Tìm x biết
a) \(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}\)
b) \(\frac{11}{x^2+7}+\frac{23}{x^2+19}=3-\frac{7}{y+3}\)
Tìm x biết
a) \(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}\)
b) \(\frac{11}{x^2+7}+\frac{23}{x^2+19}=3-\frac{7}{y+3}\)
1 giải phương trình
\(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\frac{x+2010}{2014}+\frac{x+2013}{2011}\)
2 . \(\frac{x^2+2x+1}{x^2+2x+2}+\frac{x^2+2x+2}{x^2+2x+3}=\frac{7}{6}\)
Bài 1 :
Ta có :
\(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\frac{x+2010}{2014}+\frac{x+2013}{2011}\)
\(\Rightarrow\left(\frac{x+2011}{2013}+1\right)+\left(\frac{x+2012}{2012}+1\right)=\left(\frac{x+2010}{2014}+1\right)\)
\(+\left(\frac{x+2013}{2011}+1\right)\)
\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}=\frac{x+4024}{2014}+\frac{x+4024}{2011}\)
\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}-\frac{x+4024}{2014}-\frac{x+4024}{2011}=0\)
\(\Rightarrow\left(x+4024\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)=0\)
\(\Rightarrow x+4024=0\)
\(\Rightarrow x=-4024\)
Bài 2 :
Đặt \(x^2+2x+1=a\Rightarrow a=\left(x+1\right)^2\ge0\)
=> Phương trình trở thành
\(\frac{a}{a+1}+\frac{a+1}{a+2}=\frac{7}{6}\)
\(\Rightarrow\frac{a}{a+1}.6\left(a+1\right)\left(a+2\right)+\frac{a+1}{a+2}.6\left(a+1\right)\left(a+2\right)=\frac{7}{6}.6\left(a+1\right)\left(a+2\right)\)
\(\Rightarrow6a\left(a+2\right)+6\left(a+1\right)^2=7\left(a+1\right)\left(a+2\right)\)
\(\Rightarrow12a^2+24a+6=7a^2+21a+14\)
\(\Rightarrow5a^2+3a-8=0\)
\(\Rightarrow\left(a-1\right)\left(5a+8\right)=0\)
Vì \(a\ge0\Rightarrow a=1\)
\(\Rightarrow x^2+2x+1=1\)
\(x^2+2x=0\)
\(\Rightarrow x\left(x+2\right)=0\)
\(\Rightarrow x\in\left\{-2,0\right\}\)
bài 1: tìm x biết: a) \(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}\)
b)\(\frac{11}{x^2+7}+\frac{23}{x^2+19}=3-\frac{7}{x^2+3}\)
bài 2: tìm x sao cho:a) 2x-3 có gía trị dương b) 13-5x có giá trị âm c) \(\frac{x+3}{2x-1}>0\)
d) \(\frac{x+7}{x+3}< 1\)
NHỚ CÓ LỜI GIẢI NHA, MÌNH SẼ TÍCH CHO BẠN NÀO NHANH NHẤT NHA (thứ 6 mình nộp rồi, nhanh lên nha các bạn)
x+7/2010+x+6/2011=x+5/2012+x+4/2013
((x+7/2010)-1)+((x+6/2011)-1)=(x+5/2012)-1)+(x+4/2013)-1)
x+2017/2010+x+2017/2011-x+2017/2012-x+2017/2013=0
x+2017(1/2010+1/2011-1/2012-1/2013)=0
x+2017=0(vì 1/2010+1/2011-1/2012-1/2013<0)
x=-2017
vậy.......
tk mk nha bn
Tìm số hữu tỉ x biết:
a) \(\frac{x+4}{2009}+\frac{x+3}{2010}=\frac{x+2}{2011}+\frac{x+1}{2012}\)
b) \(\frac{x-2011}{2010}+\frac{x-2011}{2011}+\frac{x-2011}{2012}=\frac{x-2011}{2013}+\frac{x-2011}{2014}\)
a) \(\frac{x+4}{2009}+1+\frac{x+3}{2010}+1=\frac{x+2}{2011}+1+\frac{x+1}{2012}\)
\(\frac{x+4+2009}{2009}+\frac{x+3+2010}{2010}=\frac{x+2+2011}{2011}+\frac{x+2+2012}{2012}\)
\(\frac{x+2013}{2009}+\frac{x+2013}{2010}-\frac{x+2013}{2011}-\frac{x+2013}{2012}=0\)
\(\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\) (1)
Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)
Nên biểu thức (1) xảy ra khi \(x+2013=0\)
\(x=-2013\)
b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\) (2)
Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)
Nên biểu thức (2) xảy ra khi \(x-2011=0\)
\(x=2011\)
tìm x: (làm đầy đủ nhaaa)
a) 2x. (x-\(\frac{1}{3}\)) = 0
b)\(\frac{x+8}{2007}+\frac{x+7}{2008}=\frac{x+6}{2009}+\frac{x+5}{2010}\)
c)\(\frac{11}{x^2+7}+\frac{23}{x^2+19}=3-\frac{7}{x^2+3}\)
Giải phương trình
\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\)
\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2013}-1\right)+\left(\frac{x-2}{2012}-1\right)+\left(\frac{x-3}{2011}-1\right)=\left(\frac{x-4}{2010}-1\right)+\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-6}{2008}-1\right)\)
\(\Leftrightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2013}{2011}=\frac{x-2014}{2010}+\frac{x-2014}{2009}+\frac{x-2014}{2008}\)
\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
tự làm nốt~
kudo shinichi làm sai ở chỗ:
\(\frac{x-2013}{2011}\)phải là \(\frac{x-2014}{2011}\)mới đúng nhé
Tìm x biết:
a) \(^{2^x+2^{x+1}+2^{x+2}+2^{x+3}=480}\)
b) \(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right).x=\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{2}{2011}+\frac{1}{2012}\)
a)
\(2^x\left(1+2+2^2+2^3\right)=480\)
\(2^x.15=480\Rightarrow2^x=\frac{480}{15}=32=2^5\Rightarrow x=5\)
Chính Xác 100% là X=5
k cho mink nhé các pạn
Tìm x :
a) \(\text{}\frac{x+5}{2011}+\frac{x+4}{2012}=\frac{x+3}{2013}+\frac{x+2}{2014}\)
b) \(\frac{5x-1.45}{6}+\frac{5x-1.45}{7}+\frac{5x-1.45}{8}=\frac{5x-1.45}{9}+\frac{5x-1.45}{10}\)
a, Bạn cộng mỗi tỉ số với 1 rồi chuyển vế phải sang vế trái, ta được:
(x+2016)(1/2011 +1/2012 -1/2013 -1/2014) =0
Ta thấy thừa số thứ hai lớn hơn 0 nên x+2016=0
Vậy x=-2016
b, Bạn chuyển vế phải sang vế trái, ta có:
(5x-1,45)(1/6 +1/7 +1/8 -1/9 +1/10)=0
Thừa số thứ 2 lớn hơn 0 do đó: 5x -1,45 =0
5x =1,45
x =0,29
Vậy x =0,29
Mong bạn hiểu cách giải của mình.
Chúc bạn học tốt.
Giải phương trình sau:
\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\)