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\(A=\frac{1}{4}-\left(x-\sqrt{x}+\frac{1}{4}\right)=\frac{1}{4}-\left(\sqrt{x}-\frac{1}{2}\right)^2\le\frac{1}{4}\)
\(\Rightarrow A_{max}=\frac{1}{4}\) khi \(x=\frac{1}{4}\), \(A_{min}\) ko tồn tại
\(B=\sqrt{2-\left(9x^2+6x+1\right)}-5=\sqrt{2-\left(3x+1\right)^2}-5\)
Do \(0\le\sqrt{2-\left(3x+1\right)^2}\le\sqrt{2}\)
\(\Rightarrow B_{max}=\sqrt{2}-5\) khi \(x=-\frac{1}{3}\)
\(B_{min}=-5\) khi \(\left(3x+1\right)^2=2\Rightarrow x=\frac{-1\pm\sqrt{2}}{3}\)
\(C=\sqrt{x-2}+\sqrt{4-x}\ge\sqrt{x-2+4-x}=\sqrt{2}\)
\(\Rightarrow C_{min}=\sqrt{2}\) khi \(\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
\(C\le\sqrt{\left(1+1\right)\left(x-2+4-x\right)}=2\)
\(\Rightarrow C_{max}=2\) khi \(x-2=4-x\Leftrightarrow x=3\)
\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
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 \(\Leftrightarrow-3\le x\le-2\)
a) \(\sqrt{x}-x=-\left(x-\sqrt{x}\right)\)
\(=-\left[\left(\sqrt{x}\right)^2-2.\frac{1}{2}\sqrt{x}+\frac{1}{4}\right]+\frac{1}{4}\)
\(=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Vậy GTLN của bt là \(\frac{1}{4}\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)