tìm x
a) (x+1/5)mũ 2 +17/25=26/25
b)-1 5/27 - (3x-7/9)mux3= -24/27
c) -2/7<x/3<11/4
tính nhanh
a)(1999/2011-2011/1999)-(-12/1999-12/2011)
b) 67/111+2/33-15/117).(1/3-1/4-1/12)
Tính nhanh:
\(\left(\frac{1999}{2011}-\frac{2011}{1999}\right)-\left(\frac{-12}{1999}-\frac{12}{2011}\right)\)
\(\left(\frac{1999}{2011}-\frac{2011}{199}\right)-\left(-\frac{12}{1999}-\frac{12}{2001}\right)\)
=\(\frac{1999}{2011}-\frac{2011}{1999}+\frac{12}{1999}+\frac{12}{2011}\)
=\(\left(\frac{1999}{2011}+\frac{12}{2011}\right)+\left(-\frac{2011}{1999}+\frac{12}{1999}\right)\)
=\(\frac{2011}{2011}+\frac{1999}{1999}\)
=1+1
=2
\(\left(\frac{1999}{2011}-\frac{2011}{1999}\right)-\left(\frac{-12}{1999}-\frac{12}{2011}\right)\)
=\(\frac{1999}{2011}-\frac{2011}{1999}+\frac{12}{1999}+\frac{12}{2011}\)
=\(\left(\frac{1999}{2011}+\frac{12}{2011}\right)+\left(\frac{-2011}{1999}+\frac{12}{1999}\right)\)
=1-1
=-2
tính nhanh: ( phân tích rõ ràng nha, bài kiểm tra hk2 đó,please,mk sẽ like cho bài nào rõ ràng nhất )
a) -9/10*5/14*1/10*(-9/2)+1/7*(-9/10)
b)(1999/2011-2011/1999)-(-12/1999-12/2011)
A.-9/10*5/14*1/10*(-9/2)+1/7*(-9/10)
=-9/10*(5/14+1/7)*1/10*(-9/2)
=-9/10*1/2*1/10*(-9/2)
=-9/20*1/10*(-9/10)
=-9/200*-9/10
=81/2000
B. (1999/2011-2011/1999)-(-12/1999-12/2011)
=1999/2011-2011/1999+12/1999+12/2011
=(1999/2011+12/2011)-(2011/1999-12/1999)
= 1-1
=0
Tính
a, \(\left(\dfrac{1999}{2011}-\dfrac{2011}{1999}\right)-\left(\dfrac{-12}{1999}-\dfrac{12}{2011}\right)\)
b, \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).....\left(1-\dfrac{1}{2015}\right)\)
Tính nhanh : a, 5/6 +11/12+19/20+29/30+41/42+55/56+71/72+89/90
b, ính tổng của 10 phân số trog phép cộng sau :
1/2 + 5/6+ 11/12 + 19/20 + 29/30 + 41/42 + 55/56 + 71/72 + 89/90 + 109/110
a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)=7,6\)
b) Bạn làm tương tự.
