Tính B= (3-1/5).(3-2/5).(3-3/5)...(3-29/5)
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- Tính giá trị biểu thức:
a) (2/5 x 25/29) + (3/5 x 25/29)
= (50/145) + (75/145)
= 125/145
b) (5/2 x 3/7) - (3/14 : 6/7)
= 15/14 - (3/14 x 7/6)
= 15/14 - 1/2
= (30/28) - (14/28)
= 16/28
= 4/7
c) (15/4 : 5/12) - (6/5 : 11/15)
= (15/4 x 12/5) - (6/5 x 15/11)
= 180/20 - 90/55
= 9 - 18/11
= (99/11) - (18/11)
= 81/11
= 7 4/11
- Tính giá trị biểu thức:
a) (2/3) + (20/21 x 3/2 x 7/5)
= 2/3 + (60/210)
= 2/3 + 2/7
= (14/21) + (6/21)
= 20/21
b) (5/17 x 21/32 x 47/24 x 0)
= 0
c) (11/3 x 26/7) - (26/7 x 8/3)
= (286/21) - (208/21)
= 78/21
= 3 9/21
= 3 3/7
- Tìm x:
a) (25/8) : x = 5/16
=> (25/8) x (16/5) = x
=> 4 = x
b) x + (7/15) = 6/15
=> x = (6/15) - (7/15)
=> x = -1/15
c) x : (28/49) = 7/12
=> x x (49/28) = 7/12
=> x = (7/12) x (28/49)
=> x = 1/2
- Tìm x:
a) 6 x x = (5/8) : (3/4)
=> 6x = (5/8) x (4/3)
=> 6x = 20/24
=> 6x = 5/6
=> x = (5/6) / 6
=> x = 5/36
câu,b,không,đủ,thông,tin,nhan,bạn.
ý mình là 38 phần 3 trừ trong ngoặc 29 phần 5 trừ 10 phần 3 cộng 39 phần 5 trong ngoặc phần 44 phần 5 trừ trong ngoặc 2 phần 3 cộng 27 phần 5 trong ngoặc trừ 1 phần 3
\(1,\\ c,\dfrac{17}{6}-\dfrac{3}{7}-\dfrac{5}{6}=\dfrac{17\times7-3\times6-5\times7}{6\times7}=\dfrac{119-18-35}{42}=\dfrac{66}{42}=\dfrac{11}{7}\\ 2,\\ a,\dfrac{6}{9}+\dfrac{3}{15}+\dfrac{7}{15}\\ =\dfrac{6:3}{9:3}+\dfrac{7+3}{15}\\ =\dfrac{2}{3}+\dfrac{10}{15}\\ =\dfrac{2}{3}+\dfrac{2}{3}\\ =\dfrac{2+2}{3}=\dfrac{4}{3}\)
\(b,\dfrac{7}{2}+\dfrac{29}{12}-\dfrac{11}{12}\\ =\dfrac{7}{2}+\left(\dfrac{29}{12}-\dfrac{11}{12}\right)\\ =\dfrac{7}{2}+\dfrac{29-11}{12}\\ =\dfrac{7}{2}+\dfrac{18}{12}\\ =\dfrac{7}{2}+\dfrac{18:6}{12:6}\\ =\dfrac{7}{2}+\dfrac{3}{2}\\ =\dfrac{7+3}{2}=\dfrac{10}{2}=5\)
\(c,\dfrac{26}{25}-\dfrac{3}{5}-\dfrac{2}{5}\\ =\dfrac{26}{25}-\dfrac{3\times5}{5\times5}-\dfrac{2\times5}{5\times5}\\ =\dfrac{26-15-10}{25}\\ =\dfrac{1}{25}\)
Bài 1:
\(\dfrac{17}{6}-\dfrac{3}{7}-\dfrac{5}{6}=\left(\dfrac{17}{6}-\dfrac{5}{6}\right)-\dfrac{3}{7}=2-\dfrac{3}{7}=\dfrac{14}{7}-\dfrac{3}{7}=\dfrac{11}{7}\)
Bài 2:
a/\(\dfrac{6}{9}+\dfrac{3}{15}+\dfrac{7}{15}\)
\(=\dfrac{2}{3}+\left(\dfrac{3}{15}+\dfrac{7}{15}\right)\)
\(=\dfrac{2}{3}+\dfrac{2}{3}\)
\(=\dfrac{4}{3}\)
b/\(\dfrac{7}{2}+\dfrac{29}{12}-\dfrac{11}{12}\)
\(=\dfrac{7}{2}+\left(\dfrac{29}{12}-\dfrac{11}{12}\right)\)
\(=\dfrac{7}{2}+\dfrac{3}{2}\)
\(=5\)
c/\(\dfrac{26}{25}-\dfrac{3}{5}-\dfrac{2}{5}\)
\(=\dfrac{26}{25}-\left(\dfrac{3}{5}+\dfrac{2}{5}\right)\)
\(=\dfrac{26}{25}-1\)
\(=\dfrac{26}{25}-\dfrac{25}{25}\)
\(=\dfrac{1}{25}\)
#Đang Bận Thở
a) \(\left(\dfrac{3}{29}-\dfrac{1}{5}\right)\cdot\dfrac{29}{3}\)
\(=\dfrac{3}{29}\cdot\dfrac{29}{3}-\dfrac{1}{5}\cdot\dfrac{29}{3}\)
\(=1-\dfrac{29}{15}\)
\(=\dfrac{15-29}{15}\)
\(=-\dfrac{14}{15}\)
b) \(\dfrac{3}{4}\cdot\dfrac{7}{9}+\dfrac{1}{4}\cdot\dfrac{7}{9}\)
\(=\dfrac{7}{9}\cdot\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(=\dfrac{7}{9}\cdot1\)
\(=\dfrac{7}{9}\)
