Tính nhanh
A=13/29+1/2+13/29×1/3-13/28×5/6
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Giải:
\(A=\dfrac{13}{29}.\dfrac{1}{2}+\dfrac{13}{29}.\dfrac{1}{3}-\dfrac{13}{29}.\dfrac{5}{6}\) (Sửa đề)
\(\Leftrightarrow A=\dfrac{13}{29}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{5}{6}\right)\)
\(\Leftrightarrow A=\dfrac{13}{29}.0\)
\(\Leftrightarrow A=0\)
Vậy ...
Sửa đề: \(\dfrac{13}{29}.\dfrac{1}{2}+\dfrac{13}{29}.\dfrac{1}{3}+\dfrac{13}{29}.\dfrac{5}{6}\)
Giải:
Đặt:
\(A=\dfrac{13}{29}.\dfrac{1}{2}+\dfrac{13}{29}.\dfrac{1}{3}+\dfrac{13}{29}.\dfrac{5}{6}\)
\(\Leftrightarrow A=\dfrac{13}{29}\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{5}{6}\right)\)
\(\Leftrightarrow A=\dfrac{13}{29}.0\)
\(\Leftrightarrow A=0\)
Vậy ...
a)29 . (19 – 13) – 19 . (29 – 13)
= 29 . 6 – 19 . 16
= 174 – 304
= –130.
b)31.(-18)+31.(-81)-31
= 31. [-18 + (-81) - 1 ]
= 31. (-100)
= -3100
c)(7.3-3):(-6)
(7.3-3):(-6)
= (21-3):(-6)
= 18 : (-6)
= 3
d)72:[(-6).2+4)]
= 72 : ( -12 + 4 )
= 72 : -8
= -9
\(E=\frac{\frac{-6}{7}+\frac{6}{13}-\frac{6}{29}}{\frac{9}{7}-\frac{9}{13}+\frac{9}{29}}=\frac{-6.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{29}\right)}{9.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{29}\right)}=\frac{-6}{9}=\frac{-2}{3}\)
\(F=\frac{\frac{2}{15}-\frac{2}{21}+\frac{2}{39}}{0,25-\frac{5}{28}+\frac{5}{52}}=\frac{\frac{2}{3}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{13}\right)}{\frac{1}{4}-\frac{5}{28}+\frac{5}{52}}\)
\(=\frac{\frac{2}{3}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{13}\right)}{\frac{5}{4}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{13}\right)}=\frac{2}{3}:\frac{5}{4}=\frac{2}{3}.\frac{4}{5}=\frac{8}{15}\)
Bạn soyeon ơi! Hình như câu đầu bạn làm sai rồi thì phải. Họ ra đề là + đến - nhưng bạn làm - đến +.
a: \(\dfrac{5}{13}\left(\dfrac{6}{29}-\dfrac{26}{39}\right)-\dfrac{6}{29}\cdot\left(\dfrac{5}{13}-\dfrac{29}{6}\right)\)
\(=\dfrac{5}{13}\cdot\dfrac{6}{29}-\dfrac{5}{13}\cdot\dfrac{26}{39}-\dfrac{6}{13}\cdot\dfrac{5}{13}+\dfrac{6}{29}\cdot\dfrac{29}{6}\)
\(=\dfrac{-5}{39}\cdot2+1=1-\dfrac{10}{39}=\dfrac{29}{39}\)
b: \(\dfrac{1\cdot198+2\cdot197+3\cdot196+...+198\cdot1}{1\cdot2+2\cdot3+...+198\cdot199}\)
\(=\dfrac{1\left(199-1\right)+2\left(199-2\right)+...+198\cdot\left(199-198\right)}{1\left(1+1\right)+2\left(1+2\right)+...+198\left(1+198\right)}\)
\(=\dfrac{199\left(1+2+...+198\right)-\left(1^2+2^2+...+198^2\right)}{\left(1+2+...+198\right)+\left(1^2+2^2+...+198^2\right)}\)
\(=\dfrac{199\cdot\dfrac{198\cdot199}{2}-\dfrac{198\cdot\left(198+1\right)\cdot\left(2\cdot198+1\right)}{6}}{198\cdot\dfrac{199}{2}+\dfrac{198\left(198+1\right)\left(2\cdot198+1\right)}{6}}\)
\(=\dfrac{3\cdot198\cdot199^2-198\cdot199\cdot397}{6}:\dfrac{3\cdot198\cdot199+198\cdot199\cdot397}{6}\)
\(=\dfrac{198\cdot199\left(3\cdot199-397\right)}{198\cdot199\left(3+397\right)}\)
\(=\dfrac{200}{400}=\dfrac{1}{2}\)
a) 26 + 27 +28 +29 +30 +31 +32 +33 = (26+33) + (27+32) + (28+31) + (29+30)
= 59 +59 +59 +59 = 59.4 = 236
b) 2 + 4 + 6 + 8 + … + 1998 + 2000
Số các số hạng là: (2000 – 2):2 + 1 = 1000 (số hạng)
Tổng có giá trị là: (2000 + 2).1000 : 2 = 1001000
c) 5 + 9 +13 +… + 1997 + 2001
Số các số hạng là: (2001 – 5) : 5 +1 = 500 (số hạng)
Tổng có giá trị là: (5 + 2001).500 : 2 = 501500
Sửa đề: \(\dfrac{13}{29}.\dfrac{1}{2}+\dfrac{13}{29}.\dfrac{1}{3}-\dfrac{13}{29}.\dfrac{5}{6}\)
Giải:
Đặt:
\(A=\dfrac{13}{29}.\dfrac{1}{2}+\dfrac{13}{29}.\dfrac{1}{3}-\dfrac{13}{29}.\dfrac{5}{6}\)
\(\Leftrightarrow A=\dfrac{13}{29}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{5}{6}\right)\)
\(\Leftrightarrow A=\dfrac{13}{29}.0\)
\(\Leftrightarrow A=0\)
Vậy ...