X x 1515 + X x4545 =2
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\(=\dfrac{15}{23}\times\dfrac{69}{54}\times\dfrac{27}{15}\\ =\dfrac{3}{2}\)
Đáp án A
x : 5 + 1515 = 9938 × 2
x : 5 + 1515 = 19876
x : 5 = 19876 − 1515
x : 5 = 18361
x = 18361 × 5
x = 91805
Vậy x có giá trị bằng 91805.
a. \(\left(\dfrac{17}{2}+\dfrac{1}{2}+1\right).\dfrac{4}{7}\)
\(=\dfrac{19}{2}.\dfrac{4}{7}=\dfrac{38}{7}\)
b. \(\dfrac{2727}{2323}.\dfrac{69}{54}\)
\(=\dfrac{27}{23}.\dfrac{23.3}{27.2}=\dfrac{3}{2}\)
2. Chiều cao hình bình hành là: 1200 : 60 = 20 cm
a, \(\dfrac{17}{2}\) \(\times\) \(\dfrac{4}{7}\) + \(\dfrac{4}{7}\) \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{4}{7}\) b, \(\dfrac{1515}{2323}\) \(\times\) \(\dfrac{69}{54}\) \(\times\) \(\dfrac{2727}{1515}\)
= \(\dfrac{17}{2}\) \(\times\) \(\dfrac{4}{7}\) + \(\dfrac{4}{7}\) \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{4}{7}\) \(\times\) 1 = \(\dfrac{15}{23}\) \(\times\) \(\dfrac{23}{18}\) \(\times\) \(\dfrac{27}{15}\)
= \(\dfrac{4}{7}\) \(\times\) ( \(\dfrac{17}{2}\) + \(\dfrac{1}{2}\) + 1) = \(\dfrac{27}{18}\)
= \(\dfrac{4}{7}\) \(\times\) ( 9 + 1) = \(\dfrac{3}{2}\)
= \(\dfrac{4}{7}\times10\)
= \(\dfrac{40}{7}\)
Bài 2: Chiều cao của hình bình hành là:
1200 : 60 = 20 cm
Đổi 20 cm = 2 dm
Đáp số: 2 dm
Lúc nãy, cô còn dạy học nên giờ cô mới giảng cho em được nhé.
B = (1 - \(\dfrac{1}{2}\))\(\times\)(1 - \(\dfrac{1}{3}\))\(\times\)(1 - \(\dfrac{1}{4}\))\(\times\)(1-\(\dfrac{1}{5}\))\(\times\)...\(\times\)(1- \(\dfrac{1}{2003}\))\(\times\)(1-\(\dfrac{1}{2004}\))
B = \(\dfrac{2-1}{2}\)\(\times\)\(\dfrac{3-1}{3}\)\(\times\)\(\dfrac{4-1}{4}\)\(\times\)\(\dfrac{5-1}{5}\)\(\times\)...\(\times\)(\(\dfrac{2003-1}{2003}\))\(\times\)(\(\dfrac{2004-1}{2004}\))
B = \(\dfrac{1}{2}\)\(\times\)\(\dfrac{2}{3}\)\(\times\)\(\dfrac{3}{4}\)\(\times\)\(\dfrac{4}{5}\)\(\times\)...\(\times\)\(\dfrac{2002}{2003}\)\(\times\)\(\dfrac{2003}{2004}\)
B = \(\dfrac{2\times3\times4\times...\times2003}{2\times3\times4\times...\times2003}\)\(\times\) \(\dfrac{1}{2004}\)
B = \(\dfrac{1}{2004}\)
Tìm x :
\(\left(\dfrac{2}{3}x-\dfrac{4}{9}\right).\left(\dfrac{1}{2}+\dfrac{3}{7}:x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(\dfrac{2}{3}x-\dfrac{4}{9}\right)=0\\\left(\dfrac{1}{2}+\dfrac{3}{7}:x\right)=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{9}\\\dfrac{3}{7}:x=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{6}{7}\end{matrix}\right.\)
Vậy.....
Tính :
\(\dfrac{1313}{1515}+\left(-\dfrac{1011}{5055}\right)=\dfrac{13}{15}-\dfrac{1}{5}=\dfrac{2}{3}\)
ta rút gọn thì sẽ đc: 13/21 x 15/13 x 14/5=(13x15x14)/(21x13x5).rút gọn 13 ;7 và 5 trên cả tử và mẫu ta đc ps còn lại là 1/2
\(x\times1515+x\times4545=2\)
\(x\times\left(1515+4545\right)=2\)
\(x\times6060=2\)
\(x=\dfrac{2}{6060}=\dfrac{1}{3030}\)
\(x.1515+x.4545=2\)
\(\Rightarrow x.\left(1515+4545\right)=2\)
\(\Rightarrow x.6060=2\)
\(\Rightarrow x=2:6060\)
\(\Rightarrow x=\dfrac{1}{3030}\)