Tìm min \(B=\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)
GTNN của P = \(\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)
\(P=\sqrt[]{9x^2-6x+1}+\sqrt[]{25-30x+9x^2}\)
\(\Leftrightarrow P=\sqrt[]{\left(3x-1\right)^2}+\sqrt[]{\left(5-3x\right)^2}\)
\(\Leftrightarrow P=\left|3x-1\right|+\left|5-3x\right|\)
\(\Leftrightarrow P=\left|3x-1\right|+\left|5-3x\right|\ge\left|3x-1+5-3x\right|=4\)
Vậy \(GTNN\left(P\right)=4\)
BT: Tìm gtnn của bt:
\(A=\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)
Tìm giá trị nhỏ nhất của biểu thức:\(A=\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)
\(A=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(5-3x\right)^2}\)
\(A=3x-1+5-3x=4\)
\(A\)có giá trị ko phụ thuộc vào biến x
Tìm giá trị nhỏ nhất
B= \(\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)
\(B=\left|3x-1\right|+\left|5-3x\right|>=\left|3x-1+5-3x\right|=4\)
Dấu '=' xảy ra khi (3x-1)(3x-5)<=0
=>1/3<=x<=5/3
Tìm GTNN của biểu thức Q = \(\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}+2011\)
\(Q=\sqrt{9x^2-6x+1}+\sqrt{25-30+9x^2}+2011\)
\(Q=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(5-3x\right)^2}+2011\)
\(Q=\left|3x-1\right|+\left|5-3x\right|+2011\)
Đặt \(Q'=\left|3x-1\right|+\left|5-3x\right|\ge\left|3x-1+5-3x\right|=4\)
Đẳng thức xảy ra \(\Leftrightarrow\left(3x-1\right)\left(5-3x\right)\ge0\)
\(\Leftrightarrow\frac{1}{3}\le x\le\frac{5}{3}\)
\(\Rightarrow Min_Q=Min_{Q'}+2011=4+2011=2015\)
Q = \(\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}+2011\)
Q = \(\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-5\right)^2}+2011\)
Q = \(3x-1+3x-5+2011\)
Q = \(6x+2005\)
\(Q=\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}+2011\)
\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-5\right)^2}+2011\)
\(=\left|3x-1\right|+\left|3x-5\right|+2011\)
Áp dụng BĐT \(\left|x\right|+\left|y\right|\ge\left|x+y\right|\)
\(\left|3x-1\right|+\left|3x-5\right|\ge\left|\left(3x-1\right)+\left(5-3x\right)\right|=4\)
(Dấu "="\(\Leftrightarrow\left(3x-1\right)\left(5-3x\right)\ge0\)
\(TH1:\hept{\begin{cases}3x-1\ge0\\5-3x\ge0\end{cases}}\Leftrightarrow\frac{1}{3}\le x\le\frac{5}{3}\)
\(TH2:\hept{\begin{cases}3x-1\le0\\5-3x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{1}{3}\\x\ge\frac{3}{5}\end{cases}}\left(L\right)\))
\(\Rightarrow Q\ge2015\)
(Dấu "="\(\Leftrightarrow\frac{1}{3}\le x\le\frac{5}{3}\))
Vậy \(Q_{min}=2015\Leftrightarrow\frac{1}{3}\le x\le\frac{5}{3}\)
\(\sqrt{9x^2-6x+2}+\sqrt{45x^2-30x+9}=\sqrt{6x-9x^2+8}\)Tìm X
Ta có :
\(\sqrt{9x^2-6x+2}=\sqrt{\left(9x^2-6x+1\right)+1}=\sqrt{\left(3x-1\right)^2+1}\ge\sqrt{1}=1\)
\(\sqrt{45x^2-30x+9}=\sqrt{5\left(9x^2-6x+1\right)+4}=\sqrt{5\left(3x-1\right)^2+4}\ge\sqrt{4}=2\)
\(\sqrt{6x-9x^2+8}=\sqrt{-\left(9x^2-6x+1\right)+9}=\sqrt{-\left(3x-1\right)^2+9}\le3\)
\(\Rightarrow VT\ge3\ge VP\)
mÀ đề lại cho \(VT=VP\) \(\Rightarrow\hept{\begin{cases}\sqrt{\left(3x-1\right)^2+1}=1\\\sqrt{\left(3x-1\right)^2+4}=2\\\sqrt{-\left(3x-1\right)^2+9}=3\end{cases}\Rightarrow x=\frac{1}{3}}\)
Vậy \(x=\frac{1}{3}\)
Giúp mình với :
Giải phương trình : \(\sqrt{9x^2-6x+2}+\sqrt{45x^2-30x+9}=\sqrt{6x-9x^2+8}\)
\(\sqrt{9x^2-6x+2}+\sqrt{45x^2-30x+9}=\sqrt{6x-9x^2+8}\)
Tìm x
\(\sqrt{x-2+2\sqrt{ }x-3}\) + \(\sqrt{x+6+6+\sqrt{x-3}}\)
Tìm GTNN
a) \(\sqrt{9x^2-6x+1}\) + \(\sqrt{25-30x+9x^2}\)
b) \(\sqrt{x^2-6x+1}\) + \(\sqrt{x^2+10x+25}\)
\(\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)
\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(5-3x\right)^2}\)
\(=\left|3x-1\right|+\left|5-3x\right|\)
\(\ge\left|3x-1+5-3x\right|=4\)