Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a.\frac{19}{5}\cdot\frac{4}{7}+\frac{3}{7}\cdot\frac{19}{5}-\frac{4}{5}\)
\(=\frac{19}{5}\cdot\left(\frac{4}{7}+\frac{3}{7}\right)-\frac{4}{5}\)
\(=\frac{19}{5}\cdot1-\frac{4}{5}\)
\(=\frac{19}{5}-\frac{4}{5}=\frac{15}{5}=3\)
\(b.2\frac{2}{7}\cdot5\frac{2}{5}+\frac{16}{7}\cdot1\frac{3}{5}+\frac{1}{2}\)
\(=\frac{16}{7}\cdot\frac{27}{5}+\frac{16}{7}\cdot\frac{8}{5}+\frac{1}{2}\)
\(=\frac{16}{7}\cdot\left(\frac{27}{5}+\frac{8}{5}\right)+\frac{1}{2}\)
\(=\frac{16}{7}\cdot7+\frac{1}{2}\)
\(=16+\frac{1}{2}=\frac{33}{2}\)
\(c.\frac{3}{7}\cdot3\frac{3}{4}-\frac{3}{7}\cdot\frac{5}{4}-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\frac{15}{4}-\frac{3}{7}\cdot\frac{5}{4}-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\left(\frac{15}{4}-\frac{5}{4}\right)-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\frac{5}{2}-\frac{1}{4}\)
\(=\frac{15}{14}-\frac{1}{4}=\frac{23}{28}\)
Chú ý: \(\cdot:\times\)
3/25 x ( 15/7 - 2/7 ) + 3/7 x 1/25
= 3/25 x 13/7 + 3/7 x 1/25
= (3 x 13/7 + 3/7 ) x 1/25
= 42/7 x 1/25
= 6 x 1/25
= 6/25
\(\dfrac{3}{25}\times\dfrac{15}{7}+\dfrac{3}{7}\times\dfrac{1}{25}-\dfrac{2}{7}\times\dfrac{3}{25}\)
\(=\dfrac{3}{25}\times\left(\dfrac{15}{7}-\dfrac{2}{7}\right)+\dfrac{3}{7}\times\dfrac{1}{25}\)
\(=\dfrac{3}{25}\times\dfrac{13}{7}+\dfrac{3}{7}\times\dfrac{1}{25}\)
\(=\dfrac{3\times13}{25\times7}+\dfrac{3\times1}{7\times25}\)
\(=\dfrac{39}{175}+\dfrac{3}{175}\)
\(=\dfrac{39+3}{175}\)
\(=\dfrac{42}{175}\)
\(=\dfrac{6}{25}\)
\(B=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x...x\frac{2002}{2003}x\frac{2003}{2004}\)
\(B=\frac{1x2x3x4x...x2002x2003}{2x3x4x5x...x2003x2004}\)
Rút gọn các thừa số ở tử và mẫu ta được:
\(B=\frac{1}{2004}\)
Đ/S:\(\frac{1}{2004}\)
Ta có:
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{2002}{2003}.\frac{2003}{2004}\)
\(=\frac{1.2....2002.2003}{2.3....2003.2004}\)
Đơn giản hết sẽ là:
\(=\frac{1}{2004}\)
\(\frac{5}{7}+y\times\frac{1}{2}=\frac{8}{7}:\frac{3}{2}\)
=> \(\frac{5}{7}+y\times\frac{1}{2}=\frac{8}{7}\times\frac{2}{3}\)
=> \(\frac{5}{7}+y\times\frac{1}{2}=\frac{16}{21}\)
=> \(y\times\frac{1}{2}=\frac{16}{21}-\frac{5}{7}=\frac{1}{21}\)
=> \(y=\frac{1}{21}:\frac{1}{2}=\frac{1}{21}\cdot2=\frac{2}{21}\)
5/7 + Y x 1/2 = 8/7 : 3/2
=> 5/7 + Y x 1/2 = 16/21
=> Y x 1/2 = 16/21 - 5/7
=> Y x 1/2 = 1/21
=> Y = 1/21 : 1/2
=> Y = 2/21
Hok tốt ^^
2/3 . ( X + 3/2 ) - 1/3 ( x + 1 ) = 2/5
=> 2/3 . (X + 1) + 2/3 . 1/2 - 1/3 (x+1) = 2/5
=> (2/3 - 1/3) . (x + 1) + 1/3 = 2/5
=> 1/3 . (x + 1) + 1/3 = 2/5
=> 1/3 (x+1 + 1) = 2/5
=> x + 2 = 2/5 : 1/3
=> x + 2 = 6/5
=> x = 6/5 - 2
=> x = -4/5
Câu 2
( 3/5 : X - 7 ) ( 7/3 . X - 5 ) = 0
- TH1: 3/5 : X - 7 = 0
=> 3/5 : X = 7
=> X = 3/35
- TH2: 7/3 . X - 5 = 0
=> 7/3 . X = 5
=> X = 15/7
Vậy x = 15/7 hoặc 3/35
câu 1 :
\(\dfrac{2}{3}.\left(x+\dfrac{3}{2}\right)-\dfrac{1}{3}.\left(x+1\right)=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{2}{3}.x-1-\dfrac{1}{3}.x-\dfrac{1}{3}=\dfrac{2}{5}\)
\(\Rightarrow\left(\dfrac{2}{3}.x-\dfrac{1}{3}.x\right)-\left(1+\dfrac{1}{3}\right)=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{3}.x-\dfrac{4}{3}=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{3}.x=\dfrac{26}{15}\)
\(\Rightarrow x=\dfrac{26}{5}\)
Câu 2 :
\(\left(\dfrac{3}{5}:x-7\right)\left(\dfrac{7}{3}.x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{5}:x-7=0\\\dfrac{7}{3}.x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{3}{5}:x=7\\\dfrac{7}{3}.x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{35}\\x=\dfrac{7}{15}\end{matrix}\right.\)