\(x^2+\left(x+1\right)^2=y^4+\left(y+1\right)^4\Leftrightarrow x^2+\left(x^2+2x+1\right)=y^4+\left(y^4+4y^3+6y^2+4y+1\right)\)\(\Leftrightarrow x^2+x=y^4+2y^3+3y^2+2y\Leftrightarrow x^2+x+1=\left(y^2+y+1\right)^2\)

\(\text{⋄}\)Xét \(x\ge0\)thì \(\hept{\begin{cases}\left(x^2+x+1\right)-x^2=x+1>0\\\left(x^2+x+1\right)-\left(x+1\right)^2=-x\le0\end{cases}}\Rightarrow x^2< x^2+x+1\le\left(x+1\right)^2\)\(\Rightarrow x^2+x+1=\left(x+1\right)^2\Leftrightarrow x=0\Rightarrow y\in\left\{0;-1\right\}\)(Do x nguyên và \(x^2+x+1\)là số chính phương)

\(\text{⋄}\)Xét \(x=-1\)thì \(y\in\left\{0;-1\right\}\)

\(\text{⋄}\)Xét \(x< -1\)thì \(\hept{\begin{cases}\left(x^2+x+1\right)-x^2=x+1< 0\\\left(x^2+x+1\right)-\left(x+1\right)^2=-x>0\end{cases}}\Rightarrow\left(x+1\right)^2< x^2+x+1< x^2\)(Không có nghiệm nguyên) 

Vậy ta có 4 cặp nghiệm nguyên (x,y) = {(0;0) ; (0;-1) ; (-1;0) ; (-1;-1)}

a) Với \(x\ne1\)ta có:

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

\(=\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x}{x^2+x+1}-\frac{1}{x-1}\right].\frac{2}{x-1}\)

\(=\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right].\frac{2}{x-1}\)

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

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

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

b) \(A=\frac{2}{3}\)\(\Leftrightarrow\frac{2}{x^2+x+1}=\frac{2}{3}\)

\(\Leftrightarrow x^2+x+1=3\)\(\Leftrightarrow x^2+x-2=0\)

\(\Leftrightarrow x^2-x+2x-2=0\)\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\left(ktmĐKXĐ\right)\\x=-2\left(tmĐKXĐ\right)\end{cases}}\)

Vậy \(A=\frac{2}{3}\)\(\Leftrightarrow x=-2\)

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

Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

\(\Rightarrow x^2+x+1\ge\frac{3}{4}\forall x\)\(\Rightarrow\frac{1}{x^2+x+1}\le\frac{4}{3}\forall x\)

\(\Rightarrow\frac{2}{x^2+x+1}\le\frac{8}{3}\forall x\)\(\Rightarrow A\le\frac{8}{3}\)

Dấu " = " xảy ra \(\Leftrightarrow x+\frac{1}{2}=0\)\(\Leftrightarrow x=-\frac{1}{2}\)( thỏa mãn ĐKXĐ )

Vậy \(maxA=\frac{8}{3}\Leftrightarrow x=-\frac{1}{2}\)

a, \(\left(2x+1\right)\left(x^2+2\right)=0\)

TH1 : \(x=-\frac{1}{2}\); TH2 : \(x^2=-2\)vô lí vì \(x^2\ge0\forall x;-2< 0\)

b, \(\left(x^2+4\right)\left(7x-3\right)=0\)

TH1 : \(x^2=-4\)vô lí vì \(x^2\ge0\forall x;-4< 0\)

TH2 : \(x=\frac{3}{7}\)

c, \(\left(x^2+x+1\right)\left(6-2x\right)=0\)

TH1 : \(x^2+x+1\ne0\)vì \(x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

TH2 : \(2x=6\Leftrightarrow x=3\)

d, \(\left(8x-4\right)\left(x^2+2x+2\right)=0\)

TH1 : \(x=\frac{1}{2}\)

TH2 : \(x^2+2x+2\ne0\)vì \(x^2+2x+1+1=\left(x+1\right)^2+1>0\)

dễ

vãi                       b b b b b b

Vì x = 5 là nghiệm của phương trình nên 

Thay x = 5 vào biểu thức trên ta được : 

\(15-2=5a+18\)

\(\Leftrightarrow5a+18=13\Leftrightarrow5a=-5\Leftrightarrow a=-1\)

Vậy a = -1 

thay x=5  vào phương trình ta được:

    3.5-2=a.5+18

\(\Rightarrow\)15-2=5a+18

\(\Rightarrow\)13=5a+18

\(\Rightarrow\)5a=18-13

\(\Rightarrow\)5a=5

\(\Rightarrow\)a=1

          Vâỵ tham số a của phương trình là 5

                              OK  Xong rồi                                 

làm TT

Vì x = 4 là nghiệm của phương trình

nên thay x = 4 vào phương trình trên ta được :

\(-20-5=4a-9\)

\(\Leftrightarrow4a-9=-25\Leftrightarrow4a=-16\Leftrightarrow a=-4\)

Vậy a = -4

\(\left(2x-1\right)^2+2.\left(2x-1\right).\left(x+1\right)+\left(x+1\right)^2\)

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

\(=\left(3x\right)^2\)

\(=9x^2\)

\(4x^2-4x-5\left|2x-1\right|-5=0\)

\(\Leftrightarrow-5\left|2x-1\right|=5-4x^2+4x\)

\(\Leftrightarrow\left|2x-1\right|=\frac{-4x^2+4x+5}{-5}\)

\(\Leftrightarrow\left|2x-1\right|=\frac{4x\left(x-1\right)}{5}-1\)

TH1 : \(2x-1=\frac{4x\left(x-1\right)}{5}-1\Leftrightarrow2x=\frac{4x\left(x-1\right)}{5}\)

\(\Leftrightarrow10x=4x^2-4x\Leftrightarrow14x-4x^2=0\)

\(\Leftrightarrow-2x\left(2x-7\right)=0\Leftrightarrow x=0;x=\frac{7}{2}\)

TH2 : \(2x-1=-\left(\frac{4x\left(x-1\right)}{5}-1\right)\Leftrightarrow2x-1=-\frac{4x\left(x-2\right)}{5}+1\)

\(\Leftrightarrow2x-2=-\frac{4x\left(x-2\right)}{5}\Leftrightarrow10x-10=-4x^2+8x\)

\(\Leftrightarrow2x-10+4x^2=0\Leftrightarrow2\left(2x^2+x-5\ne0\right)=0\)tự chứng minh 

Vậy tập nghiệm của phương trình là S = { 0 ; 7/2 }

bằng2

1+1=2
#hỏi ngu thía!...:((
 

\(\frac{1}{x^2}+\frac{1}{9y^2}\ge2\sqrt{\frac{1}{x^2}.\frac{1}{9y^2}}=\frac{2}{3xy}=\frac{2}{3}\)

Dấu \(=\)xảy ra khi \(\hept{\begin{cases}\frac{1}{x^2}=\frac{1}{9y^2}\\xy=1\end{cases}}\Rightarrow\hept{\begin{cases}x=\sqrt{3}\\y=\frac{1}{\sqrt{3}}\end{cases}}\).