5}=\dfrac{25+12}{15}=\dfrac{37}{15}\)

b: \(\dfrac{12}{7}-y=\dfrac{1}{3}+2\)

=>\(\dfrac{12}{7}-y=\dfrac{7}{3}\)

=>\(y=\dfrac{12}{7}-\dfrac{7}{3}=\dfrac{36-49}{21}=\dfrac{-13}{21}\)

c: \(y:\dfrac{5}{6}=1+\dfrac{7}{2}\)

=>\(y:\dfrac{5}{6}=\dfrac{9}{2}\)

=>\(y=\dfrac{9}{2}\times\dfrac{5}{6}=\dfrac{45}{12}=3,75\)

d: \(3,6\times y=2,52:10\)

=>\(3,6\times y=0,252\)

=>y=0,252:3,6=0,07

e: 1,05:y+1,4=3,5

=>1,05:y=2,1

=>y=1,05:2,1=0,5

f: y-4,7=24,9+1,6

=>y-4,7=26,5

=>y=26,5+4,7=31,2

gr

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