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Có :
A = (2014/1-x + 2014/1+x) + 4028/1+x^2 + 8056/1+x^4 + 16112/1+x^8 + 2,1314
= 4028/1-x^2 + 4028/1+x^2 + 8056/1+x^4 + 16112/1+x^8 + 2,1314
= 8056/1-x^4 + 8056/1+x^4 + 16112/1+x^8 + 2,1314
= 16112/1-x^8 + 16112/1+x^8 + 2,1314
= 32224/1-x^16 + 2,1314
Tk mk nha
Đề bài là gì vậy bạn
Sửa lại đề đi rùi báo cho mk để mk làm cho
Nhớ đó nha
\(\dfrac{1}{x+1}\)-\(\dfrac{5}{x-2}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\)\(\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}\)-\(\dfrac{5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\)x-2-5(x+1)=15
\(\Leftrightarrow\) x-2-5x-5=15
\(\Leftrightarrow\)x-5x=15+2+5
\(\Leftrightarrow\)-4x=22
\(\Leftrightarrow\)x=-\(\dfrac{11}{2}\)
vậy
a.
\(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)
\(\Leftrightarrow\dfrac{x+1}{2}+\dfrac{x+3}{4}=3-\dfrac{x+2}{3}\)
\(\Leftrightarrow\dfrac{\left(x+1\right).6}{12}+\dfrac{\left(x+3\right).3}{12}=\dfrac{36}{12}-\dfrac{\left(x+2\right).4}{12}\)
\(\Leftrightarrow6x+6+3x+9=36-4x-8\)
\(\Leftrightarrow9x+15=28-4x\)
\(\Leftrightarrow9x+4x=28-15\)
\(\Leftrightarrow13x=13\)
\(\Leftrightarrow x=1\)
a) \(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+3\left(x+3\right)}{12}=\dfrac{36-4\left(x+2\right)}{12}\)
\(\Leftrightarrow6\left(x+1\right)+3\left(x+3\right)=36-4\left(x+2\right)\)
\(\Leftrightarrow6x+6+3x+9=36-4x-8\)
\(\Leftrightarrow9x+15=-4x+28\)
\(\Leftrightarrow9x+4x=28-15\)
\(\Leftrightarrow13x=13\)
\(\Leftrightarrow x=1\)
Vậy ................................
\(\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}=5\)
\(\Leftrightarrow\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-5=0\)
\(\Leftrightarrow\dfrac{x}{2012}-1+\dfrac{x+1}{2013}-1+\dfrac{x+2}{2014}-1+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-1=0\)
\(\Leftrightarrow\dfrac{x-2012}{2012}+\dfrac{x-2012}{2013}+\dfrac{x-2012}{2014}+\dfrac{x-2012}{2015}+\dfrac{x-2012}{2016}=0\)
\(\Leftrightarrow\left(x-12\right).\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x-12=0\)
\(\Leftrightarrow x=12\)
a: ĐKXĐ: x<>1; x<>2; x<>-2; x<>-1
\(P=\dfrac{2017x+2017-2016x+2016-2014x-2016}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2015x+2017}{x^2-4}\)
1,
\(A=\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x^2+x-2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2-4}{\left(x-2\right)\left(x+2\right)}\)
\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)
2.
\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
3.
Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)
4.
\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)
\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)
5.
\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)
\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)
\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1+\dfrac{x+3}{2010}+1=\dfrac{x-1}{2014}+1+\dfrac{x-2}{2015}+1+\dfrac{x-3}{2016}+1\)
=>x+2013=0
hay x=-2013
\(\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1+\dfrac{x+3}{2010}+1=\dfrac{x-1}{2014}+1+\dfrac{x-2}{2015}+1+\dfrac{x-3}{2016}+1\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2022}+\dfrac{1}{2011}+\dfrac{2}{2010}-\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}\ne0\right)=0\Leftrightarrow x=-2013\)
a: \(P=\left(\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+2\right)\left(x-1\right)}-\dfrac{x}{x+1}\right):\dfrac{x-1+x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\left(\dfrac{x+1}{x-1}-\dfrac{x}{x+1}\right)\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{2x}\)
\(=\dfrac{x^2+2x+1-x^2+x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{2x}=\dfrac{3x+1}{2x}\)
b: 1/P-x+1/8>=1
=>\(\dfrac{2x}{3x+1}-\dfrac{x+1}{8}>=1\)
=>\(\dfrac{16x-3x^2-4x-1-24x-8}{8\left(3x+1\right)}>=0\)
=>\(\dfrac{-3x^2-24x-9}{8\left(3x+1\right)}>=0\)
=>\(\dfrac{x^2+8x+3}{3x+1}< =0\)
TH1: x^2+8x+3<=0 và 3x+1>0
=>x>-1/3 và \(-4-\sqrt{13}< =x< =-4+\sqrt{13}\)
=>Loại
TH2: x^2+8x+3>=0 và 3x+1<0
=>x<-1/3 và (x<=-4-căn 13 hoặc x>=-4+căn 13)
=>x<=-4-căn 13
\(A=\dfrac{x-2014}{\dfrac{x^2-4x+4-x^2-2x-1}{\left(x+1\right)\left(x-2\right)}:\dfrac{x^2-4x+4+x^2+2x+1}{\left(x+1\right)\left(x-2\right)}}\)
\(=\dfrac{x-2014}{\dfrac{-6x+3}{\left(x+1\right)\left(x-2\right)}\cdot\dfrac{\left(x+1\right)\left(x-2\right)}{2x^2-2x+5}}\)
\(=\left(x-2014\right)\cdot\dfrac{2x^2-2x+5}{-6x+3}\)
Để A>=0 thì \(\left(x-2014\right)\left(-6x+3\right)>=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-2014\right)< =0\)
=>1/2<x<=2014
bạn có thể ghi rõ câu hỏi k
A.Mình ghi thiếu đề bài đó .Phần tiếp theo là rút gọn và tính giá trị của Akhi x=1,1