a\(\left(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right):\frac{1}{4}:\frac{1}{6}\)
b, \(\frac{2006\cdot2005-1}{2004\cdot2006+2005}\)
ý thứ 2 giả cả 2 cách hộ mình nhé
Những câu hỏi liên quan
Bài 4 :
a) Tính giá trị của biểu thức :
\(A=\left(\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right)\cdot\frac{31}{50}\)
b) Chứng tỏ rằng : \(B=1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{3^2}-...-\frac{1}{2004^2}>\frac{1}{2004}\)
\(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\cdot\cdot\cdot\left(1+\frac{1}{2004\cdot2006}\right)\)
\(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{2004\cdot2006}\right)\)
\(=\frac{4}{1\cdot3}+\frac{9}{2\cdot4}+\frac{16}{3\cdot5}+...+\frac{420025}{2004\cdot2006}\)
\(=\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2005\cdot2005\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2004\cdot2006\right)}\)
\(=\frac{\left(2\cdot3\cdot4\cdot...\cdot2005\right)\left(2\cdot3\cdot4\cdot...\cdot2005\right)}{\left(1\cdot2\cdot3\cdot...\cdot2004\right)\left(3\cdot4\cdot5\cdot...\cdot2006\right)}\)
\(=\frac{2005\cdot2}{1\cdot2006}\)
\(=\frac{4010}{2006}\)
Đúng 0
Bình luận (0)
\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)...\left(1+\frac{1}{2004.2006}\right)\)
\(=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}....\frac{2004.2006+1}{2004.2006}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}....\frac{2005^2}{2004.2006}\)
\(=\frac{2.3....2005}{1.2....2004}.\frac{2.3...2005}{3.4....2006}\)
\(=2005.\frac{2}{2006}=\frac{2005}{1003}\)
Đúng 0
Bình luận (0)
\(=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right)...\left(1+\frac{1}{2004.2006}\right)\)
\(=\left(\frac{2.2}{1.3}\right).\left(\frac{3.3}{2.4}\right)....\left(\frac{2005.2005}{2004.2006}\right)\)
\(=\frac{\left(2.3.4....2005\right).\left(2.3.4....2005\right)}{\left(1.2.3...2004\right).\left(3.4.5....2006\right)}\)
\(=\frac{2005}{2006}\)
Đúng 0
Bình luận (0)
A=\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)........\left(1-\frac{1}{2004}\right)\)
\(B=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}\cdot4\frac{1}{2}-2\cdot2\frac{1}{3}\right):\frac{7}{4}\)
A = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)\)
A = \(\left(\frac{2}{2}-\frac{1}{2}\right).\left(\frac{3}{3}-\frac{1}{3}\right).\left(\frac{4}{4}-\frac{1}{4}\right)....\left(\frac{2004}{2004}-\frac{1}{2004}\right)\)
A = \(\frac{1}{2}\)x\(\frac{2}{3}.\)\(\frac{3}{4}....\)\(\frac{2003}{2004}\)
A = \(\frac{1}{2004}\)
Đúng 0
Bình luận (0)
\(\frac{2006\cdot2005-1}{2004\cdot2006+2005}\)
=2006×(2004+1)-1/2004×2006+2005
=2006×2004+2006×1-1/2004×2006+2005
=2006×2004+2005/2004×2006+2005
=1
Đúng 0
Bình luận (0)
Tính
a)\(\left(-\frac{1}{4}\right)^2+\frac{3}{8}\cdot\left(-\frac{1}{6}\right)-\frac{3}{16}:\left(-\frac{1}{2}\right)\)
b)\(-\frac{1}{2}:\left(1-\frac{3}{4}\right)^2-\frac{2}{3}:\frac{9}{8}-\left(\frac{9}{8}\right)^0\)
c)\(4\cdot\left(-\frac{1}{2}\right)^3+2\cdot\left(-\frac{1}{2}\right)^2-3\cdot\left(-\frac{1}{2}\right)+2006^0\)
a) \(\frac{\left(-1\right)}{4}^2+\frac{3}{8}.\left(\frac{-1}{6}\right)-\frac{3}{16}:\left(\frac{-1}{2}\right)=\left(\frac{-1}{4}\right)^2+\left(\frac{-3}{68}\right)-\left(\frac{-3}{8}\right)=\left(\frac{1}{16}\right)+\left(\frac{-3}{68}\right)-\left(\frac{-3}{8}\right)=\frac{5}{272}-\left(\frac{-3}{8}\right)=\frac{107}{272}\)
Đúng 0
