Câu 1:

Tìm max:

Áp dụng BĐT Bunhiacopxky ta có:

\(y^2=(3\sqrt{x-1}+4\sqrt{5-x})^2\leq (3^2+4^2)(x-1+5-x)\)

\(\Rightarrow y^2\leq 100\Rightarrow y\leq 10\)

Vậy \(y_{\max}=10\)

Dấu đẳng thức xảy ra khi \(\frac{\sqrt{x-1}}{3}=\frac{\sqrt{5-x}}{4}\Leftrightarrow x=\frac{61}{25}\)

Tìm min:

Ta có bổ đề sau: Với $a,b\geq 0$ thì \(\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}\)

Chứng minh:

\(\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}\)

\(\Leftrightarrow (\sqrt{a}+\sqrt{b})^2\geq a+b\)

\(\Leftrightarrow \sqrt{ab}\geq 0\) (luôn đúng).

Dấu "=" xảy ra khi $ab=0$

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Áp dụng bổ đề trên vào bài toán ta có:

\(\sqrt{x-1}+\sqrt{5-x}\geq \sqrt{(x-1)+(5-x)}=2\)

\(\sqrt{5-x}\geq 0\)

\(\Rightarrow y=3(\sqrt{x-1}+\sqrt{5-x})+\sqrt{5-x}\geq 3.2+0=6\)

Vậy $y_{\min}=6$

Dấu "=" xảy ra khi \(\left\{\begin{matrix} (x-1)(5-x)=0\\ 5-x=0\end{matrix}\right.\Leftrightarrow x=5\)

Bài 2:

\(A=\sqrt{(x-1994)^2}+\sqrt{(x+1995)^2}=|x-1994|+|x+1995|\)

Áp dụng BĐT dạng \(|a|+|b|\geq |a+b|\) ta có:

\(A=|x-1994|+|x+1995|=|1994-x|+|x+1995|\geq |1994-x+x+1995|=3989\)

Vậy \(A_{\min}=3989\)

Đẳng thức xảy ra khi \((1994-x)(x+1995)\geq 0\Leftrightarrow -1995\leq x\leq 1994\)

Bài 3:

Ta thấy:

\(2x-x^2+7=8-(x^2-2x+1)=8-(x-1)^2\leq 8, \forall x\in\mathbb{R}\)

\(\Rightarrow 2+\sqrt{2x-x^2+7}\leq 2+\sqrt{8}=2+2\sqrt{2}\)

\(\Rightarrow B=\frac{3}{2+\sqrt{2x-x^2+7}}\geq \frac{3}{2+2\sqrt{2}}\)

Vậy GTNN của $B$ là \(\frac{3}{2+2\sqrt{2}}\).

Đẳng thức xảy ra tại \((x-1)^2=0\Leftrightarrow x=1\)

1.

6x + 1 ≥0

<=>6x≥-1

<=>x≥-1/6

2.

3x - 5 > 0 

<=> 3x > 5

<=> x > 5/3

3.

x - 7 > 0

<=> x > 7

4. 

-3x ≥0

<=>x≤0

5.

√5 - √3 . x ≥0

<=> √3 . x ≤ √5

<=> x ≤ √5/3 = (√15)/3

a) \(P=\dfrac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

\(=\dfrac{3x+3\sqrt{x}-9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(=\dfrac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)

b) \(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+2}{\sqrt{x}+2}=3+\dfrac{2}{\sqrt{x}+2}\)

Để \(P\in Z\Rightarrow2⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2=2\left(\sqrt{x}+2\ge2\right)\)

\(\Rightarrow x=0\)

c) Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\Rightarrow3+\dfrac{2}{\sqrt{x}+2}\le4\)

\(\Rightarrow P_{max}=4\) khi \(x=0\)

\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)

Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)

Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)