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a: |2x-3|=1

=>2x-3=1 hoặc 2x-3=-1

=>x=1(nhận) hoặc x=2(loại)

KHi x=1 thì \(A=\dfrac{1+1^2}{2-1}=2\)

b: ĐKXĐ: x<>-1; x<>2

\(B=\dfrac{2x^2-4x+3x+3-2x^2-1}{\left(x-2\right)\left(x+1\right)}=\dfrac{-x+2}{\left(x-2\right)\left(x+1\right)}=\dfrac{-1}{x+1}\)

a, Thay x = 1/4 vào A ta được : 

\(A=\dfrac{\dfrac{1}{2}+1}{\dfrac{1}{2}-3}=\dfrac{\dfrac{3}{2}}{-\dfrac{5}{2}}=-\dfrac{3}{5}\)

b, Với x >= 0 ; x khác 1 ; 9 

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

Hai biểu thức này chỉ có min thui bạn nhé.

1.

\(N=\frac{2x+5}{\sqrt{x}+1}=\frac{2\sqrt{x}(\sqrt{x}+1)-2(\sqrt{x}+1)+7}{\sqrt{x}+1}=2\sqrt{x}-2+\frac{7}{\sqrt{x}+1}\)

\(=2(\sqrt{x}+1)+\frac{7}{\sqrt{x}+1}-4\)

\(=\frac{7}{16}(\sqrt{x}+1)+\frac{7}{\sqrt{x}+1}+\frac{25}{16}(\sqrt{x}+1)-4\)

\(\geq 2\sqrt{\frac{7}{16}.7}+\frac{25}{16}(\sqrt{9}+1)-4=\frac{23}{4}\) (theo BĐT AM-GM)

Vậy $N_{\min}=\frac{23}{4}$ khi $x=9$

2.

\(F=\frac{x+3}{\sqrt{x}+1}=\frac{\sqrt{x}(\sqrt{x}+1)-(\sqrt{x}+1)+4}{\sqrt{x}+1}=\sqrt{x}-1+\frac{4}{\sqrt{x}+1}\)

\(=\frac{4}{9}(\sqrt{x}+1)+\frac{4}{\sqrt{x}+1}+\frac{5\sqrt{x}}{9}-\frac{13}{9}\)

\(\geq 2\sqrt{\frac{4}{9}.4}+\frac{5\sqrt{4}}{9}-\frac{13}{9}=\frac{7}{3}\)

Vậy $F_{\min}=\frac{7}{3}$ khi $x=4$

1 quy đồng lên ra được

2 \(A=\dfrac{1}{x-2\sqrt{x-5}+3}\le\dfrac{1}{5-2.0+3}=\dfrac{1}{8}\)

dấu"=" xảy ra<=>x=5

a) Thay x=4 vào biểu thức \(B=\dfrac{3}{\sqrt{x}-1}\), ta được:

\(B=\dfrac{3}{\sqrt{4}-1}=\dfrac{3}{2-1}=3\)

Vậy: Khi x=4 thì B=3

b) Ta có: P=A-B

\(\Leftrightarrow P=\dfrac{6}{x-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{3}{\sqrt{x}-1}\)

\(\Leftrightarrow P=\dfrac{6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Leftrightarrow P=\dfrac{6+x-\sqrt{x}-3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Leftrightarrow P=\dfrac{x-\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Leftrightarrow P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Leftrightarrow P=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Leftrightarrow P=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

Lời giải:

$\frac{\sqrt{x}+1}{\sqrt{x}+4}=\frac{\sqrt{x}+4-3}{\sqrt{x}+4}=1-\frac{3}{\sqrt{x}+4}$

Vì $\sqrt{x}\geq 0$ nên $\sqrt{x}+4\geq 4$
$\Rightarrow \frac{3}{\sqrt{x}+4}\leq \frac{3}{4}$

$\Rightarrow \frac{\sqrt{x}+1}{\sqrt{x}+4}=1-\frac{3}{\sqrt{x}+4}\geq 1-\frac{3}{4}=\frac{1}{4}$

Vậy $M=\frac{1}{4}$

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$N=\frac{\sqrt{x}+5}{\sqrt{x}+2}=1+\frac{3}{\sqrt{x}+2}$

Do $\sqrt{x}\geq 0$ nên $\sqrt{x}+2\geq 2$

$\Rightarrow \frac{3}{\sqrt{x}+2}\leq \frac{3}{2}$

$\Rightarrow \frac{\sqrt{x}+5}{\sqrt{x}+2}\leq 1+\frac{3}{2}=\frac{5}{2}$

Vậy $N=\frac{5}{2}$

$\Rightarrow 2M+N =2.\frac{1}{4}+\frac{5}{2}=3$

Đáp án C.

1/  196

2/  5/4

3/  1/3