tính giá trị biểu thức \(\left(\frac{215}{2010}-\frac{120}{2011}\right)\)x \(\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)
Bài 1 : tính giá trị biểu thức :
a) \(\left(\frac{215}{2010}-\frac{120}{2011}\right)\)x \(\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)
b) ( 45 x 46 +47 x 48 ) ( 45 x 128 - 90 x 64 ) ( 2009 x 2010 + 2011 x 2012 )
Cho 3 số x y z thỏa mãn x + y + z = 2010 và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2010}\)
Tính giá trị biểu thức P= \(\left(x^{2007}+y^{2007}\right)\left(y^{2009}+z^{2009}\right)\left(z^{2011}+x^{2011}\right)\)
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)
\(=1006\)
Tính giá trị của biểu thức: \(\left(-2\right)\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2010}\right)\)
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:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)
\(=\frac{1}{2004}\)
Tính giá trị biểu thức :
\(A=\frac{\left(x^2+x-3\right)^{2011}}{\left(x^5+x^4-x^3-2\right)^{2011}}+\left(x^5+x^4-x^3+1\right)^{2011}\)
Với \(x=\frac{\sqrt{5}-1}{2}\)
Tính giá trị biểu thức:
\(B=\left(1-\frac{1}{^{2^2}}\right).\left(1-\frac{1}{3^2}\right).....\left(1-\frac{1}{2010^2}\right)\)
\(B=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)...\left(\frac{2010^2-1}{2010^2}\right)\)
\(B=\left(\frac{\left(2-1\right)\left(2+1\right)}{2^2}\right)...\left(\frac{\left(2010-1\right)\left(2010+1\right)}{2010^2}\right)\)
\(B=\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{2009.2011}{2010.2010}\)
\(B=\left(\frac{1}{2}.\frac{2}{3}...\frac{2009}{2010}\right)\left(\frac{3}{2}.\frac{4}{3}...\frac{2011}{2010}\right)\)
\(B=\frac{1}{2010}.\frac{2011}{2}\)
\(B=\frac{2011}{4020}\)
\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)
\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)
\(\frac{1}{2011}.x=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2009}{2010}.\frac{2010}{2011}\)
\(\frac{1}{2011}.x=\frac{1.2.3...2009.2010}{2.3.4...2010.2011}\)\(=\frac{1}{2011}\)
\(x=\frac{1}{2011}:\frac{1}{2011}=1\)
Vậy x=1
\(\frac{1}{2011}.x=\frac{1}{2}.\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)......\left(\frac{2010}{2011}\right)\)
\(\frac{1}{2011}.x=\frac{2}{4}.\left(\frac{4}{6}\right).\left(\frac{6}{8}\right).......\left(\frac{4018}{4020}\right).\left(\frac{4020}{4022}\right)\)
\(\frac{1}{2011}.x=\frac{2.4.6.8.....4018.4020}{4.6.8.10.....4020.4022}\)
\(\frac{1}{2011}.x=\frac{2}{4022}\)
\(\Rightarrow\)\(x=\frac{2}{4022}:\frac{1}{2011}=1\)
Ai thấy đún thì ủng hộ mink nha !!!
Thanks you very much !!
Chúc các bạn luôn học giỏi !!!
Tính giá trị của các biểu thức sau
1)\(\left(3\frac{5}{12}-1\frac{8}{9}\right):\left(0,25-2\frac{1}{12}\right)\)
2)\(\left(1\frac{4}{7}-2\frac{2}{5}\right):\left(1\frac{1}{8}-4\frac{3}{4}\right)\)