Tính giá trị biểu thức:
\(B=\left(1-\frac{1}{^{2^2}}\right).\left(1-\frac{1}{3^2}\right).....\left(1-\frac{1}{2010^2}\right)\)
Tính giá trị của biểu thức: \(\left(-2\right)\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2010}\right)\)
Tính giá trị biểu thức:
(-2).\(\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right)...\left(-1\frac{1}{2009}\right).\left(-1\frac{1}{2010}\right)\)
=(-2)(-3/2)(-4/3)...(-1010/2009)(-2011/2010)
=2011
Tính giá trị biểu thức
a, \(A=2010^{2010}.\left(7^{10}:7^8-3.16-2^{2010}:2^{2010}\right)\)
b, \(B=\left(\frac{1}{7}+\frac{1}{23}-\frac{1}{1009}\right):\left(\frac{1}{23}+\frac{1}{7}-\frac{1}{1009}+\frac{1}{7}.\frac{1}{23}.\frac{1}{1009}\right)+1:\left(30.1009-160\right)\)
a,
A = 20102010.[710:78-3.16-22010:22010]
= 20102010.[72-48-1]
= 20102010.0 = 0
b,
B = 1
\(A=2010^{2010}.\left[7^{10}:7^8-3.16-2^{2010}:2^{2010}\right]\)
\(A=2010^{2010}.\left[7^2-48-1\right]\)
\(A=2010^{2010}.0\)
\(Vay\)\(A=0\)
A= 20102010(72 - 48 - 1)
A=20102010(49-48-1)
A=20102010.0
A=0
Tính giá trị biểu thức:
B=\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).......\left(\frac{1}{98^2}-1\right).\left(\frac{1}{99^2}-1\right)\)
\(B=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{98^2}-1\right)\left(\frac{1}{99^2}-1\right)\)
\(=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right).....\left(1-\frac{1}{98^2}\right)\left(1-\frac{1}{99^2}\right)\)
\(=\frac{3}{2^2}.\frac{8}{3^2}......\frac{9603}{98^2}.\frac{9800}{99^2}\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{97.99}{98^2}.\frac{98.100}{99^2}\)
\(=\frac{1.2.4...97.98}{2.3....98.99}.\frac{3.4...99.100}{2.3....98.99}\)
\(=\frac{1}{99}.\frac{100}{2}\)
\(=\frac{50}{99}\)
bn viết sai 1 chỗ nhưng ko s ^^ tks nhoa
Tính giá trị biểu thức: A=\(\frac{\left(1+17\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right)....\left(1+\frac{17}{19}\right)}{\left(1+19\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right)...\left(1+\frac{19}{17}\right).}\)
Cho \(f\left(x\right)=\frac{x^3}{1-3x+3x^2}\)hãy tính giá trị biểu thức
\(A=f\left(\frac{1}{2012}\right)+f\left(\frac{2}{2012}\right)+...+f\left(\frac{2010}{2012}\right)+f\left(\frac{2011}{2012}\right)\)
Ta xét : \(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{3x^2-3x+1}=\frac{\left(x+1-x\right)\left(x^2+x^2-2x+1+x^2-x\right)}{3x^2-3x+1}=\frac{3x^2-3x+1}{3x^2-3x+1}=1\)
Áp dụng ta có :
\(A=\left[f\left(\frac{1}{2012}\right)+f\left(\frac{2011}{2012}\right)\right]+\left[f\left(\frac{2}{2012}\right)+f\left(\frac{2010}{2012}\right)\right]+...+\left[f\left(\frac{1006}{2012}\right)+f\left(\frac{1006}{2012}\right)\right]\)
\(=1+1+...+1\)(Có tất cả 1006 số 1)
\(=1006\)
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2012}\right)\)Tính giá trị biểu thức
Tính giá trị các biểu thức sau:
a) \({\left( {\frac{3}{4}} \right)^{ - 2}}{.3^2}{.12^0}\);
b) \({\left( {\frac{1}{{12}}} \right)^{ - 1}}.{\left( {\frac{2}{3}} \right)^{ - 2}}\);
c) \({\left( {{2^{ - 2}}{{.5}^2}} \right)^{ - 2}}:\left( {{{5.5}^{ - 5}}} \right)\).
a) \(\left(\dfrac{3}{4}\right)^{-2}\cdot3^2\cdot12^0=16\)
b) \(\left(\dfrac{1}{12}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-2}=27\)
c) \(\left(2^{-2}\cdot5^2\right)^{-2}:\left(5\cdot5^{-5}\right)=16\)
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!
Bài 2:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)
\(=\frac{1}{2004}\)