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a) \(A=\sqrt{4x^2+4x+2}=\sqrt{4x^2+4x+1+1}=\sqrt{\left(2x+1\right)^2+1}\)
Vì \(\left(2x+1\right)^2\ge0\forall x\)\(\Rightarrow\left(2x+1\right)^2+1\ge1\forall x\)
\(\Rightarrow A\ge\sqrt{1}=1\)
Dấu " = " xảy ra \(\Leftrightarrow2x+1=0\)\(\Leftrightarrow2x=-1\)\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy \(minA=1\Leftrightarrow x=\frac{-1}{2}\)
b) \(B=\sqrt{2x^2-4x+5+1}=\sqrt{2x^2-4x+2+3+1}=\sqrt{2\left(x^2-2x+1\right)+4}\)
\(=\sqrt{2\left(x-1\right)^2+4}\)
Vì \(\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-1\right)^2+4\ge4\forall x\)
\(\Rightarrow B\ge\sqrt{4}=2\)
Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)
Vậy \(minB=2\Leftrightarrow x=1\)
\(M=\sqrt{x^2-4x+4}+2014\sqrt{x^2-6x+9}+\sqrt{x^2-10x+25}\)
\(M=\left|x-2\right|+2014\left|x-3\right|+\left|x-5\right|\)
\(M=\left|x-2\right|+\left|5-x\right|+2014\left|x-3\right|\)
\(M\ge\left|x-2+5-x\right|+2014\left|x-3\right|=3+2014\left|x-3\right|\ge3\)
\("="\Leftrightarrow x=3\)
Ngại làm lần 2 quá bạn ơi
Câu hỏi của Chuột yêu Gạo - Toán lớp 9 | Học trực tuyến
\(A=\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)
\(A=\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}\)
\(A=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(A=\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|\)
\(A=\left|1-\sqrt{x-1}\right|+\left|\sqrt{x-1}+1\right|\ge\left|1-\sqrt{x-1}+\sqrt{x-1}+1\right|=\left|2\right|=2\)
Dấu "=" xảy ra \(\Leftrightarrow1\le x\le2\)
\(B=\sqrt{x^2+4x+4}+\sqrt{x^2+6x+9}\)
\(B=\sqrt{\left(x+2\right)^2}+\sqrt{\left(x+3\right)^2}\)
\(B=\left|x+2\right|+\left|x+3\right|\)
\(B=\left|-x-2\right|+\left|x+3\right|\ge\left|-x-2+x+3\right|=\left|1\right|=1\)
Dấu "=" xảy ra \(\Leftrightarrow-3\le x\le-2\)
Lời giải:
Ta có:
\(A=\sqrt{-9x^2+6x+3}=\sqrt{4-(9x^2-6x+1)}=\sqrt{4-(3x-1)^2}\)
Ta thấy \((3x-1)^2\geq 0\Rightarrow 4-(3x-1)^2\leq 4\Rightarrow A=\sqrt{4-(3x-1)^2}\leq 2\)
Vậy \(A_{\max}=2\Leftrightarrow (3x-1)^2=0\Leftrightarrow x=\frac{1}{3}\)
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Ta thấy:
\(2x^2\geq 0\Rightarrow 3-2x^2\leq 3\Rightarrow B=\sqrt{3-2x^2}\leq \sqrt{3}\)
Vậy \(B_{\max}=\sqrt{3}\Leftrightarrow x^2=0\Leftrightarrow x=0\)
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\(\sqrt{-4x^2-4x}=\sqrt{1-(4x^2+4x+1)}=\sqrt{1-(2x+1)^2}\)
Ta thấy \((2x+1)^2\geq 0\Rightarrow 1-(2x+1)^2\leq 1\)
\(\Rightarrow C=5+\sqrt{1-(2x+1)^2}\leq 5+\sqrt{1}=6\)
Vậy \(C_{\max}=6\Leftrightarrow (2x+1)^2=0\Leftrightarrow x=\frac{-1}{2}\)
Lời giải:
Ta có:
\(A=\sqrt{-9x^2+6x+3}=\sqrt{4-(9x^2-6x+1)}=\sqrt{4-(3x-1)^2}\)
Ta thấy \((3x-1)^2\geq 0\Rightarrow 4-(3x-1)^2\leq 4\Rightarrow A=\sqrt{4-(3x-1)^2}\leq 2\)
Vậy \(A_{\max}=2\Leftrightarrow (3x-1)^2=0\Leftrightarrow x=\frac{1}{3}\)
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Ta thấy:
\(2x^2\geq 0\Rightarrow 3-2x^2\leq 3\Rightarrow B=\sqrt{3-2x^2}\leq \sqrt{3}\)
Vậy \(B_{\max}=\sqrt{3}\Leftrightarrow x^2=0\Leftrightarrow x=0\)
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\(\sqrt{-4x^2-4x}=\sqrt{1-(4x^2+4x+1)}=\sqrt{1-(2x+1)^2}\)
Ta thấy \((2x+1)^2\geq 0\Rightarrow 1-(2x+1)^2\leq 1\)
\(\Rightarrow C=5+\sqrt{1-(2x+1)^2}\leq 5+\sqrt{1}=6\)
Vậy \(C_{\max}=6\Leftrightarrow (2x+1)^2=0\Leftrightarrow x=\frac{-1}{2}\)