S= \(\sqrt{x-2}\) + \(\sqrt{y-3}\) biết x+y = 6
Tìm GTNN của S= \(\sqrt{x-2}+\sqrt{y-3}\)biết x+y=6
ĐKXĐ : \(x\ge2;y\ge3\)
\(\Rightarrow S=\sqrt{x-2}+\sqrt{y-3}\ge1\)
Dấu " = " xảy ra \(\Leftrightarrow\orbr{\begin{cases}x=2;y=4\\y=3;x=3\end{cases}}\)
Tìm GTLN của :
S = \(\sqrt{x-2}+\sqrt{y-3}\) ( biết x + y =6 )
a/d bunhiacopxki co:
\(S^2=\left(\sqrt{x-2}+\sqrt{y-3}\right)^2\le\left(1^2+1^2\right)\left(x-2+y-3\right)=2\cdot1=2\)
\(\Rightarrow S\le\sqrt{2}\)
Dấu ''='' xảy ra khi \(x=\dfrac{5}{2};y=\dfrac{7}{2}\)
Vậy GTLN của S = \(\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{7}{2}\end{matrix}\right.\)
Tìm GTNN của \(S=\sqrt{x-2}+\sqrt{y-3}\)
biết x+y=6
\(S=\sqrt{x-2}+\sqrt{y-3}\)
\(\Rightarrow S^2=\left(\sqrt{x-2}+\sqrt{y-3}\right)^2\)
\(\Rightarrow S^2=x-2+2\sqrt{\left(x-2\right)\left(y-3\right)}+y-3\)
\(\Rightarrow S^2=x+y-5+2\sqrt{\left(x-2\right)\left(y-3\right)}\)
\(\Rightarrow S^2=1+2\sqrt{\left(x-2\right)\left(y-3\right)}\)
Vì \(2\sqrt{\left(x-2\right)\left(y-3\right)}\ge0\)
\(\Rightarrow1+2\sqrt{\left(x-2\right)\left(y-3\right)}\ge1\)
\(\Rightarrow S^2\ge1\Leftrightarrow\orbr{\begin{cases}S\ge1\left(tm\right)\\S\le-1\left(ktm\right)\end{cases}}\)
\(\Rightarrow S_{min}=1\Leftrightarrow2\sqrt{\left(x-2\right)\left(y-3\right)}=0\)
TH1 : \(x-2=0\Leftrightarrow x=2\Rightarrow y=6-2=4\)
Th2 : \(y-3=0\Rightarrow y=3\Rightarrow x=6-3=3\)
Vậy \(S_{min}=1\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)hoặc \(x=y=3\)
Áp dụng bđt Bu-nhi-a-cốp-xki ta có
\(S^2=\left(\sqrt{x-2}+\sqrt{y-3}\right)^2\le\left(1+1\right)\left(x+y-5\right)=2\left(6-5\right)=2\)(vì \(x+y=6\) )
\(\Rightarrow S^2\le2\)
\(\Leftrightarrow-\sqrt{2}\le S\le\sqrt{2}\)
\(\Rightarrow minS=-\sqrt{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{\sqrt{x-2}}{1}=\frac{\sqrt{y-3}}{1}\\x+y=6\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=2,5\\y=3,5\end{cases}}\)
So sánh x và y, biết rằng
a) \(x=2.\sqrt{7}\) và \(y=3\sqrt{3}\)
b) \(x=6\sqrt{2}\) và \(y=5\sqrt{3}\)
c) \(x=\sqrt{31}-\sqrt{13}\) và \(y=6-\sqrt{11}\)
a: \(x=2\sqrt{7}=\sqrt{28}>\sqrt{27}=y\)
b: \(x=6\sqrt{2}=\sqrt{72}< \sqrt{75}=y\)
tính giá trị lớn nhất của biểu thức S=\(\sqrt{x-2}+\sqrt{y-3}\) ,biết x+y=6
Cách khác:
Đk: \(x\ge2,y\ge3\)
Với a,b\(\ge\) 0có:
\(a+b\le\sqrt{2\left(a^2+b^2\right)}\) <=> \(a^2+2ab+b^2\le2a^2+2b^2\) <=> \(0\le a^2-2ab+b^2\)
<=>\(0\le\left(a-b\right)^2\)
Dấu "=" xảy ra <=>a=b>0
Áp dụng bđt trên có:
\(S=\sqrt{x-2}+\sqrt{y-3}\le\sqrt{2\left(x-2+y-3\right)}=\sqrt{2\left(6-2-3\right)}\)(do x+y=6)
