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Lời giải:
\(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{x+y-z}\Leftrightarrow \frac{x+y}{xy}=\frac{1}{z}+\frac{1}{x+y-z}=\frac{x+y}{z(x+y-z)}\)
\(\Leftrightarrow (x+y)(\frac{1}{xy}-\frac{1}{z(x+y-z)})=0\)
\(\Leftrightarrow (x+y).\frac{z(x+y-z)-xy}{xyz(x+y-z)}=0\)
\(\Leftrightarrow (x+y).\frac{(z-x)(y-z)}{xyz(x+y-z)}=0\)
\(\Leftrightarrow (x+y)(z-x)(y-z)=0\)
Xét các TH sau:
TH1: $x+y=0$. TH này loại do ĐKXĐ $x,y>0$
TH2: $z-x=0\Leftrightarrow z=x$
$\Leftrightarrow \frac{1}{y}=\frac{2020}{2021}$
\(M=\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{y}}=\frac{2}{\sqrt{y}}=2\sqrt{\frac{2020}{2021}}\)
TH3: $y-z=0$ tương tự TH2, ta có \(M=2\sqrt{\frac{2020}{2021}}\)
Ta có :
\(\dfrac{1}{\sqrt{x+1}+\sqrt{x+2}}=\dfrac{\sqrt{x+1}-\sqrt{x+2}}{\left(\sqrt{x+1}+\sqrt{x+2}\right)\left(\sqrt{x+1}-\sqrt{x+2}\right)}=\dfrac{\sqrt{x+1}-\sqrt{x+2}}{-1}=-\sqrt{x+1}+\sqrt{x+2}\)
Tương tự :
\(\dfrac{1}{\sqrt{x+2}+\sqrt{x+3}}=-\sqrt{x+2}+\sqrt{x+3}\)
\(\dfrac{1}{\sqrt{x+3}+\sqrt{x+4}}=-\sqrt{x+3}+\sqrt{x+4}\)
....
\(\dfrac{1}{\sqrt{x+2019}+\sqrt{x+2010}}=-\sqrt{x+2019}+\sqrt{x+2010}\)
Từ những ý trên , pt trở thành :
\(-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}-\sqrt{x+3}+\sqrt{x+4}-.....-\sqrt{x+2019}+\sqrt{x+2020}=11\)
\(\Leftrightarrow\sqrt{x+2020}-\sqrt{x+1}=11\)
\(\Leftrightarrow x+2020-2\sqrt{\left(x+2020\right)\left(x+1\right)}+x+1=121\)
\(\Leftrightarrow2x+1900=2\sqrt{\left(x+1\right)\left(x+2020\right)}\)
\(\Leftrightarrow x+950=\sqrt{\left(x+1\right)\left(x+2020\right)}\)
\(\Leftrightarrow x^2+1900x+902500=x^2+2021x+2020\)
\(\Leftrightarrow121x-900480=0\)
\(\Leftrightarrow x=\dfrac{900480}{121}\)
\(x+y+z-6046=2\sqrt{x-2019}+4\sqrt{y-2020}+6\sqrt{z-2021}\)
\(\left(x-2019\right)+\left(x-2020\right)+\left(x-2021\right)+1+4+9\)\(=2\sqrt{x-2019}+4\sqrt{y-2020}+6\sqrt{z-2021}\)
đặt :\(\hept{\begin{cases}\sqrt{x-2019}=a\\\sqrt{y-2020}=b\\\sqrt{z-2021}=c\end{cases}\left(đk:a,b,c\ge0\right)}\)
PT <=> \(a^2+b^2+c^2+1+4+9=2a+4b+6c\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-2\right)^2+\left(c-6\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a-1=0\\b-2=0\\c-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=2\\c=3\end{cases}\left(tm\right)}}\)
\(\Rightarrow\hept{\begin{cases}x=2020\\y=2024\\z=2030\end{cases}}\)
bài 3:
a, đặt x12=y9=z5=kx12=y9=z5=k
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: x5=y7=z3=x225=y249=z29x5=y7=z3=x225=y249=z29
A/D tính chất dãy tỉ số bằng nhau ta có:
x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
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 ...