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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 \(S_{ABD}=\dfrac{2}{3}\cdot S_{BED}\)
=>\(S_{BED}=8:\dfrac{2}{3}=12\left(cm^2\right)\)
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: \(\dfrac{x^5+x+1}{x^3-x}\)
\(=\dfrac{x^5-x^3+x^3-x+2x+1}{x^3-x}\)
\(=\dfrac{x^2\left(x^3-x\right)+\left(x^3-x\right)+2x+1}{x^3-x}\)
\(=x^2+1+\dfrac{2x+1}{x^3-x}\)
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: \(y-\dfrac{4}{5}=2-\dfrac{1}{3}\)
=>\(y-\dfrac{4}{5}=\dfrac{5}{3}\)
=>\(y=\dfrac{5}{3}+\dfrac{4}{5}=\dfrac{25+12}{15}=\dfrac{37}{15}\)
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
\(673\times14+327\times14=\left(673+327\right)\times14=1000\times14=14000\)
\(\dfrac{11}{4}\times\dfrac{8}{14}\times\dfrac{7}{11}\times\dfrac{5}{3}=\left(\dfrac{11}{4}\times\dfrac{7}{11}\right)\times\left(\dfrac{8}{14}\times\dfrac{5}{3}\right)=\dfrac{7}{4}\times\dfrac{20}{21}=\dfrac{5}{3}\)
673 . 14 + 327 . 14
= 14. (673 + 327)
= 14. 1000
= 14000.
11/4 . 8/14 . 7/11 . 5/3
= 11/7 . 35/33
= 5/3.
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)