Tìm GTLN của biểu thức \(A=\frac{-7x^2+6x+3}{x^2+2}\)
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Answer:
Mình chỉ làm câu Đại thôi nhé, còn bài Hình thì nhờ cao nhân khác.
\(A=\frac{x+2\sqrt{x}+1}{x-1}+\frac{x-\sqrt{x}}{x-2\sqrt{x}+1}-\frac{x-2\sqrt{x}}{x-\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}-\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+3}{\sqrt{x}+1}\)
Để cho \(A=2\) thì \(\frac{\sqrt{x}+3}{\sqrt{x}+1}=2\Rightarrow\sqrt{x}+3=2\sqrt{x}+2\Rightarrow\sqrt{x}=1\Rightarrow x=1\) (Loại)
đkxđ \(x-2\ge0\Leftrightarrow x\ge2\)
phương trình đã cho \(\Leftrightarrow\left[\left(x-8\right)\sqrt{x-2}\right]^2=4\)\(\Leftrightarrow\left(x^2-16x+64\right)\left(x-2\right)=4\)
\(\Leftrightarrow x^3-2x^2-16x^2+32x+64x-128=4\)
\(\Leftrightarrow x^3-18x^2+96x-132=0\)
Tới đây bạn bấm máy Casio giải được rồi.
In my opinion, Vietnamese teenagers should learn and prace. Among them, I think communication skill is one of the most important ones as this affects their future. Actually, many youngsters lack basic abilities to interact with people in society. For example, they do not know how to have a successful conversation with other people, especially the elder. Besides, they should boost their representation skill in public. In fact, Vietnamese students are quite shy and lack confidence when speaking and giving their opinions in front of many others, which limits their creativity. Consequently, they cannot get good jobs after graduation although they graduate with flying colours.
Căn (35 + 12.căn 6)
= căn(27 + 12.căn6 + 8)
= căn(3.căn3 + 2.căn2)²
= 3.căn3 + 2.căn2
\(\sqrt{35+2\sqrt{6^2\times6}}\)=\(\sqrt{8+2\sqrt{8}\sqrt{27}+27}\)=\(\sqrt{\left(2\sqrt{2}+3\sqrt{3}\right)^2}\)=\(2\sqrt{2}+3\sqrt{3}\)
\(Ax^2-2A=-7x^2+6x+3!\)
\(x^2\left(A+7\right)-6x-2A-3=0\)
\(\text{Δ}=3^2=\left(2A+3\right)\left(A+7\right)>0\)
\(\orbr{\begin{cases}A< -6\\A>\frac{5}{2}\end{cases}}\)
A không có max và min
NHẦM
\(A=\frac{-7x^2+6x+3}{x^2+2}\)
\(=\frac{2\left(x^2+2\right)-9x^2-6x-1}{x^2+2}\)
\(=\frac{2\left(x^2+2\right)-\left(9x^2-6x-1\right)}{x^2+2}\)
\(=\frac{2\left(x^2+2\right)-\left(3x-1\right)^2}{x^2+2}\)
\(=2-\frac{\left(3x-1\right)^2}{x^2+2}\)
Vì \(-\left(3x-1\right)^2< 0\text{∀}x\)
\(x^2+2>0\text{∀}x\)
\(-\frac{\left(3x-1\right)^2}{x^2+2}< 0\)
\(2-\frac{\left(3x-1^2\right)}{x^2+2}< 2-0=2\)
Vậy GTLN của \(A\)là \(2\)khi : \(\left(3x-1\right)^2=0\)
\(x=\frac{1}{3}\)