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a) Ta có: \(A=\sqrt{4x^2+4x+2}=\sqrt{\left(4x^2+4x+1\right)+1}\)
\(=\sqrt{\left(2x+1\right)^2+1}\ge\sqrt{1}=1\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(2x+1\right)^2=0\Rightarrow x=-\frac{1}{2}\)
Vậy Min(A) = 1 khi x = -1/2
b) Ta có: \(B=\sqrt{2x^2-4x+5}=\sqrt{\left(2x^2-4x+2\right)+3}\)
\(=\sqrt{2\left(x-1\right)^2+3}\ge\sqrt{3}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-1\right)^2=0\Rightarrow x=1\)
Vậy Min(B) = \(\sqrt{3}\) khi x = 1
c)\(C=5+\sqrt{-4x^2-4x}\)
\(C=5+\sqrt{1-\left(4x^2+4x+1\right)}\)
\(C=5+\sqrt{1-\left(2x+1\right)^2}\)
Ta có: \(-\left(2x+1\right)^2\le0\)
\(\sqrt{1-\left(2x+1\right)^2}\le1\)
\(\sqrt{1-\left(2x+1\right)^2}+5\le6\Leftrightarrow C\le6\)
Vậy \(C_{max}=6\) khi \(2x+1=0\Leftrightarrow x=-\frac{1}{2}\)
f) \(F=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)
\(F=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)
\(F=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x+1+3-2x\right|=4\)
\(F_{min}=4\) khi \(\left(2x-1\right)\left(3-2x\right)\ge0\Leftrightarrow\frac{1}{2}\le x\le\frac{3}{2}\)
Mấy còn lại tương tự =)))
\(E=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)
\(=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)
\(=2x-1+2x-3\)
\(=4x-4\)
Làm nốt
a) \(\sqrt{x^2+2x+1}=9\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x}+1\right)^2}=9\)
\(\Leftrightarrow\left|\sqrt{x}+1\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=9\\x+1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-10\end{matrix}\right.\)
b)\(\sqrt{1-4x+4x^2}=5\)
\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\)
\(\Leftrightarrow\left|1-2x\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}1-2x=5\\1-2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
c)\(\sqrt{x^2-2x\sqrt{2}+2}=5\)
\(\Leftrightarrow\sqrt{\left(x-\sqrt{2}\right)^2}=5\)
\(\Leftrightarrow\left|x-\sqrt{2}\right|=5\)
\(\left[{}\begin{matrix}x-\sqrt{2}=5\\x-\sqrt{2}=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5+\sqrt{2}\\x=-5+\sqrt{2}\end{matrix}\right.\)
Mình giải tới đây thôi
a) Do VT >=0 nên VP >=0 nên \(x\ge4\)
\(PT\Leftrightarrow\left(x-2\right)-\sqrt{x-2}-2=0\)
Đặt \(\sqrt{x-2}=t\ge\sqrt{4-2}=\sqrt{2}\) thì \(t^2-t-2=0\)
\(\Leftrightarrow t=2\left(loại t = -1 vì nó không thỏa mãn đk\right)\Leftrightarrow x-2=4\Leftrightarrow x=6\)
c/ \(C=\sqrt{x^2-6x+9}+\sqrt{x^2+10x+25}\)
\(=\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+5\right)^2}\)
\(=|3-x|+|x+5|\ge|3-x+x+5|=8\)
d/ \(D=\sqrt{x^2-6x+9}+\sqrt{4x^2+24x+36}\)
\(=\sqrt{\left(x-3\right)^2}+\sqrt{4\left(x+3\right)^2}\)
\(=|3-x|+|x+3|+|x+3|\ge|3-x+x+3|+0=6\)
e/ \(2E=\sqrt{x^2}+2\sqrt{x^2-2x+1}\)
\(=\sqrt{x^2}+2\sqrt{\left(x-1\right)^2}\)
\(=|x|+|1-x|+|x-1|\ge|x+1-x|+0=1\)
\(\Rightarrow E\ge\frac{1}{2}\)
c, Ta có: \(C^2=x^2+4x+4+1=\left(x+2\right)^2+1\)
Vì \(\left(x+2\right)^2\ge0\Rightarrow C^2\ge1\Rightarrow C\ge1\)
Dấu ''='' xảy ra khi x=-2
Vậy...
câu d,e bình lên là ra ,tương tự câu này nhé.
d, Ta có: \(\sqrt{x+2}\ge0\) \(\Rightarrow D\ge1\)
Dấu ''='' xảy ra khi x=-2.
Vậy ....
e, \(E^2=x^2+1\ge1\Rightarrow E\ge1\)
Dấu ''='' xảy ra khi x=0
Vậy .....