Ta có: \(\frac{2011113}{2011114}=\frac{2011114-1}{2011114}=1-\frac{1}{2011114}\)

           \(\frac{2011114}{2011115}=\frac{2011115-1}{2011115}=1-\frac{1}{2011115}\)

Vì \(\frac{1}{2011114}>\frac{1}{2011115}\) nên \(1-\frac{1}{2011114}< 1-\frac{1}{2011115}\)

Suy ra: \(\frac{2011113}{2011114}< \frac{2011114}{2011115}\)

a) 

\(1+tan^2a=\frac{1}{cos^2a}\)           

\(1+3^2=\frac{1}{cos^2a}\)  

\(10=\frac{1}{cos^2a}\)     

\(cos^2a=\frac{1}{10}\) 

\(cosa=\pm\sqrt{\frac{1}{10}}=\pm\frac{1}{\sqrt{10}}\)    

\(sin^2a+cos^2a=1\)                                                             

\(sin^2a+\frac{1}{10}=1\)   

\(sin^2a=\frac{9}{10}\)     

\(sina=\pm\sqrt{\frac{9}{10}}=\pm\frac{3}{\sqrt{10}}\)   

Vì tan = 3 nên M có 2 trường hợp : 

TH1 : 

sin và cos cùng dương 

\(\Rightarrow M=\frac{\frac{1}{\sqrt{10}}+\frac{3}{\sqrt{10}}}{\frac{1}{\sqrt{10}}-\frac{3}{\sqrt{10}}}\)     

\(=\frac{\frac{4}{\sqrt{10}}}{-\frac{2}{\sqrt{10}}}\)                        

= -2 

TH2 : 

Cả sin và cos cùng âm 

\(\Rightarrow M=\frac{-\frac{1}{\sqrt{10}}+\left(-\frac{3}{\sqrt{10}}\right)}{-\frac{1}{\sqrt{10}}-\left(-\frac{3}{\sqrt{10}}\right)}\)            

=\(\frac{-\frac{4}{\sqrt{10}}}{\frac{2}{\sqrt{10}}}\)                 

= -2 

b) 

\(B=\frac{sin15+cos15}{cos15}-cot75\)         

=\(\frac{sin15}{cos15}+\frac{cos15}{cos15}-cot75\)          

=\(tan15+1-cot75\)     

=\(cot75+1-cot75\)    

= 1 

Chữ số thứ 2004 là chữ số 4

Từ 1 đến 5 có 5 số hạng

Ta có : 2004 : 5 dư 4

Vậy số thứ 2004 là chữ số 4

Từ \(\frac{3x+y}{47}=\frac{x+y}{-17}=\frac{-2}{x^2}=\frac{-xz^2-yz^2}{z^2+1}\)(1)

=> \(\frac{x+y}{-17}=\frac{-xz^2-yz^2}{z^2+1}\Rightarrow\frac{x+y}{-17}=\frac{-z^2\left(x+y\right)}{z^2+1}\)

=> (z² + 1)(x + y)  = 17z²(x + y)

=> z² + 1 = 17z²

=> 16z² = 1

=> \(z^2=\frac{1}{16}\Rightarrow\orbr{\begin{cases}z=\frac{1}{4}\\z=-\frac{1}{4}\end{cases}}\)

Từ (1) => \(\frac{3x+y}{47}=\frac{x+y}{-17}=\frac{3x+y-x-y}{47+17}=\frac{2x}{64}=\frac{x}{32}\)

Kết hợp với đề bài => \(\frac{x}{32}=\frac{-2}{x^2}\Rightarrow x^3=-64\Rightarrow x=-4\)

\(\frac{3x+y}{47}=\frac{x+y}{-17}\Rightarrow-17\left(3x+y\right)=47\left(x+y\right)\)

=> - 51x - 17y = 47x + 47y

=> -51x - 47x = 17y + 47y

=> -98x = 64y

=> -49x = 32y

=> -49 x (-4) = 32y

=> 196 = 32y

=> y = 6,125

Vậy các cặp (x;y;z) thỏa mãn là (-4 ;  6,125 ; -1/4) ; (-4 ; 6,125 ; 1/4)

a) đkxđ: \(a>0;a\ne1\)

Ta có:

\(P=\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(1-\frac{1}{\sqrt{a}}\right)\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)

\(P=\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\frac{\sqrt{a}-1}{\sqrt{a}}.\frac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(P=\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}+\frac{2a+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)

\(P=\frac{2\sqrt{a}\left(\sqrt{a}+1\right)+2a+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)

\(P=\frac{2a+2\sqrt{a}+2a+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)

\(P=\frac{4a+2\sqrt{a}+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)

b) \(P=7\)

\(\Leftrightarrow\frac{4a+2\sqrt{a}+2}{\left(\sqrt{a}+1\right)\sqrt{a}}=7\)

\(\Leftrightarrow4a+2\sqrt{a}+2=7a+7\sqrt{a}\)

\(\Leftrightarrow3a+5\sqrt{a}-2=0\)

\(\Leftrightarrow\left(3a-\sqrt{a}\right)+\left(6\sqrt{a}-2\right)=0\)

\(\Leftrightarrow\left(3\sqrt{a}-1\right)\sqrt{a}+2\left(3\sqrt{a}-1\right)=0\)

\(\Leftrightarrow\left(3\sqrt{a}-1\right)\left(\sqrt{a}+2\right)=0\)

Mà \(\sqrt{a}+2\ge2\left(\forall a\right)\)

\(\Rightarrow3\sqrt{a}-1=0\Leftrightarrow3\sqrt{a}=1\)

\(\Leftrightarrow\sqrt{a}=\frac{1}{3}\Rightarrow a=\frac{1}{9}\)

Ta có : x⁴ + x² + 1

= x⁴ + x² + x² + 1 - x²

= (x² + 1)² - x²

= (x² + 1 - x)(x² + 1 + x)

x⁴ + x² + 1

= x⁴ + 2x² + 1 - x²

= ( x² + 1 )² - x²

= ( x² - x + 1 )( x² + x + 1 )

a) đkxđ: \(x\ne\pm1\)

Ta có:

\(P=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)\div\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)

\(P=\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\div\frac{x-1+x\left(x+1\right)+2}{\left(x-1\right)\left(x+1\right)}\)

\(P=\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\div\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)

\(P=\frac{4x}{\left(x-1\right)\left(x+1\right)}\div\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\)

\(P=\frac{4x}{\left(x+1\right)^2}\)

b) Ta có: \(x=\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

=> \(P=\frac{4\left(\sqrt{3}+1\right)}{\left(\sqrt{3}+1+1\right)^2}=\frac{4\sqrt{3}+4}{\left(\sqrt{3}+2\right)^2}=\frac{4+4\sqrt{3}}{7+4\sqrt{3}}\)

c) \(P=-3\)

\(\Leftrightarrow\frac{4x}{\left(x+1\right)^2}=-3\)

\(\Leftrightarrow-3\left(x^2+2x+1\right)=4x\)

\(\Leftrightarrow-3x^2-6x-3=4x\)

\(\Leftrightarrow3x^2+10x+3=0\)

\(\Leftrightarrow\left(3x^2+x\right)+\left(9x+3\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{3}\\x=-3\end{cases}}\)