a , Phân tích \(a^4+4\)thành nhân tử
b, Hãy tính \(B=\dfrac{3^4+4}{5^4+4}\times\dfrac{7^4+4}{9^4+4}....\dfrac{19^4+4}{21^4+4}\)
K
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18 tháng 2 2022
a: =-1/3+1/3=0
b: \(=\dfrac{4}{11}\left(-\dfrac{2}{7}-\dfrac{4}{7}-\dfrac{1}{7}\right)=\dfrac{4}{11}\cdot\left(-1\right)=-\dfrac{4}{11}\)
c: \(=10+\dfrac{5}{9}-3-\dfrac{5}{7}-4-\dfrac{5}{9}=3-\dfrac{5}{7}=\dfrac{16}{7}\)
d: \(=\dfrac{1}{3}+\dfrac{7}{4}-\dfrac{7}{4}+\dfrac{4}{5}=\dfrac{1}{3}+\dfrac{4}{5}=\dfrac{5+12}{15}=\dfrac{17}{15}\)
PH
18 tháng 2 2022
a: =-1/3+1/3=0
b: =411(−27−47−17)=411⋅(−1)=−411=411(−27−47−17)=411⋅(−1)=−411
c: =10+59−3−57−4−59=3−57=167=10+59−3−57−4−59=3−57=167
d: =13+74−74+45=13+45=5+1215=1715
H9
HT.Phong (9A5)
CTVHS
24 tháng 8 2023
a) \(\dfrac{2}{5}\times?=\dfrac{3}{10}\)
\(?=\dfrac{3}{10}:\dfrac{2}{5}=\dfrac{3}{4}\)
b) \(\dfrac{1}{8}:?=\dfrac{1}{5}\)
\(?=\dfrac{1}{8}:\dfrac{1}{5}=\dfrac{5}{8}\)
24 tháng 8 2023
a: Phân số cần tìm là: \(\dfrac{3}{10}:\dfrac{2}{5}=\dfrac{3}{10}\cdot\dfrac{5}{2}=\dfrac{15}{20}=\dfrac{3}{4}\)
b: Phân số cần tìm là \(\dfrac{1}{8}:\dfrac{1}{5}=\dfrac{5}{8}\)
VT
31 tháng 12 2022
\(C=\dfrac{4}{9}\times\dfrac{13}{17}+\dfrac{4}{17}\times\dfrac{4}{9}+\dfrac{2}{9}\\ =\dfrac{4}{9}\times\left(\dfrac{13}{17}+\dfrac{4}{17}\right)+\dfrac{2}{9}\\ =\dfrac{4}{9}\times1+\dfrac{2}{9}\\ =\dfrac{4}{9}+\dfrac{2}{9}\\ =\dfrac{6}{9}=\dfrac{2}{3}\)
\(D=\dfrac{8}{19}\times\dfrac{5}{11}+\dfrac{7}{11}\times\dfrac{8}{19}+\dfrac{12}{11}\times\dfrac{11}{19}\\ =\dfrac{8}{19}\times\left(\dfrac{5}{11}+\dfrac{7}{11}\right)+\dfrac{12}{11}\times\dfrac{11}{19}\\ =\dfrac{8}{19}\times\dfrac{12}{11}+\dfrac{12}{11}\times\dfrac{11}{19}\\ =\dfrac{12}{11}\times\left(\dfrac{8}{19}+\dfrac{11}{19}\right)\\ =\dfrac{12}{11}\times19\\ =\dfrac{12}{11}\)
31 tháng 12 2022
\(C=\dfrac{4}{9}\cdot\dfrac{13}{17}+\dfrac{4}{17}\cdot\dfrac{4}{9}+\dfrac{2}{9}\)
\(C=\dfrac{4}{9}\cdot\left(\dfrac{13}{17}+\dfrac{4}{17}\right)+\dfrac{2}{9}\)
\(C=\dfrac{4}{9}\cdot\dfrac{13+4}{17}+\dfrac{2}{9}\)
\(C=\dfrac{4}{9}\cdot\dfrac{17}{17}+\dfrac{9}{2}\)
\(C=\dfrac{4}{9}\cdot1+\dfrac{2}{9}\)
\(C=\dfrac{4}{9}+\dfrac{2}{9}\)
\(C=\dfrac{4+2}{9}\)
\(C=\dfrac{6}{9}\)
\(C=\dfrac{2}{3}\)
\(D=\dfrac{8}{19}\cdot\dfrac{5}{11}+\dfrac{7}{11}\cdot\dfrac{8}{19}+\dfrac{12}{11}\cdot\dfrac{11}{19}\)
\(D=\dfrac{8}{19}\cdot\left(\dfrac{5}{11}+\dfrac{7}{11}\right)+\dfrac{12}{11}\cdot\dfrac{11}{19}\)
\(D=\dfrac{8}{19}\cdot\dfrac{12}{11}+\dfrac{12}{11}\cdot\dfrac{11}{19}\)
\(D=\dfrac{12}{11}\cdot\left(\dfrac{8}{19}+\dfrac{11}{19}\right)\)
\(D=\dfrac{12}{11}\cdot\dfrac{19}{19}\)
\(D=\dfrac{12}{11}\cdot1\)
\(D=\dfrac{12}{11}\)
15 tháng 3 2022
\(a,\dfrac{3}{5}+\dfrac{-5}{9}=\dfrac{27-25}{45}=\dfrac{2}{49}.\)
\(c,\dfrac{-27}{23}+\dfrac{5}{21}+\dfrac{4}{23}+\dfrac{16}{21}+\dfrac{1}{2}=\dfrac{-23}{23}+\dfrac{21}{21}+\dfrac{1}{2}=-1+1+\dfrac{1}{2}=\dfrac{1}{2}.\)
\(d,\dfrac{-8}{9}+\dfrac{1}{9}.\dfrac{2}{9}+\dfrac{1}{9}.\dfrac{7}{9}=\dfrac{-8}{9}+\dfrac{1}{9}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)=\dfrac{-8}{9}+\dfrac{1}{9}.1=\dfrac{-8+1}{9}=\dfrac{-7}{9}.\)
25 tháng 1 2022
