\(x+\dfrac{1}{x}=a\Rightarrow\left(x+\dfrac{1}{x}\right)^3=a^3\)

\(\Leftrightarrow x^3+\dfrac{1}{x^3}+3.x^2.\dfrac{1}{x}+3x.\dfrac{1}{x^2}=a^3\)

\(\Leftrightarrow x^3+\dfrac{1}{x^3}+3x\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)=a^3\)

\(\Leftrightarrow x^3+\dfrac{1}{x^3}+3a=a^3\)

\(\Rightarrow x^3+\dfrac{1}{x^3}=a^3-3a\)

có thì có thật , nhưng cho bạn kiểu j

là 29 bạn ạ
bây giờ mình off nên k thể ghi cách giải ra được 
Lần sau tớ sẽ tl rõ ràng ra
Tick tớ nhé
 

\(a,14x^2+3x+9=14\left(x^2+\dfrac{14}{3}x+\dfrac{49}{9}\right)-\dfrac{605}{9}\ge\dfrac{-605}{9}\)(câu a âm mà)

Câu b cũng thế !

\(x^2+8x+16=\left(x+4\right)^2\ge0\)

Vậy ....

Bài 1:ĐKXĐ:

\(x\ne0;x\ne-1\)

\(a,\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)

\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x+1\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)}-\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}=0\)\(\Rightarrow x^2+3x+x^2-2x+x-2-2x^2-2x=0\)\(\Leftrightarrow-2=0\)(vô lí)

\(A=x^2-10x+26\)

\(=\left(x^2-10x+25\right)+1\)

\(=\left(x-5\right)^2+1\ge1\)

Vậy \(Min_A=1\) khi \(x-5=0\Rightarrow x=5\)

\(B=x^2+7x+10=\left(x^2+7x+\dfrac{49}{4}\right)-\dfrac{9}{4}=\left(x+\dfrac{7}{2}\right)^2-\dfrac{9}{4}\ge\dfrac{-9}{4}\)Vậy \(Min_B=\dfrac{-9}{4}\) khi \(x+\dfrac{7}{2}=0\Rightarrow x=\dfrac{-7}{2}\)

\(C=4x^2+8x+15=4\left(x^2+2x+1\right)+11=4\left(x+1\right)^2+11\ge11\)Vậy \(Min_C=11\) khi \(x+1=0\Rightarrow x=-1\)

\(D=3x^2-7x+20=3\left(x^2-\dfrac{7}{3}x+\dfrac{49}{36}\right)+\dfrac{191}{12}=3\left(x-\dfrac{7}{6}\right)^2+\dfrac{191}{12}\ge\dfrac{191}{12}\)Vậy \(Min_D=\dfrac{191}{12}\) khi \(x-\dfrac{7}{6}=0\Rightarrow x=\dfrac{7}{6}\)

\(E=x^2-4xy+5y^2-22y+8\)

\(=\left(x^2-4xy+4y^2\right)+\left(y^2-22y+121\right)-113\)\(=\left(x-2y\right)^2+\left(y-11\right)^2-113\ge-113\)

Vậy \(Min_E=-113\) khi \(\left[{}\begin{matrix}x-2y=0\\x-11=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}11-2y=0\\x=11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2y=11\\x=11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=11\end{matrix}\right.\)

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

\(=x^3+1-x^3-3x-1=-3x\)

\(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\) \(=\left(x-y+z\right)^2-2\left(x-y+z\right)\left(z-y\right)+\left(z-y\right)^2\)\(=\left(x-y+z-z+y\right)^2=x^2\)

x + y = x . y

=> x = x . y - y = y.(x - 1)

Do đó \(\frac{x}{y}=\frac{y.\left(x-1\right)}{y}=x-1\)

=> x + y = x - 1

=> y = x - 1 - x

=> y = -1

=> x = -1 . (x - 1) = -x - (-1) = -x + 1

=> x - (-x) = 1

=> 2x = 1

=> x = 0,5

Vậy x = 0,5 và y = -1

\(a,\left(x+5\right)^2-\left(x-5\right)^2-20x+2\)

\(=\left(x+5+x-5\right)\left(x+5-x+5\right)-20x+2\)\(=2x.10-20x+2\)

\(=20x-20x+2=2\)

Vậy....

\(b,\left(x-3\right)\left(x+5\right)-\left(x-1\right)^2\)

\(=x^2-5x+3x-15-x^2+2x-1\)

\(=-15-1=-16\)

Vậy ....