c) \(\dfrac{1}{7}\cdot\dfrac{5}{9}+\dfrac{5}{9}\cdot\dfrac{1}{7}+\dfrac{5}{9}\cdot\dfrac{3}{7}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}\cdot\dfrac{5}{7}\)
\(=\dfrac{25}{63}\)
d) \(4\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)
\(=\left(4\cdot\dfrac{3}{4}\right)\cdot\left(11\cdot\dfrac{9}{121}\right)\)
\(=3\cdot\dfrac{9}{11}\)
\(=\dfrac{27}{11}\)
a: \(=\dfrac{-5}{7}-\dfrac{2}{7}+\dfrac{3}{4}+\dfrac{1}{4}-\dfrac{1}{5}=-\dfrac{1}{5}\)
b: \(=\dfrac{-3}{31}-\dfrac{28}{31}+\dfrac{-6}{17}-\dfrac{11}{17}+\dfrac{1}{29}-\dfrac{1}{5}=\dfrac{-24}{145}\)
\(B=\dfrac{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{2017}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{2017}}:\dfrac{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2018}}{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2018}}\)
\(B=\dfrac{2\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}{3\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}:\dfrac{4\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}{5\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}\)
\(B=\dfrac{2}{3}:\dfrac{4}{5}\) ( Do \(\left\{{}\begin{matrix}1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\ne0\\1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\ne0\end{matrix}\right.\))
\(B=\dfrac{2}{3}\cdot\dfrac{5}{4}=\dfrac{2\cdot5}{3\cdot4}=\dfrac{5}{6}\)
\(B=\dfrac{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{2017}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{2017}}:\dfrac{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2018}}{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2018}}\)
\(\Rightarrow\)\(B=\dfrac{2-\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}{3\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}:\dfrac{4\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}{5\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}\)
\(\Rightarrow B=\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{10}{12}=\dfrac{5}{6}\)
\(A=-1,6:\left(1+\frac{2}{3}\right)\)
\(A=\frac{-8}{5}:\frac{5}{3}\)
\(A=\frac{-24}{25}\)
\(B=1,4.\frac{15}{29}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
\(B=\frac{7}{5}.\frac{15}{29}-\frac{22}{15}:\frac{11}{5}\)
\(B=\frac{21}{29}-\frac{22}{15}.\frac{5}{11}\)
\(B=\frac{21}{29}-\frac{2}{3}\)
\(B=\frac{5}{87}\)
\(\frac{3}{5}\times\frac{6}{7}+\frac{3}{5}:7+\frac{6}{5}\)
\(=\frac{3}{5}\times\frac{6}{7}+\frac{3}{5}\times\frac{1}{7}+\frac{6}{5}\)
\(=\frac{3}{5}\times\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{6}{5}\)
\(=\frac{3}{5}\times1+\frac{6}{5}\)
\(=\frac{3}{5}+\frac{6}{5}=\frac{9}{5}\)
~ Hok tốt ~
\(\frac{29}{12}:\frac{1}{2}-\frac{5}{12}:\frac{1}{2}-\frac{1}{2}\)
\(=\left(\frac{29}{12}-\frac{5}{12}\right):\frac{1}{2}-\frac{1}{2}\)
\(=2:\frac{1}{2}-\frac{1}{2}\)
\(=4-\frac{1}{2}=\frac{7}{2}\)
~ Hok tốt ~
\(B=\left(3-\dfrac{1}{5}\right)\left(3-\dfrac{2}{5}\right)\cdot...\cdot\left(3-\dfrac{29}{5}\right)\)
\(=\left(3-\dfrac{15}{5}\right)\left(3-\dfrac{1}{5}\right)\cdot...\cdot\left(3-\dfrac{29}{5}\right)\)
\(=\left(3-3\right)\left(3-\dfrac{1}{5}\right)\cdot...\cdot\left(3-\dfrac{29}{5}\right)=0\)