Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{x-28-124}{2011}+\frac{x-124-2011}{28}+\frac{x-2011-28}{124}\)\(=3\)
\(\Leftrightarrow\frac{x-152}{2011}+\frac{x-2135}{28}+\frac{x-2039}{124}=3\)\(\Leftrightarrow\left(\frac{x-152}{2011}-1\right)+\left(\frac{x-2135}{28}-1\right)+\left(\frac{x-2039}{124}-1\right)=3-1-1-1\)
\(\Leftrightarrow\frac{x-2163}{2011}+\frac{x-2163}{28}+\frac{x-2163}{124}=0\)
\(\Leftrightarrow\left(x-2163\right)\left(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\right)=0\)
Vì \(\left(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\right)>0\)
\(\Rightarrow x-2163=0\)
\(\Leftrightarrow x=2163\)
Vậy \(x=2163\)
=10/17+-5/13+7/17+-8/13+-11/25
=1+-1+-11/25
=11/25
b ,
= 1+-2+3+-4+5+-6+......2011+-2012
=1+1+1+1+1.........+-2012
=1.2011+-2012
=2011+-2012
=-1
2.
2/3-x = 5/4
=>x=2/3-5/4
=>x=-7/12
b,
[124-(20-4x)]:30+7=11
=>124-(20-4x)] :30 =11-7
=>[124-(20-4x)]:30=4
=>124-(20-4x)=4x30
=>124-(20-4x)=120
=>20-4x=120-124
=>20-4x=4
=>4x=20-4
=>4x=16
=>x=16:4
=>x=4
bài 1
a, ghép cặp phân số 10/17 + 7/17 và 5/13 + -8/13
b, 1-2+3-4+5-6+.....+2011-2012
= ( 1-2) + ( 3-4) + ( 5-6) +....+(2011-2012)
= (-1) + (-1) + (-1) + ....+ (-1)
< từ 1 đến 2012 có 2012 số số hạng, suy ra có 1006 cặp mà mỗi cặp có giá trị = (-1) Suy ra tổng trên = (-1) * 1006=-1006
\(\frac{2011^{2011}+2}{2011^{2011}-1}=\frac{2011^{2011}-1+3}{2011^{2011}-1}=1+\frac{3}{2011^{2011}-1}\)
\(\frac{2011^{2011}}{2011^{2011}-3}=\frac{2011^{2011}-3+3}{2011^{2011}-3}=1+\frac{3}{2011^{2011}-3}\)
Vì 1 = 1 và \(\frac{3}{2011^{2011}-1}>\frac{3}{2011^{2011}-3}\)nên \(1+\frac{3}{2011^{2011}-1}>1+\frac{3}{2011^{2011}-3}\)hay \(A>B\)
Vậy \(A>B\)
-So sánh B với A
B>1 ( tử lớn hơn mẫu ) => B> 20112011+2/20112011-3+2 => B> 20112011+2/20112011+ -1 =A
=> B > A => A < B.
(Bài trên dựa vào cách sau : Nếu a/b > 1 => a/b > a+m/b+m (m thuộc N*) và nếu a/b <1 thì ngược lại)
\(\left(x-2011\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\left(x-2011\right)\cdot\frac{2}{9}=\frac{16}{9}\)
\(x-2011=8\)
\(x=2019\)
\(\frac{x-2011}{12}+\frac{x-2011}{20}+\frac{x-2011}{30}+\frac{x-2011}{42}+\frac{x-2011}{56}+\frac{x-2011}{72}=\frac{16}{9}\)
\(\left(x-2011\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\left(x-2011\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\frac{16}{9}\)
\(\left(x-2011\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2011\right)\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2011\right)\frac{2}{9}=\frac{16}{9}\)
\(x-2011=8\Rightarrow x=2019\)
A)=>x + 1/2011+ x + 1/2012 - x + 1/2013 - x + 1/2014 =0
<=>(x + 1) . ( 1/2011+ 1/2012-1/2013 - 1/2014) = 0
=>x + 1 = 0 (vì 1/2011+ 1/2012 - 1/2013 - 1/2014 khác 0)
=>x = -1
vậy x = -1
B)x-100/24+x-98/26+x-96/28=3
<=>x - 100/24 - 1 + x - 98/26 - 1 + x - 96/28 =0
<=>x - 124/24 + x - 124/26 + x - 124/28 = 0
<=>(x-124).( 1/24 + 1/26 + 1/28 ) = 0
mà 1/24 + 1/26 +1/28 khác 0
=>x - 124 = 0
<=>x = 124
Ta có: \(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}=\frac{x+4}{2011}+\frac{x+5}{2010}+\frac{x+6}{2009}\)
\(\Rightarrow\frac{x+1}{2014}+1+\frac{x+2}{2013}+1+\frac{x+3}{2012}+1=\frac{x+4}{2011}+1+\frac{x+5}{2010}+1+\frac{x+6}{2009}+1\)
\(\Rightarrow\frac{2015+x}{2014}+\frac{2015+x}{2013}+\frac{2015+x}{2012}=\frac{2015+x}{2011}+\frac{2015+x}{2010}+\frac{2015+x}{2009}\)
\(\Rightarrow\left(2015+x\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
=> 2015 + x = 0
=> x = -2015
\(\frac{x-28-124}{2011}+\frac{x-124-2011}{28}+\frac{x-2011-28}{124}=3\)
<=> \(\left(\frac{x-152}{2011}-1\right)+\left(\frac{x-2135}{28}-1\right)+\left(\frac{x-2039}{124}-1\right)=0\)
<=> \(\frac{x-2163}{2011}+\frac{x-2163}{28}+\frac{x-2163}{124}=0\)
<=> \(\left(x-2163\right)\left(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\right)=0\)
<=> \(x-2163=0\)(vì \(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\ne0\)
<=> \(x=2163\)
\(\frac{x-28-124}{2011}+\frac{x-124-2011}{28}+\frac{x-2011-28}{124}=3\)
<=> \(\frac{x-28-124}{2011}+\frac{x-124-2011}{28}+\frac{x-2011-28}{124}-3=0\)
<=> \(\left(\frac{x-28-124}{2011}-1\right)+\left(\frac{x-124-2011}{28}-1\right)+\left(\frac{x-2011-28}{124}-1\right)=0\)
<=> \(\frac{x-28-124-2011}{2011}+\frac{x-124-2011-28}{28}+\frac{x-2011-28-124}{124}=0\)
<=> \(\frac{x-2163}{2011}+\frac{x-2163}{28}+\frac{x-2163}{124}=0\)
<=> \(\left(x-2163\right)\left(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\right)=0\)
Vì \(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\ne0\)
=> \(x-2163=0\)
=> \(x=2163\)