Ta có:

\(4\sqrt{8-x}+4\sqrt{8-y}+4\sqrt{8-z}\)

\(\le8-x+4+8-y+4+8-z+4\)

\(=36-x-y-z\)

\(=48-\left(x+4\right)-\left(y+4\right)-\left(z+4\right)\)

\(\le48-4\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)

\(=48-4.6=24\)

\(\Rightarrow\sqrt{8-x}+\sqrt{8-y}+\sqrt{8-z}\le6\)

Dấu = xảy ra khi \(x=y=z=4\)     

bạn tham khảo nhé:

Vì \(x,y,z\ge0\)không mất tính tổng quát ta giả sử \(x\ge y\ge z\)

hệ \(\Leftrightarrow\hept{\begin{cases}3\sqrt{x}=6\\3\sqrt{8-x}=6\end{cases}\Leftrightarrow3\sqrt{x}=3\sqrt{8-x}\Leftrightarrow x=4}\)

\(\Rightarrow4\ge y\ge z\)

Nếu \(x=1\)thì \(\sqrt{8-x}=\sqrt{7}\left(L\right)\)

nếu \(x=2\)thì \(\sqrt{x}=\sqrt{2}\left(L\right)\)

\(\)nếu \(x=3\)thì \(\sqrt{x}=\sqrt{3}\left(L\right)\)

Loại vì các số vô tỉ không thẻ nào cộng lại là 1 số nguyên

Vậy \(\left(x;y;z\right)\)là \(\left(4;4;4\right)\)

\(4\sqrt{8-x}+4\sqrt{8-y}+4\sqrt{8-z}\)

\(\le8-x+4+8-y+4+8-z+4\)

\(=36-x-y-z\)

\(=48-\left(x+4\right)-\left(y+4\right)-\left(z+4\right)\)

\(\le48-4\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)

\(=48-4\cdot6=24\)

\(\Rightarrow\sqrt{8-x}+\sqrt{8-y}+\sqrt{8-z}\)

6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)

Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)

Phương trình sẽ trở thành là: a^2+a-42=0

=>(a+7)(a-6)=0

=>a=-7(loại) hoặc a=6(nhận)

=>2x^2+3x+9=36

=>2x^2+3x-27=0

=>2x^2+9x-6x-27=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)

3 an 2 phuong trinh cai nay toan Dai Hoc ma

1.

ĐKXĐ: $x\geq 1; y\geq 2; z\geq 3$

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

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

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

\(\Rightarrow \sqrt{x-1}-1=\sqrt{y-2}-2=\sqrt{z-3}-3=0\)

\(\Leftrightarrow \left\{\begin{matrix} x=2\\ y=6\\ z=12\end{matrix}\right.\)

2.

ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow \sqrt{x+1}=1-\sqrt{x}$

$\Rightarrow x+1=(1-\sqrt{x})^2=x+1-2\sqrt{x}$

$\Leftrightarrow 2\sqrt{x}=0$

$\Leftrightarrow x=0$

Thử lại thấy thỏa mãn 

Vậy $x=0$

3.

ĐKXĐ: $x\geq -1$

PT \(\Leftrightarrow (1+\sqrt{x^2+4033}).\frac{(x+2016)-(x+1)}{\sqrt{x+2016}+\sqrt{x+1}}=2015\)

\(\Leftrightarrow 1+\sqrt{x^2+4033}=\sqrt{x+2016}+\sqrt{x+1}\)

\(\Leftrightarrow (1+\sqrt{x^2+4033})^2=(\sqrt{x+2016}+\sqrt{x+1})^2\)

Áp dụng BĐT Bunhiacopxky:

\(\text{VP}\leq 2(x+2016+x+1)=4x+4034\)

\(\text{VP}=x^2+4034+2\sqrt{x^2+4033}\geq x^2+4034+2\sqrt{4033}>x^2+4034+5\)

Mà: $x^2+4034+5-(4x+4034)=(x-2)^2+1> 0$

$\Rightarrow x^2+4034+5> 4x+4034$

$\Rightarrow \text{VP}> \text{VT}$

Do đó pt vô nghiệm.

a,

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

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

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

tham khảo:

 ĐK: 

Do 

Dấu = xảy ra khi 

Vậy 

ĐK: \(x\ge1;y\ge2;z\ge3\)

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-2}=2\\\sqrt{z-3}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\left(tm\right)\\y=6\left(tm\right)\\z=12\left(tm\right)\end{matrix}\right.\)

Mình chia thành 2 phần lời giải để thuận tiện trong việc quan sát nhé!

a. \(a+b+c=2\sqrt{a}+2\sqrt{b-3}+2\sqrt{c}\left(ĐK:a\ne0;b\ne3;c\ne0\right)\\ \Leftrightarrow a-2\sqrt{a}+1+b-3-2\sqrt{b-3}+1+c-2\sqrt{c}+1=0\\ \Leftrightarrow\left(\sqrt{a}-1\right)^2+\left(\sqrt{b-3}-1\right)^2+\left(\sqrt{c}-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}a=1\\b=4\\c=1\end{matrix}\right.\)

Vậy \(\left(a;b;c\right)=\left(1;4;1\right)\)

b. \(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\left(ĐK:x\ne1;y\ne2;z\ne3\right)\\ x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{y-3}+9=0\\ \Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(2;4;6\right)\)

P/s: Trước khi kết luận, kiểm tra lại điều kiện thấy thỏa mãn rồi nên mình kết luận luôn nhé. Còn trong bài làm bạn nên ghi kết quả kiểm tra điều kiện cạnh giá trị mới tìm được nhé.

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/ \(2x-8\sqrt{2x-3}+9=0\)

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

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

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

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