\(\frac{2}{7}\)+ \(\frac{5}{14}\)+\(\frac{1}{7}\)+\(\frac{3}{14}\)
\(\frac{1995×1997-1}{1996×1995+1994}\)
469 × 281 + 489 × 719
Giúp mình nha . Mai mình nội bài rồi :((
Những câu hỏi liên quan
65 + x + 95784 = 101569
87 : x + 93 : x - 66 : x = 38
x : 2 + x × 4 = 7
( x : 42 ) + 457 = 625
Bài 4:
\(\frac{2}{7}\)+ \(\frac{5}{14}\)+ \(\frac{1}{7}\)+ \(\frac{3}{14}\)
\(\frac{1995×1997-1}{1996×1995+1994}\)
469 × 281 + 469 × 719
( 158 + 154 ) × 32 - 284 × 35
( 873 - 586 ) × 25 + ( 315 + 289 )
Bài 2 :
65 + x + 95784 = 101569
87 : x + 93 : x - 66 : x = 38
x : 2 + x × 4 = 7
( x : 42 ) + 457 = 625
Bài 3 :
7/2 + 5/14 + 1/7 + 3/14
1995 × 1997 - 1 / 1996 × 1995 + 1994
469 × 281 + 469 × 718
( 158 + 154 ) × 32 - 284 × 35
( 873 - 586 ) × 25 + ( 315 + 289 )
Bài 3 :
65 + x + 95784 = 101569
87 : x + 93 : x - 66 : x = 38
x : 2 + x × 4 = 7
( x : 42 ) + 457 = 625
Bài 4 : 7/2 + 5/14 + 1/7 + 3/14
19995 × 1997 - 1 / 1996 × 1995 + 1994
469 × 281 + 469 × 719
Bài 2 :
( 158 + 154 ) × 32 - 284 × 35
( 873 - 586 ) × 25 + ( 315 + 289 )
Bài 3 :
65 + x + 95784 = 101569
87 : x + 93 : x - 66 : x = 38
x : 2 + x × 4 = 7
( x : 42 ) + 457 = 625
Bài 4 :
7/2 + 5/14 + 1/7 + 3/14
19995 × 1997 - 1 / 1996 × 1995 + 1994
469 × 281 + 469 × 719
So sánh:
E=\(\frac{1995^3+1}{1995^2-1994}\)và F=\(\frac{1996^3-1}{^{1996^2}+1997}\)
Giúp mk nha
# Thầy Soobin
Số nào lớn hơn trong hai số : \(E=\frac{1995^3}{1995^2-1994};F=\frac{1996^3-1}{1996^2+1997}\)
\(F=\frac{1996^3-1}{1996^2+1997}=\frac{\left(1996-1\right)\left(1996^2+1996+1\right)}{1996^2+1997}=\frac{1995.\left(1996^2+1997\right)}{1996^2+1997}=1995\)
E = \(\frac{1995^3}{1995^2-1994}=\frac{1995^3+1-1}{1995^2-1994}=\frac{\left(1995+1\right)\left(1995^2-1995+1\right)-1}{1995^2-1994}\)
=\(\frac{1996\left(1995^2-1994\right)-1}{1995^2-1994}=1996-\frac{1}{1995^2-1994}\)
Vì \(1995^2-1994>0\) => \(\frac{1}{1995^2-1994}-1\) => \(1996-\frac{1}{1995^2-1994}>1996-1\)
HAy E > F
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\(\frac{x-1991}{9}+\frac{x-1993}{7}+\frac{x-1995}{5}+\frac{x-1997}{3}+\frac{x-1999}{1}=\frac{x-9}{1991}+\frac{x-7}{1993}+\frac{x-5}{1995}+\frac{x-3}{1997}+\frac{x-1}{1999}\)
Ta có : \(\frac{x-1991}{9}+\frac{x-1993}{7}+\frac{x-1995}{5}+\frac{x-1997}{3}+\frac{x-1999}{1}\)\(=\frac{x-9}{1991}+\frac{x-7}{1993}+\frac{x-5}{1995}+\frac{x-3}{1997}+\frac{x-1}{1999}\)
\(\Rightarrow\left(\frac{x-1991}{9}-1\right)+\left(\frac{x-1993}{7}-1\right)+\left(\frac{x-1995}{5}-1\right)+\left(\frac{x-1997}{3}-1\right)+\left(\frac{x-1999}{1}-1\right)\)
\(=\left(\frac{x-9}{1991}-1\right)+\left(\frac{x-7}{1993}-1\right)+\left(\frac{x-5}{1995}-1\right)+\left(\frac{x-3}{1997}-1\right)+\left(\frac{x-1}{1999}\right)\)
\(\Rightarrow\frac{x-2000}{9}+\frac{x-2000}{7}+\frac{x-2000}{5}+\frac{x-2000}{3}\)
\(=\frac{x-2000}{1991}+\frac{x-2000}{1993}+\frac{x-2000}{1995}+\frac{x-2000}{1997}+\frac{x-2000}{1999}\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)=0\)
\(\Rightarrow\left(x-2000\right)\left[\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\right]=0\)
Vì \(\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\ne0\)
=> x - 2000 = 0
=> x = 2000
Tìm x, biết
\(\frac{x+5}{1995}+\frac{x+4}{1996}+\frac{x+3}{1997}=\frac{x+1995}{5}+\frac{x+1996}{4}+\frac{x+1997}{3}\)
ta có \(1+\frac{x+5}{1995}+1+\frac{x+4}{1996}+1+\frac{x+3}{1997}=1+\frac{x+1995}{5}+1+\frac{x+1996}{4}+1+\frac{x+1997}{3}\)
\(=\frac{x+2000}{1995}+\frac{x+2000}{1996}+\frac{x+2000}{1997}=\frac{x+2000}{5}+\frac{x+2000}{4}+\frac{x+2000}{3}\)
\(=\left(x+2000\right)\left(\frac{1}{1995}+\frac{1}{1996}+\frac{1}{1997}\right)=\left(x+2000\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}\right)\) (1)
Xét \(\frac{1}{1995}+\frac{1}{1996}+\frac{1}{1997}\ne\frac{1}{5}+\frac{1}{4}+\frac{1}{3}vàx+2000=x+2000\) (2)
từ \(\left(1\right)\Leftrightarrow x+2000=0\) ( để (1) là đúng )
\(\Rightarrow x=2000\)
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Bình luận (0)
\(\frac{1}{2}\times\frac{1}{3}\times\frac{1}{4}\times...\times\frac{1995}{1996}\times\frac{1996}{1997}\)
