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

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

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

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

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

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)

Điều kiện: \(x\ge2012;y\ge2013;z\ge2014\)

Áp dụng bất đẳng thức Cauchy, ta có:

\(\left\{{}\begin{matrix}\dfrac{\sqrt{x-2012}-1}{x-2012}=\dfrac{\sqrt{4\left(x-2012\right)}-2}{2\left(x-2012\right)}\le\dfrac{\dfrac{4+x-2012}{2}-2}{2\left(x-2012\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{y-2013}-1}{y-2013}=\dfrac{\sqrt{4\left(y-2013\right)}-2}{2\left(y-2013\right)}\le\dfrac{\dfrac{4+y-2013}{2}-2}{2\left(y-2013\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{z-2014}-1}{z-2014}=\dfrac{\sqrt{4\left(z-2014\right)}-2}{2\left(z-2014\right)}\le\dfrac{\dfrac{4+z-2014}{2}-2}{2\left(z-2014\right)}=\dfrac{1}{4}\end{matrix}\right.\)

Cộng vế theo vế, ta được:

\(\dfrac{\sqrt{x-2012}-1}{x-2012}+\dfrac{\sqrt{y-2013}-1}{y-2013}+\dfrac{\sqrt{z-2014}-1}{z-2014}\le\dfrac{3}{4}\)

Đẳng thức xảy ra khi \(x=2016;y=2017;z=2018\)

Vậy....

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/ \(2x-8\sqrt{2x-3}+9=0\)

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

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

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

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

ĐKXĐ : \(\left\{{}\begin{matrix}x>2019\\y>2020\\z>2021\end{matrix}\right.\)

Đặt \(\sqrt{x-2019}=a,......\)

Ta được PT : \(\dfrac{1-a}{a^2}+\dfrac{1-b}{b^2}+\dfrac{1-c}{c^2}+\dfrac{3}{4}=0\)

\(\Leftrightarrow\dfrac{1}{a^2}-\dfrac{1}{a}+\dfrac{1}{4}+\dfrac{1}{b^2}-\dfrac{1}{b}+\dfrac{1}{4}+\dfrac{1}{c^2}-\dfrac{1}{c}+\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{2}\right)^2=0\)

- Thấy : \(\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2\ge0,......\)

\(\Rightarrow\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{2}\right)^2\ge0\)

- Dấu " = " xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{2}\\\dfrac{1}{b}=\dfrac{1}{2}\\\dfrac{1}{c}=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)

- Thay lại a. b. c ta được : \(\left\{{}\begin{matrix}\sqrt{x-2019}=2\\\sqrt{y-2020}=2\\\sqrt{z-2021}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2019=4\\y-2020=4\\z-2021=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2023\\y=2024\\z=2025\end{matrix}\right.\) ( TM )

Vậy ...

1/

ĐKXĐ: \(x\ge\frac{2}{3}\)

\(\Leftrightarrow2x^2+2-2\sqrt{x}-2\sqrt{3x-2}=0\)

\(\Leftrightarrow2x^2-4x+2+\left(x+1-2\sqrt{x}\right)+\left(3x-1-2\sqrt{3x-2}\right)=0\)

\(\Leftrightarrow2\left(x-1\right)^2+\frac{\left(x-1\right)^2}{x+1+2\sqrt{x}}+\frac{9\left(x-1\right)^2}{3x-1+2\sqrt{3x-2}}=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(2+\frac{1}{x+1+2\sqrt{x}}+\frac{9}{3x-1+2\sqrt{3x-2}}\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\) (ngoặc to luôn dương)

\(\Leftrightarrow x=1\)

2/

ĐKXĐ: \(x\ge-\frac{1}{2}\)

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

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

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

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

\(\Leftrightarrow x-4=0\) (ngoặc to luôn dương)

\(\Leftrightarrow x=4\)

3/

ĐKXĐ: \(x\ge1\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=0\\x\sqrt{x-1}=-2\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow x-1=0\Rightarrow x=1\)

4/

Bạn coi lại đề, nghiệm pt này rất xấu (là nghiệm của pt bậc 3 không phân tích được về hằng đẳng thức, chương trình Việt Nam ko học)