e/ \(\sqrt{x-2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\)

\(\Leftrightarrow4+2\sqrt{\left(x-2\right)\left(6-x\right)}=x^2-8x+24\)

\(\Leftrightarrow2\sqrt{-x^2+8x-12}=x^2-8x+20\)

Đặt \(\sqrt{-x^2+8x-12}=a\left(a\ge0\right)\)thì pt thành

\(2a=-a^2+8\)

\(\Leftrightarrow a^2+2a-8=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-4\left(l\right)\\a=2\end{cases}}\)

\(\Leftrightarrow\sqrt{-x^2+8x-12}=2\)

\(\Leftrightarrow-x^2+8x-12=4\)

\(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x=4\)

a/ \(4x^2+3x+3-4x\sqrt{x+3}-2\sqrt{2x-1}=0\)

\(\Leftrightarrow\left(4x^2-4x\sqrt{x+3}+x+3\right)+\left(2x-1-2\sqrt{2x-1}+1\right)=0\)

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

\(\Leftrightarrow\hept{\begin{cases}2x=\sqrt{x+3}\\1=\sqrt{2x-1}\end{cases}\Leftrightarrow}x=1\)

b/ \(2x-8\sqrt{2x-3}+9=0\)

\(\Leftrightarrow\left(2x-3-2.4.\sqrt{2x-3}+16\right)-4=0\)

\(\Leftrightarrow\left(4-\sqrt{2x-3}\right)^2-4=\)

\(\Leftrightarrow\left(2-\sqrt{2x-3}\right)\left(6-\sqrt{2x-3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2=\sqrt{2x-3}\\6=\sqrt{2x-3}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{39}{2}\end{cases}}}\)

Lời giải:
ĐKXĐ: $x\geq 0; y\geq 1; z\geq 2$

PT \(\Leftrightarrow (x-2\sqrt{x}+1)+[(y-1)-4\sqrt{y-1}+4]+[(z-2)-6\sqrt{z-2}+9]=0\)

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

Vì \((\sqrt{x}-1)^2, (\sqrt{y-1}-2)^2, (\sqrt{z-2}-3)^2\geq 0\) nên để tổng của chúng bằng $0$ thì:
\((\sqrt{x}-1)^2=(\sqrt{y-1}-2)^2=(\sqrt{z-2}-3)^2=0\)

$\Leftrightarrow x=1; y=5; z=11$

ĐK: \(x\ge1,y\ge2,z\ge3\).

\(x+y+z-2\sqrt{x-1}-4\sqrt{y-2}-6\sqrt{z-3}+8=0\)

\(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)

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

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=12\end{cases}}\)(tm) 

\(a,\) Sửa đề: \(\sqrt{3x^2-12x+16}+\sqrt{y^2-4y+13}=5\)

Ta thấy \(3x^2-12x+16=3\left(x-2\right)^2+4\ge4\Leftrightarrow\sqrt{3x^2-12x+16}\ge\sqrt{4}=2\)

\(y^2-4y+13=\left(y-2\right)^2+9\ge9\Leftrightarrow\sqrt{y^2-4y+13}\ge\sqrt{9}=3\)

Cộng vế theo vế 2 BĐT trên:

\(\sqrt{3x^2-12x+16}+\sqrt{y^2-4y+13}\ge2+3=5\)

Dấu \("="\Leftrightarrow x=y=2\)

Vậy pt có nghiệm \(\left(x;y\right)=\left(2;2\right)\)

\(b,x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\\ \Leftrightarrow x+y+z+4-2\sqrt{x-2}-4\sqrt{y-3}-6\sqrt{z-5}=0\\ \Leftrightarrow\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5+6\sqrt{z-5}+9\right)=0\\ \Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}-1=0\\\sqrt{y-3}-2=0\\\sqrt{z-5}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\y-3=4\\z-5=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\\z=14\end{matrix}\right.\)

Bạn xem lại đề câu b và c nhé !

a) \(\sqrt{x^2+2x+4}\ge x-2\) \(\left(ĐK:x\ge2\right)\)

