\(\dfrac{x-1}{5}+\dfrac{x+1}{7}+\left(x-1\right)\left(x+1\right)=\left(x+1\right)^2\)

\(\Leftrightarrow\dfrac{7\left(x-1\right)+5\left(x+1\right)}{35}+x^2-1=x^2+2x+1\)

\(\Leftrightarrow\dfrac{12x-2}{35}=2x+2\)

\(\Leftrightarrow\dfrac{6x-1}{35}=x+1\)

\(\Leftrightarrow35x+35=6x-1\)

\(\Leftrightarrow x=-\dfrac{36}{29}\)

mik cũng ko chắc đâu nếu sai thì thôi nhé:
 

ĐKXĐ : $-x^2+6x-9\ge 0$

$\iff$$-\left(-x^2+6x-9\right)\le 0$

$\iff$$x^2-6x+9\le 0$

$\iff$${\left(x-3\right)}^2\le 0$

Mà $\left(x-3\right)\ge 0$

Suy ra : ${\left(x-3\right)}^2=0$

$\iff$$x-3=0$

$\iff$$x=3$

ĐKXĐ: \(^{x^2}\)- 6x + 9 ≥0 với mọi x

\(ĐKXĐ:\left\{{}\begin{matrix}x\ge-1\\x^2-3x-1\ge0\end{matrix}\right.\)

Ta có \(\sqrt{x^3+1}=x^2-3x-1\)

\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=x^2-x+1-2\left(x+1\right)\)

Đặt \(\sqrt{x+1}=a;\sqrt{x^2-x+1}=b\left(a\ge0;b>0\right)\)

Khi đó ab = b² - 2a² 

<=> b² - ab - 2a² = 0 

<=> (b + a)(b - 2a) = 0 

<=> b - 2a = 0 (vì \(a\ge0;b>0\Rightarrow a+b>0\)) 

<=> b = 2a 

<=> \(\sqrt{x^2-x+1}=2\sqrt{x+1}\)

<=> \(x^2-x+1=4\left(x+1\right)\)

<=> \(x^2-5x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{37}}{2}\\x=\dfrac{5-\sqrt{37}}{2}\end{matrix}\right.\)(tm)

Vậy tập nghiệm \(S=\left\{\dfrac{5\pm\sqrt{37}}{2}\right\}\)

- Bổ sung đề: CMR \(\dfrac{1+3x}{1+y^2}+\dfrac{1+3y}{1+z^2}+\dfrac{1+3z}{1+x^2}\ge6\).

- Ta có: \(\dfrac{1+3x}{1+y^2}=\left(1+3x\right)-\dfrac{y^2\left(1+3x\right)}{1+y^2}\ge\left(1+3x\right)-\dfrac{y^2\left(1+3x\right)}{2y}=1+3x-\dfrac{y\left(1+3x\right)}{2}=1+3x-\dfrac{y}{2}-\dfrac{3xy}{2}\left(1\right)\)

- Tương tự, ta cũng có:

\(\dfrac{1+3y}{1+z^2}=1+3y-\dfrac{z}{2}-\dfrac{3yz}{2}\left(2\right)\), \(\dfrac{1+3z}{1+x^2}=1+3z-\dfrac{x}{2}-\dfrac{3zx}{2}\left(3\right)\)

- Lấy \(\left(1\right)+\left(2\right)+\left(3\right)\), ta được:

\(\dfrac{1+3x}{1+y^2}+\dfrac{1+3y}{1+z^2}+\dfrac{1+3z}{1+x^2}\ge3+\dfrac{5}{2}\left(x+y+z\right)-\dfrac{3}{2}\left(xy+yz+zx\right)=3+\dfrac{5}{2}\left(x+y+z\right)-\dfrac{3}{2}.3=\dfrac{5}{2}\left(x+y+z\right)-\dfrac{3}{2}\left(4\right)\)

- Mặt khác: \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)=3.3=9\Rightarrow x+y+z\ge3\left(5\right)\)

- Từ (4), (5) ta có:

\(\dfrac{1+3x}{1+y^2}+\dfrac{1+3y}{1+z^2}+\dfrac{1+3z}{1+x^2}\ge\dfrac{5}{2}.3-\dfrac{3}{2}=6\left(đpcm\right)\)

- Dấu "=" xảy ra khi \(x=y=z=1\)