mnbvcxzasdf
Tính nhanh
a, 1/2 + 5/6 + 11/ 12 + 19/20 + 29/30 + 41/42 + 55/56 + 71/72 + 89/90
b, (1+1/2).(1+1/3)+......+(1+1/2011)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+....+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)=9-\frac{9}{10}=\frac{81}{10}\)
b.=3/2.4/3....2012/2011
=3.4....2012/2.3....2011=2012/2=1006
Bài 1:
a) tính nhanh :
A= 3333*6666
B=1/90+1/72+1/56+1/42+1/30+1/20+1/12+1/6+1/2
b) Cho M=1/2+1/3+1/4+...1/2000 và N=1/1999+2/1998+3/1997+...+1999/1 .Tìm tỉ số của M và N
Cho \(f(x)=\frac{x^3}{1-3x+3x^2}\)Hãy tính giá trị của biểu thức sau
\(A=f(\frac{1}{2012})+f(\frac{2}{2012})+...+f(\frac{2010}{2012})+f(\frac{2011}{2012})\)
\(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3x+3x^2}=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)
Ta có \(f\left(x\right)+f\left(1-x\right)=1\) khi đó
\(A=\left[f\left(\frac{1}{2012}\right)+f\left(\frac{2011}{2012}\right)\right]+...+\left[f\left(\frac{1005}{2012}\right)+f\left(\frac{1007}{2012}\right)\right]+f\left(\frac{1006}{2012}\right)\)
\(=1+1+...+1+f\left(\frac{1}{2}\right)=1005+\frac{\left(\frac{1}{2}\right)^3}{1-3.\frac{1}{2}+3.\left(\frac{1}{2}\right)^2}=1005+\frac{1}{2}=\frac{2011}{2}\)
Ta có: \(F\left(x\right)=\frac{x^3}{1-3x+3x^2}\)
\(\Leftrightarrow F\left(1-x\right)=1-\frac{x^3}{1-3x+3x^2}\)
\(=\frac{1-3x+3x^2-x^3}{1-3x+3x^2}\)
\(=\frac{\left(1-x\right)^3}{1-3x+3x^2}\)
Ta có: \(F\left(x\right)+F\left(1-x\right)\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3x+3x^2}\)
\(=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)
\(\Leftrightarrow F\left(\frac{1}{2012}\right)+F\left(\frac{2011}{2012}\right)=1\)
...
\(F\left(\frac{1005}{2012}\right)+F\left(\frac{1007}{2012}\right)=1\)
Do đó: \(A=F\left(\frac{1}{2012}\right)+F\left(\frac{2}{2012}\right)+...+F\left(\frac{2010}{2012}\right)+F\left(\frac{2011}{2012}\right)\)
\(=\left[F\left(\frac{1}{2012}\right)+F\left(\frac{2011}{2012}\right)\right]+\left[F\left(\frac{2}{2012}\right)+F\left(\frac{2010}{2012}\right)\right]+...+F\left(\frac{1006}{2012}\right)\)
\(=1+1+...+F\left(\frac{1}{2}\right)\)
\(=1005+\left[\left(\frac{1}{2}\right)^3:\left(1-3\cdot\frac{1}{2}+3\cdot\frac{1}{4}\right)\right]\)
\(=1005+\left[\frac{1}{8}:\left(1-\frac{3}{2}+\frac{3}{4}\right)\right]\)
\(=1005+\left(\frac{1}{8}:\frac{1}{4}\right)\)
\(=1005+\frac{1}{2}=\frac{2011}{2}\)
\(f\left(1-x\right)=\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}=\frac{1-3x+3x^2-x^3}{3x^2-3x+1}\)
\(\Rightarrow f\left(x\right)+f\left(1-x\right)=\frac{x^3}{3x^2-3x+1}+\frac{1-3x+3x^2-x^3}{3x^2-3x+1}=\frac{3x^2-3x+1}{3x^2-3x+1}=1\)
Do đó:
\(A=f\left(\frac{1}{2012}\right)+f\left(\frac{2011}{2012}\right)+...+f\left(\frac{1005}{2012}\right)+f\left(\frac{1007}{2012}\right)+f\left(\frac{1}{2}\right)\)
\(=f\left(\frac{1}{2012}\right)+f\left(1-\frac{1}{2012}\right)+...+f\left(\frac{1005}{2012}\right)+f\left(1-\frac{1005}{2012}\right)+f\left(\frac{1}{2}\right)\)
\(=1+1+...+1+f\left(\frac{1}{2}\right)\)
\(=1005+f\left(\frac{1}{2}\right)=1005+\frac{\left(\frac{1}{2}\right)^3}{1-3.\left(\frac{1}{2}\right)+3.\left(\frac{1}{2}\right)^2}=...\)
tìm đa thức bậc 4 f(x) thoả mãn: f(x)-f(x-1)=x3. trình bày sơ lược cách giải. từ đó lập công thức tính tổng quát Sn=1+23+33+43+....+n3 và tính chính xác giá trị của Sn vs n=2011
phantuananh mấy tháng nữa chắc mk cũng chả cần nữa rồi
