b) cách khác:

\(pt\Leftrightarrow11-x-4\sqrt{x+3}-2\sqrt{3-2x}=0\)

\(\Leftrightarrow3-2x-2\sqrt{3-2x}+1+x+3-4\sqrt{x+3}+4=0\)

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

\(\Leftrightarrow\sqrt{3-2x}-1=\sqrt{x+3}-2=0\)

\(\Leftrightarrow x=1\)

b liên hợp hoặc cosi, đặt ẩn cx đc

Đề sai. Sửa đề \(\sqrt{2059-x}+\sqrt{2035-x}+\sqrt{2154-x}=24\)            (1)

Điều kiện: \(x\le2035\)

\(\left(1\right)\Leftrightarrow\left(\sqrt{2059-x}-7\right)+\left(\sqrt{2035-x}-5\right)+\left(\sqrt{2154-x}-12\right)=0\)

\(\Leftrightarrow\frac{2010-x}{\sqrt{2059-x}+7}+\frac{2010-x}{\sqrt{2035-x}+5}+\frac{2010-x}{\sqrt{2154-x}+12}=0\)

\(\Leftrightarrow\left(2010-x\right)\left(\frac{1}{\sqrt{2059-x}+7}+\frac{1}{\sqrt{2035-x}+5}+\frac{1}{\sqrt{2154-x}+12}\right)=0\)

Ta thấy biếu thức \(\frac{1}{\sqrt{2059-x}+7}+\frac{1}{\sqrt{2035-x}+5}+\frac{1}{\sqrt{2154-x}+12}\)luôn dương nên \(2010-x=0\Leftrightarrow x=2010\)(TM)

Vậy ...

\(\Leftrightarrow\left(\sqrt{x+1}+\sqrt{x+16}\right)^2=\left(\sqrt{x+4}+\sqrt{x+9}\right)^2\)

\(\Leftrightarrow x+1+x+16+2.\sqrt{\left(x+1\right).\left(x+16\right)}=x+4+x+9+2.\sqrt{\left(x+4\right).\left(x+9\right)}\)

\(\Leftrightarrow2x+17+2.\sqrt{\left(x+1\right).\left(x+16\right)}=2x+13+2.\sqrt{\left(x+4\right).\left(x+9\right)}\)

\(\Leftrightarrow4+2.\sqrt{\left(x+1\right)\left(x+16\right)}=2.\sqrt{\left(x+4\right).\left(x+9\right)}\)

\(\Leftrightarrow2.\left(2+\sqrt{\left(x+1\right)\left(x+16\right)}\right)=2.\sqrt{\left(x+4\right).\left(x+9\right)}\)

\(\Leftrightarrow\sqrt{x^2+17x+16}+1=\sqrt{x^2+13x+36}\)

Bình phương 2 vế ta được 

\(x^2+17x+16+1+2.\sqrt{x^2+17x+16}=x^2+13x+36\)

\(\Leftrightarrow2.\sqrt{x^2+17x+16}=-4x+19\)

Bình phương 2 vế ta được 

\(2x^2+34x+32=16x^2-152x+361\)

\(\Leftrightarrow14x^2-186x+329=0\)

\(\Delta=\left(-186\right)^2-4.14.329=16172\)

\(x_1=\frac{186-\sqrt{16172}}{26}=2,262723898\)

\(x_2=\frac{186+\sqrt{16172}}{26}=12,04496841\)

\(\sqrt{x+1}+\sqrt{x+16}=\sqrt{x+4}+\sqrt{x+9}\) 

\(\left(\sqrt{x+1}+\sqrt{x+16}\right)^2=\left(\sqrt{x+4}+\sqrt{x+9}\right)^2\)  

\(x+1+x+16+2\sqrt{\left(x+1\right)\left(x+16\right)}=x+4+x+9+2\sqrt{\left(x+4\right)\left(x+9\right)}\)     

\(2x+17+2\sqrt{x^2+17x+16}=2x+13+2\sqrt{x^2+13x+36}\) 

\(4+2\sqrt{x^2+17x+16}=2\sqrt{x^2+13x+36}\)   

\(2+\sqrt{x^2+17x+16}=\sqrt{x^2+13x+36}\) 

\(\left(2+\sqrt{x^2+17x+16}\right)^2=\left(\sqrt{x^2+13x+36}\right)^2\)             

\(4+x^2+17x+16+4\sqrt{x^2+17x+16}=x^2+13x+36\) 

\(4\sqrt{x^2+17x+16}=-4x+16\) 

\(\sqrt{x^2+17x+16}=-x+4\)          

\(\hept{\begin{cases}-x+4\ge0\\x^2+17x+16=\left(-x+4\right)^2\end{cases}}\)    

\(\hept{\begin{cases}-x\ge-4\\x^2+17x+16=x^2-8x+16\end{cases}}\) 

\(\hept{\begin{cases}x\le4\\25x=0\end{cases}}\)  

\(\hept{\begin{cases}x\le4\\x=0\end{cases}}\)      

\(\Rightarrow x=0\) 

Đặt \(\hept{\begin{cases}\sqrt{x+1}=a\left(a\ge0\right)\\\sqrt{x-2}=b\left(b\ge0\right)\end{cases}}\)

\(\Rightarrow a^2-b^2=3\)

\(1PT\Leftrightarrow\left(a-b\right)\left(1+ab\right)=a^2-b^2\)

\(\Leftrightarrow\left(a-b\right)\left(1+ab-a-b\right)=0\)

 \(\Leftrightarrow\left(a-b\right)\left(a-1\right)\left(b-1\right)=0\)

 Tới đây tự làm tiếp nhé

có ai chơi ngọc rồng online ko