nhân cả 2 vế với 2 ta có

\(2\sqrt{x-2}+2\sqrt{y+2000}+2\sqrt{z-2001}=x+y+z\)

\(\left(x-2\right)-2\sqrt{x-2}+1+\left(y+2000\right)-2\sqrt{y+2000}+1+\left(z-2001\right)-2\sqrt{z-2001}+1=0\)

\(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2000}-1\right)^2+\left(\sqrt{z-2001}-1\right)^2=0\)

cho cả 3 cái =0 thì giả ra x=3  y=-1999  z=2002

how about the technology in the future Which things will happen Draw a picture about the technology in the future Note You can draw everything but they are different from now Please help me

hình như...

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

\(\Leftrightarrow x-3+y-3+z-3+17=2\sqrt{x-3}+4\sqrt{y-3}+6\sqrt{z-3}\)

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

\(\Leftrightarrow\left(\sqrt{x-3}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-3}-3\right)^2+3=0\) (vô nghiệm, VT >/3)

Kl: ptvn

c) là y - 2002 , z-2003 chứ 0 phải x đúng 0? (đoán thôi)

Đặt √x = t, x ≥ 0 => t ≥ 0.

Vế trái trở thành: t⁸ – t⁵ + t² – t + 1 = f(t)

Nếu t = 0, t = 1, f(t) = 1 >0

Với 0 < t <1,      f(t) = t⁸ + (t² - t⁵)+1 - t 

       t⁸ > 0, 1 - t > 0, t² - t⁵ = t³(1 – t) > 0. Suy ra f(t) > 0.

Với t > 1 thì f(t) = t⁵(t³ – 1) + t(t - 1) + 1 > 0

Vậy f(t) > 0 ∀t ≥ 0. Suy ra: x⁴ - √x⁵ + x - √x + 1 > 0, ∀x ≥ 0

\(\Leftrightarrow2\sqrt{x-2000}+2\sqrt{y-2001}+2\sqrt{z-2002}=x+y+z-6000\)

\(\Leftrightarrow z+y+z-2\sqrt{x-2000}+2\sqrt{y-2001}+2\sqrt{z-2002}-6000=0\)

\(\Leftrightarrow\left(\left(\sqrt{x-2000}\right)^2-2\sqrt{x-2000}+1\right)+\left(\left(\sqrt{y-2001}\right)^2-2\sqrt{y-2001}+1\right)+\left(\left(\sqrt{z-2002}\right)^2-2\sqrt{z-2002}+1\right)=0\)\(\Leftrightarrow\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)

\(\Leftrightarrow x=2001;y=2002;z=2003\)

Tham khảo:

đề cho thêm x nữa hén=) , p/s đưa đề đàng hoàng có thịn cảm ng làm hén , ụa mà hiện hồn là sao.-. ghét nghỉ=))

ủa cái z dị pa :)

Lời giải:

Từ \(x+y+z=xyz\Rightarrow \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)

Đặt \((\frac{1}{a}, \frac{1}{b}, \frac{1}{c})=(x,y,z)\), trong đó $a,b,c>0$ thì ta có:

\(ab+bc+ac=1\) và cần phải CMR:

\(P=\frac{\sqrt{(\frac{1}{b^2}+1)(\frac{1}{c^2}+1})-\sqrt{\frac{1}{b^2}+1}-\sqrt{\frac{1}{c^2}+1}}{\frac{1}{bc}}+\frac{\sqrt{(\frac{1}{c^2}+1)(\frac{1}{a^2}+1})-\sqrt{\frac{1}{c^2}+1}-\sqrt{\frac{1}{a^2}+1}}{\frac{1}{ac}}+\frac{\sqrt{(\frac{1}{a^2}+1)(\frac{1}{b^2}+1})-\sqrt{\frac{1}{a^2}+1}-\sqrt{\frac{1}{b^2}+1}}{\frac{1}{ab}}\)

-----------------------------------------------

Ta có:
\(\frac{\sqrt{(\frac{1}{b^2}+1)(\frac{1}{c^2}+1})-\sqrt{\frac{1}{b^2}+1}-\sqrt{\frac{1}{c^2}+1}}{\frac{1}{bc}}=\sqrt{(b^2+1)(c^2+1)}-b\sqrt{c^2+1}-c\sqrt{b^2+1}\)

\(=\sqrt{(b^2+ab+bc+ac)(c^2+ac+bc+ab)}-b\sqrt{c^2+ac+bc+ab}-c\sqrt{b^2+ab+bc+ac}\)

\(=\sqrt{(b+a)(b+c)(c+a)(c+b)}-b\sqrt{(c+a)(c+b)}-c\sqrt{(b+a)(b+c)}\)

\(=(b+c)\sqrt{(a+b)(a+c)}-b\sqrt{(c+a)(c+b)}-c\sqrt{(b+a)(b+c)}(1)\)

Tương tự:

\(\frac{\sqrt{(\frac{1}{c^2}+1)(\frac{1}{a^2}+1})-\sqrt{\frac{1}{c^2}+1}-\sqrt{\frac{1}{a^2}+1}}{\frac{1}{ac}}=(a+c)\sqrt{(b+a)(b+c)}-a\sqrt{(c+a)(c+b)}-c\sqrt{(a+b)(a+c)}(2)\)

\(\frac{\sqrt{(\frac{1}{a^2}+1)(\frac{1}{b^2}+1})-\sqrt{\frac{1}{a^2}+1}-\sqrt{\frac{1}{b^2}+1}}{\frac{1}{ab}}=(a+b)\sqrt{(c+a)(c+b)}-b\sqrt{(a+b)(a+c)}-a\sqrt{(b+c)(b+a)}(3)\)

Từ \((1);(2);(3)\Rightarrow P=(b+c-c-b)\sqrt{(a+b)(a+c)}+(a+c-c-a)\sqrt{(b+a)(b+c)}+(a+b-b-a)\sqrt{(c+a)(c+b)}\)

\(=0\)

Ta có đpcm.

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/ \(2x-8\sqrt{2x-3}+9=0\)

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

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

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

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