ai giải hộ mk cái đi!!

Đặt \(A=\sqrt{2003}+\sqrt{2005};B=2\sqrt{2004}\)

\(A^2=2003+2005+2\sqrt{\left(2003.2005\right)}\)

\(=4008+2\sqrt{\left[2004-1\left(2004+1\right)\right]}\)

\(=3008+2\sqrt{\left(2004^2-1\right)}< 2.2004+2\sqrt{\left(2004^2\right)}=4.2004=B^2\\ \Rightarrow A< B\)

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

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

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

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

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

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

mik cũng ko chắc đâu nếu sai thì thôi nhé:
 

ĐKXĐ : −x2+6x−9≥0−x2+6x−9≥0

⇔⇔−(−x2+6x−9)≤0−(−x2+6x−9)≤0

⇔⇔x2−6x+9≤0x2−6x+9≤0

⇔⇔(x−3)2≤0(x−3)2≤0

Mà (x−3)≥0(x−3)≥0

Suy ra : (x−3)2=0(x−3)2=0

⇔⇔x−3=0x−3=0

⇔⇔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 \(S=\left\{\dfrac{5\pm\sqrt{37}}{2}\right\}\)