Question

Bài1:Tính giá trị biểu thức sau:

A=\(\left(6:\frac{3}{5}-1\frac{1}{6}x\frac{6}{7}\right):\left(4\frac{1}{5}x\frac{10}{11}+5\frac{2}{11}\right)\)

Bài 2: Tính giá trị biểu thức:

B= \(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2003}\right)x\left(1-\frac{1}{2004}\right)\)

ai xong sẽ có tích , phải làm giải từng bước ra nhé!

Answer 1

Bài 2:

\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)

\(=\frac{1}{2004}\)

Answer 2

Bài 2

=1/2 x 2/3 ... x 2003/2004

=1/2004

Answer 3

2.

1/2004

sudy well

study well

Answer 4

Bài 1:

\(A=\left(6:\frac{3}{5}-1\frac{1}{6}.\frac{6}{7}\right):\left(4\frac{1}{5}.\frac{10}{11}+5\frac{2}{11}\right)\)

\(=\left(10-\frac{7}{6}.\frac{6}{7}\right):\left(\frac{21}{5}.\frac{10}{11}+\frac{57}{11}\right)\)

\(=\left(10-1\right):\left(\frac{42}{11}+\frac{57}{11}\right)\)

\(=9:9\)

\(=1\)

Answer 5

Bài 2:

\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}....\frac{2002}{2003}.\frac{2003}{2004}\)

\(=\frac{1}{2004}\)