Bình luận (0)
Thực hiện phép tính một cách hợp lí: 1) Afrac{636363cdot37-373737cdot63}{1+2+3+...+2006}2)B1frac{6}{41}cdotleft(frac{12+frac{12}{19}-frac{12}{37}-frac{12}{53}}{3+frac{1}{3}-frac{3}{37}-frac{3}{53}}:frac{4+frac{4}{17}+frac{4}{19}+frac{4}{2006}}{5+frac{5}{17}+frac{5}{19}+frac{5}{2006}}right)cdotfrac{124242423}{237373735}
Đọc tiếp
Thực hiện phép tính một cách hợp lí:
1) \(A=\frac{636363\cdot37-373737\cdot63}{1+2+3+...+2006}\)
2)\(B=1\frac{6}{41}\cdot\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2006}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2006}}\right)\cdot\frac{124242423}{237373735}\)
Mình nghĩ đề thế này mới tính hợp lí được
2 ) B = \(1\frac{6}{41}.\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2006}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2006}}\right).\frac{124242423}{237373735}\)
B = 47/41 . ( 12/3 : 4/5 ) . 123/235
B = 47/41 . ( 4 : 4/5 ) . 123/235
B = 47/41 . 5 . 123/235
B = \(\frac{47.5.123}{41.235}\)
B = 3
Đúng 0
Bình luận (0)
1 ) A = \(\frac{636363.37-373737.63}{1+2+3+...+2006}\)
A = \(\frac{63.10101.37-37.10101.63}{1+2+3+...+2006}\)
A = \(\frac{0}{1+2+3+...+2006}\)
A = 0
Đúng 0
Bình luận (0)
\(a.A=[\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}]+\frac{1890}{2005}+115\)
b.B=\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\cdot\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(42-5\frac{1}{3}\right)}\cdot\left(-1\frac{19}{93}\right)\right]\cdot\frac{31}{50}\)
Bài 1: Giả các phương trình sau:
a) left(x^2+xright)^2+4left(x^2+xright)12
b) frac{x+1}{2008}+frac{x+2}{2007}+frac{x+3}{2006}frac{x+4}{2005}+frac{x+5}{2004}+frac{x+6}{2003}
Cả nhà giúp mk nhé. Mai mk phải nộp bài rùi. Cảm ơn cả nhà trước nha
Đọc tiếp
Bài 1: Giả các phương trình sau:
a) \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
b) \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
==> Cả nhà giúp mk nhé. Mai mk phải nộp bài rùi. Cảm ơn cả nhà trước nha <==
b)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
<=>\(\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)\( \left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
mà \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)
nên phương trình đó xảy ra khi và chỉ khi x+2009=0
<=>x=-2009
Vậy phương trình có no là x=-2009
Đúng 0
Bình luận (0)
b) \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)\)=
\(\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\) \(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\)\(\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
\(\Leftrightarrow\) \(\left(x+2009\right)\)\(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\)= 0
\(\Leftrightarrow\)\(x+2009=0\)
( vì \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\) \(\ne0\))
\(\Leftrightarrow\) \(x=-2009\)
Vậy x = -2009
Đúng 0
Bình luận (0)
a)<=>(x2+x)2+2*(x2+x)*2+22=12+4
<=>(x2+x+2)2=16
<=>x2+x+2=4 hoặc x2+x+2=-4
*)x2+x+2=4 <=> x2+x-2=0 <=> x2+2x-x-2=0
<=>x(x+2)-(x+2)=0
<=>(x-1)(x+2)=0
<=>x-1=0 hoặc x+2=0
*1)x-1=0 <=>x=1
*2)x+2=0 <=>x=-2
*)x2+x+2=-4 <=> x2+x+6=0 <=> x2+2*1/2*x+1/4=-23/4 <=> (x+1/2)2=-23/4 (PTVN)
Vậy phương trình có tập no là S={1;-2}
Đúng 0
Bình luận (1)
Xem thêm câu trả lời
\(\left(1:2\frac{1}{3}+\frac{5}{14}\right):2\frac{1}{5}\) \(\frac{1}{2003}\cdot\left(1-\frac{1}{2004}\right)\cdot\left(1-\frac{1}{2006}\left(1-\frac{1}{2006}\right)\right)\cdot\)