\(\Leftrightarrow x^2+2x+4>x^2-4x+4\)

\(\Leftrightarrow6x>0\Leftrightarrow x>0\) kết hợp với ĐKXĐ

\(\Rightarrow x\ge2\) thỏa mãn đề.

d) \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)

\(ĐKXĐ:x\ge2,y\ge3,z\ge5\)

Pt tương đương :

\(\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5-6\sqrt{z-5}+9\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-3}=2\\\sqrt{z-5}=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=7\\z=14\end{cases}}\) ( Thỏa mãn ĐKXĐ )

e) \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\) (1)

\(ĐKXĐ:x\ge0,y\ge1,z\ge2\)

Phương trình (1) tương đương :

\(x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0\)

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

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

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)( Thỏa mãn ĐKXĐ )

ĐKXĐ : \(2\le x,y,z\le4\)

Từ hệ phương trình ta suy ra được

\(\Sigma x+\Sigma\sqrt{x-2}+\Sigma\sqrt{4-x}=\Sigma x^2-5\Sigma x+33\\ \Leftrightarrow\Sigma\left(x^2-6x+9\right)+6=\Sigma\left(\sqrt{x-2}+\sqrt{4-x}\right)\\ \Leftrightarrow\Sigma\left(x-3\right)^2+6=\Sigma\left(\sqrt{x-2}+\sqrt{4-x}\right)\left(1\right)\)

Áp dụng bất đẳng thức \(\sqrt{A}+\sqrt{B}\le\sqrt{2\left(A+B\right)}\)

\(\Sigma\left(\sqrt{x-2}+\sqrt{4-x}\right)\le\Sigma\sqrt{2\left(x-2+4-x\right)}=\Sigma2=6\)

\(\Rightarrow\Sigma\left(x-3\right)^2+6\le6\Rightarrow\Sigma\left(x-3\right)^2\le0\)

Mà \(\Sigma\left(x-3\right)^2\ge0\)

\(\Rightarrow\left(x-3\right)^2=\left(y-3\right)^2=\left(z-3\right)^2=0\\ \Leftrightarrow x=y=z=3\)

Thay vào ta thấy thỏa mãn -> x=y=z=3 là nghiệm hpt

Làm hơi tắt , thông cảm  ;))

Từ (1) \(\Rightarrow36=\left(x+y+z\right)^2\Leftrightarrow36=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)

          \(\Leftrightarrow36=18+2\left(xy+yz+zx\right)\Leftrightarrow xy+yz+zx=9\)(4)

Từ (3) \(\Rightarrow16=\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2\Leftrightarrow16=x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)

          \(\Leftrightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=5\Leftrightarrow\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2=25\)

         \(\Leftrightarrow xy+yz+zx+2\left(\sqrt{xy^2z}+\sqrt{xyz^2}+\sqrt{x^2yz}\right)=25\)

         \(\Leftrightarrow\sqrt{xyz}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)=8\Leftrightarrow\sqrt{xyz}=\frac{8}{4}\Leftrightarrow xyz=4\)(5)

Vậy hệ đã cho tương đương với :

\(\hept{\begin{cases}x+y+z=6\left(1\right)\\xy+yz+zx=9\left(4\right)\\xyz=4\left(5\right)\end{cases}}\)

Từ (5) \(\Rightarrow yz=\frac{4}{x}\)(Dễ thấy \(x,y,z>0\))

     (4)  \(\Leftrightarrow xy+yz+zx+x^2=9+x^2\Leftrightarrow x\left(x+y+z\right)+yz=9+x^2\)

           \(\Leftrightarrow x.6+\frac{4}{x}=9+x^2\Leftrightarrow x^3-6x^2+9x-4=0\)

           \(\Leftrightarrow\left(x-1\right)^2\left(x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=4\end{cases}.}\)

Thế vào ta suy ra hệ có các nghiệm : \(\left(x,y,z\right)=\left(1,1,4\right),\left(1,4,1\right),\left(4,1,1\right).\)

thanks bạn Đào Thu Hòa