a: \(A=\dfrac{19}{9}+\dfrac{4}{11}+\dfrac{2}{3}=\dfrac{209}{99}+\dfrac{44}{99}+\dfrac{66}{99}=\dfrac{319}{99}\)
b: \(B=\dfrac{-50}{60}+\dfrac{-35}{60}+\dfrac{12}{60}=\dfrac{-73}{60}\)
c: \(C=\dfrac{-27}{36}+\dfrac{132}{36}+\dfrac{10}{36}=\dfrac{115}{36}\)
d: \(D=\dfrac{-19}{3}+\dfrac{2}{3}-\dfrac{4}{5}=\dfrac{-17}{3}-\dfrac{4}{5}=\dfrac{-85-12}{15}=-\dfrac{97}{15}\)
TT
6 tháng 12 2017
a. \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)
\(=1+1+0,5\)
\(=2,5\)
b. \(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)
\(=\dfrac{3}{7}.\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)
\(=\dfrac{3}{7}.\left(-14\right)=-6\)
c. \(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(\dfrac{-5}{7}\right)\)
\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)\)
\(=-10:\left(-\dfrac{5}{7}\right)\)
\(=14\)
d. \(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\dfrac{-5}{21}:\dfrac{4}{5}+\dfrac{5}{21}:\dfrac{4}{5}\)
\(=\left(\dfrac{-5}{7}+\dfrac{5}{7}\right):\dfrac{4}{5}\)
\(=0:\dfrac{4}{5}\)
\(=0\)
LD
6 tháng 12 2017
a,
\(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)
\(=1+1-0,5=1,5\)
b,
\(\dfrac{3}{7}\cdot19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)
\(=\dfrac{3}{7}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{3}{7}.\left(-14\right)=-6\)
c,
\(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)\)
\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)=-10:\left(-\dfrac{5}{7}\right)=14\)
d,
\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\left(-\dfrac{2}{3}+\dfrac{3}{7}+\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\left[\left(-\dfrac{2}{3}+\dfrac{-1}{3}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\right]:\dfrac{4}{5}\)
\(=\left(-1+1\right):\dfrac{4}{5}=0:\dfrac{4}{5}=0\)
KV
5 tháng 5 2023
a) 3/4 + (-7/5) + 1/4 + (-3/5)
= (3/4 + 1/4) + (-7/5 - 3/5)
= 1 - 2
= -1
b) 4/9 . 7/11 - 4/11 . 2/9 + 6/11 . 4/9
= 4/9 . (7/11 - 2/11 + 6/11)
= 4/9 . 1
= 4/9
C
15 tháng 5 2022
8/9 x 5/4 + 3/5 = 10/9 + 3/5 = 77/45
5/8 + 1/4 = 5/9 + 2/8 = 7/8
1/2 + 5/9 = 9/18 + 5/9 = 14/9
9/8 x 3/4 - 4/7 = 27/32 - 4/7 = 61/224
O
15 tháng 5 2022
a) \(\dfrac{8}{9}:\dfrac{4}{5}+\dfrac{3}{5}=\dfrac{8}{9}\times\dfrac{5}{4}+\dfrac{2}{5}=\dfrac{10}{9}+\dfrac{2}{5}=\dfrac{50}{45}+\dfrac{18}{45}=\dfrac{68}{45}\)
b) \(\dfrac{5}{8}+\dfrac{1}{4}=\dfrac{5}{8}+\dfrac{2}{8}=\dfrac{6}{8}=\dfrac{3}{4}\)
c) \(\dfrac{7}{8}\times\dfrac{4}{7}+\dfrac{5}{9}=\dfrac{1}{2}+\dfrac{5}{9}=\dfrac{9}{18}+\dfrac{10}{18}=\dfrac{19}{18}\)
d) \(\dfrac{9}{8}:\dfrac{4}{3}-\dfrac{4}{7}=\dfrac{9}{8}\times\dfrac{3}{4}-\dfrac{4}{7}=\dfrac{27}{32}-\dfrac{4}{7}=\dfrac{61}{244}\)
a) Ta có: \(a^4+4=a^4+4a^2+4-4a^2=\left(a^2+2\right)^2-\left(2a\right)^2=\left(a^2+2-2a\right)\left(a^2+2+2a\right)\)
ta thấy n4+4=(n2-2n+2)(n2+2n+2)=\(\left[\left(n-1\right)^2-1\right]\) \(\left[\left(n+1\right)^2+1\right]\)
Do đó B=\(\dfrac{\left(2^2+1\right)\left(4^2+1\right)}{\left(4^2+1\right)\left(6^2+1\right)}.\dfrac{\left(6^2+1\right)\left(8^2+1\right)}{\left(8^2+1\right)\left(10^2+1\right)}.....\dfrac{\left(18^2+1\right)\left(20^2+1\right)}{\left(20^2+1\right)\left(22^2+1\right)}=\dfrac{2^2+1}{22^2+1}=\dfrac{5}{485}=\dfrac{1}{97}\)