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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
\(\Leftrightarrow2\sqrt{x-2000}+2\sqrt{y-2001}+2\sqrt{z-2002}=x+y+z-6000\)
\(\Leftrightarrow z+y+z-2\sqrt{x-2000}+2\sqrt{y-2001}+2\sqrt{z-2002}-6000=0\)
\(\Leftrightarrow\left(\left(\sqrt{x-2000}\right)^2-2\sqrt{x-2000}+1\right)+\left(\left(\sqrt{y-2001}\right)^2-2\sqrt{y-2001}+1\right)+\left(\left(\sqrt{z-2002}\right)^2-2\sqrt{z-2002}+1\right)=0\)\(\Leftrightarrow\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)
\(\Leftrightarrow x=2001;y=2002;z=2003\)
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
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
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/ \(4x^2+3x+3-4x\sqrt{x+3}-2\sqrt{2x-1}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{x+3}+x+3\right)+\left(2x-1-2\sqrt{2x-1}+1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)^2+\left(1-\sqrt{2x-1}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=\sqrt{x+3}\\1=\sqrt{2x-1}\end{cases}\Leftrightarrow}x=1\)
ĐK: \(x\ge2000;y\ge2001;z\ge2002\)
Biến đổi pt về dạng: \(\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\sqrt{x-2000}-1=0\\\sqrt{y-2001}-1=0\\\sqrt{z-2002}-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=2001\\y=2002\left(tm\right)\\z=2003\end{cases}}}\)
Sửa lại đề ( đề sai )
\(\sqrt{x-2000}+\sqrt{y-2001}+\sqrt{z-2002}=\frac{1}{2}\left(x+y+z\right)-3000\)
\(ĐKXĐ:\hept{\begin{cases}x\ge2000\\y\ge2001\\z\ge2002\end{cases}}\)
\(\sqrt{x-2000}+\sqrt{y-2001}+\sqrt{z-2002}=\frac{1}{2}\left(x+y+z\right)-3000\)
\(\Leftrightarrow\sqrt{x-2000}+\sqrt{y-2001}+\sqrt{z-2002}=\frac{x+y+z-6000}{2}\)
\(\Leftrightarrow2\sqrt{x-2000}+2\sqrt{y-2001}+2\sqrt{z-2002}=x+y+z-6000\)
\(\Leftrightarrow x+y+z-6000-2\sqrt{x-2000}-2\sqrt{y-2001}-2\sqrt{z-2002}=0\)
\(\Leftrightarrow\left[\left(x-2000\right)-2\sqrt{x-2000}+1\right]+\left[\left(y-2001\right)-2\sqrt{y-2001}+1\right]+\left[\left(z-2002\right)-2\sqrt{y-2002}+1\right]=0\)\(\Leftrightarrow\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)(1)
Vì \(\left(\sqrt{x-2000}-1\right)^2\ge0\), \(\left(\sqrt{y-2001}-1\right)^2\ge0\), \(\left(\sqrt{z-2002}-1\right)^2\ge0\)\(\forall x,y,z\)
\(\Rightarrow\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2\ge0\)\(\forall x,y,z\)(2)
Từ (1) và (2)
\(\Rightarrow\)Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2000}-1=0\\\sqrt{y-2001}-1=0\\\sqrt{z-2002}-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x-2000}=1\\\sqrt{y-2001}=1\\\sqrt{z-2002}=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-2000=1\\y-2001=1\\z-2002=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2001\\y=2002\\z=2003\end{cases}}\)( thỏa mãn ĐKXĐ )
Vậy \(x=2001\); \(y=2002\)và \(z=2003\)