=> \(S\le\sqrt{2}\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{y-3}\\x+y=6\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x-2=y-3\\x+y=6\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}y-x=1\\x+y=6\end{matrix}\right.\) <=> x=2,5 và y=3,5(t/m)
\(S^2\le\left(1+1\right)\left(x-2+y-3\right)=2\left(x+y-5\right)=2\)
\(\Rightarrow S\le\sqrt{2}\)
\(\Rightarrow S_{max}=\sqrt{2}\) khi \(\left\{{}\begin{matrix}x-2=y-3\\x+y=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=\frac{7}{2}\end{matrix}\right.\)
ĐKXĐ :\(x\ge2;y\ge3\)
\(S^2=x-2+y-3+2\sqrt{\left(x-2\right)\left(y-3\right)}\)
\(\ge6-2-3+0=1\)
Vì \(S>0\Rightarrow S\ge1\)
Vậy \(Max_S=1\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\y=3\end{matrix}\right.\end{matrix}\right.\)
tìm x,y,z biết
a) x+y+z+12=4\(\sqrt{x}+6\sqrt{y-1}\)
b)x+y+z+8=2\(\sqrt{x-3}+4\sqrt{y-3}+6\sqrt{z-3}\)
c)\(\sqrt{x-2001}+\sqrt{x-2002}-\sqrt{x-2003}=\dfrac{1}{2}\left(x+y+z\right)-3015\)
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)
Tìm x,y,z biết:\(\sqrt{x-1}+\sqrt{y-2}+\sqrt{z-3}=6-\frac{1}{\sqrt{x-1}}-\frac{1}{\sqrt{y-2}}-\frac{1}{\sqrt{z-3}}\)
\(ĐK:x\ge1,y\ge2,z\ge3\)
\(PT\Leftrightarrow\sqrt{x-1}+\frac{1}{\sqrt{x-1}}+\sqrt{y-2}+\frac{1}{\sqrt{y-2}}+\sqrt{z-3}+\frac{1}{\sqrt{z-3}}=6\)
Theo bđt AM-GM thì \(VT\ge6\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\sqrt{x-1}=\frac{1}{\sqrt{x-1}}=1\\\sqrt{y-2}=\frac{1}{\sqrt{y-2}}=1\\\sqrt{z-3}=\frac{1}{\sqrt{z-3}}=1\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=1\\y=3\\z=4\end{cases}}\)
Tìm các số x,y,z biết:
a,
\(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
b,
\(x+y+z+9=2\sqrt{x-2}+6\sqrt{y-3}+4\sqrt{z-9}\)
giải hộ mình vs :3
a,
\(pt\Leftrightarrow\left(x-1-2\sqrt{x-1}+1\right)+\left(y-2-4\sqrt{y-2}+4\right)+\left(z-3-6\sqrt{z-3}+9\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=12\end{cases}}\)
Bài 1: Tìm số x,y,z biết x + y + z + 11 = 2\(\sqrt{x}+4\sqrt{y-1}+6\sqrt{z-2}\)
Bài 2: Tìm GTNN Q= \(\sqrt{x^2+4x+4}+\sqrt{x^2-4x+4}\)
Bài 3: Tìm GTLN P = \(\sqrt{x-2}+\sqrt{x-3}\) biết x+y = 6
Bài 1: \(x+y+z+11=2\sqrt{x}+4\sqrt{y-1}+6\sqrt{z-2}\)
ĐKXĐ:\(x\ge0;y\ge1;z\ge2\)
\(\Leftrightarrow x-2\sqrt{x}+1+\left(y-1\right)-2\cdot\sqrt{y-1}\cdot2+4+\left(z-2\right)-2\cdot\sqrt{z-2}\cdot3+9=0\)\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-2\right)^2+\left(\sqrt{z-2}-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{y-1}=2\\\sqrt{z-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=5\\z=11\end{matrix}\right.\)
Bài 2:
Q=|x+2|+|x-2|>=|x+2+2-x|=4
Dấu = xảy ra khi (x+2)(x-2)<=0
=>-2<=x<=2