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\(B=x\left(x-3\right)\left(x+1\right)\left(x+4\right)\)

\(B=\left[x\left(x+1\right)\right]\left[\left(x-3\right)\left(x+4\right)\right]\)

\(B=\left(x^2+x\right)\left(x^2+x-12\right)\)

Đặt \(x^2+x=a\)ta được;

\(B=a\left(a-12\right)=a^2-12a=\left(a^2-2.a.6+36\right)-36\)\(=\left(a-6\right)^2-36\)

Vì \(\left(a-6\right)^2\ge0\)\(\Rightarrow\left(a-6\right)^2-36\ge-36\)

Dấu ''='' xảy ra khi \(a-6=0\Rightarrow a=6\Rightarrow x^2+x-6=0\)\(\Rightarrow\left(x^2+3x\right)-\left(2x+6\right)=0\)

\(\Rightarrow x\left(x+3\right)-2\left(x+3\right)=0\)\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

Vậy GTNN của B là B=-36 khi x=-3 hoặc x=2

9 tháng 7 2021

`B=(1/(3-sqrtx)-1/(3+sqrtx))*(3+sqrtx)/sqrtx(x>=0,x ne 9)`

`B=((3+sqrtx)/(9-x)-(3-sqrtx)/(9-x))*(3+sqrtx)/sqrtx`

`B=((3+sqrtx-3+sqrtx)/(9-x))*(3+sqrtx)/sqrtx`

`B=(2sqrtx)/((3-sqrtx)(3+sqrtx))*(3+sqrtx)/sqrtx`

`B=2/(3-sqrtx)`

`B>1/2`

`<=>2/(3-sqrtx)-1/2>0`

`<=>(4-3+sqrtx)/[2(3-sqrtx)]>0`

`<=>(sqrtx+1)/(2(3-sqrtx))>0`

Mà `sqrtx+1>=1>0`

`<=>2(3-sqrtx)>0`

`<=>3-sqrtx>0`

`<=>sqrtx<3`

`<=>x<9`

6 tháng 2 2022

ĐKXĐ: \(x\ge0;x\ne9\)

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

\(=\left[\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{1}{\sqrt{x}-3}\)

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

\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)

\(B=\dfrac{\sqrt{x}-3+3}{x-9}\cdot\left(\sqrt{x}-3\right)=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

NV
5 tháng 5 2021

\(A=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)\)

\(A=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)

\(A=\left(x^2+5x+4\right)\left(x^2+5x+4+2\right)\)

\(A=\left(x^2+5x+4\right)^2+2\left(x^2+5x+4\right)\)

\(A=\left(x^2+5x+4\right)^2+2\left(x^2+5x+4\right)+1-1\)

\(A=\left(x^2+5x+5\right)^2-1\ge-1\)

\(A_{min}=-1\) khi \(x^2+5x+5=0\)

14 tháng 7 2021

\(B=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1}{x-1}=\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{x-1}{x+1}\)

\(=\dfrac{2}{x-1}.\dfrac{x-1}{x+1}=\dfrac{2}{x+1}\)

Để \(B< 1\Rightarrow\dfrac{2}{x+1}< 1\Rightarrow1-\dfrac{2}{x+1}>0\Rightarrow\dfrac{x-1}{x+1}>0\)

mà \(x+1>0\left(x\ge0\right)\Rightarrow x-1>0\Rightarrow x>1\)

AH
Akai Haruma
Giáo viên
6 tháng 7 2021

Lời giải:

\(B=\frac{3}{x-1}\sqrt{\frac{(x-1)^2}{(3x)^2}}=\frac{3}{x-1}|\frac{x-1}{3x}|\)

\(=\frac{3}{x-1}.\frac{1-x}{3x}=\frac{-1}{x}\)

6 tháng 7 2021

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

\(=\dfrac{3}{x-1}.\left|\dfrac{x-1}{3x}\right|=\dfrac{3}{x-1}.\dfrac{1-x}{3x}=-\dfrac{1}{x}\)

17 tháng 7 2021

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

\(B=\dfrac{x-\sqrt{x}-5\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{x-6\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(B=\dfrac{\left(\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-4}{\sqrt{x}}\)

Ta có: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-8}{x-2\sqrt{x}}\)

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

\(=\dfrac{x-6\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)

29 tháng 6 2021

Đk:\(x>0;x\ne1\)

\(B=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(x-1\right)}\)

\(=\dfrac{\sqrt{x}+1+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{1}{\sqrt{x}-1}\)

\(B=\dfrac{1}{2}\Leftrightarrow\dfrac{1}{\sqrt{x}-1}=\dfrac{1}{2}\Leftrightarrow\sqrt{x}-1=2\)\(\Leftrightarrow x=9\) (tm)

Vậy..

29 tháng 6 2021

 

a) \(B=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{x-1}\right):\dfrac{x\sqrt{x}-1}{x\sqrt{x}-\sqrt{x}}\)

\(B=\dfrac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

\(B=\dfrac{1}{\sqrt{x}-1}\)

b) Với \(B=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{x}-1}=\dfrac{1}{2}\Leftrightarrow\sqrt{x}-1=2\)

\(\Leftrightarrow\sqrt{x}=3\)

\(\Leftrightarrow x=9\)

Vậy...

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