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24 tháng 6 2021

`đk:x ne +-3,x ne -2`

`B=(21/(x^2-9)-(x-4)/(3-x)-(x-1)/(3+x)):(1-1/(x+3))`

`=(21/(x^2-9)+(x-4)/(x-3)-(x-1)/(x+3)):((x+3-1)/(x+3))`

`=((21+x^2-x-12-x^2+4x-3)/((x-3)(x+3))):(x+2)/(x+3)`

`=(3x+6)/((x-3)(x+3))*(x+3)/(x+2)`

`=(3x+6)/((x-3)(x+2))`

`=3/(x-3)`

`b)|2x+1|=5`

`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=2(tm)\\x=-3(l)\end{array} \right.\) 

`=>B=3/(2-3)=-3`

`c)B=-3/5`

`<=>3/(x-3)=3/(-5)`

`<=>x-3=-5`

`<=>x=-2(l)`

`d)B<0`

`<=>3/(x-3)<0`

Mà `3>0`

`=>x-3<0<=>x<3`

24 tháng 6 2021

a) đk: \(x\ne\pm3\)

 \(B=\left[\dfrac{21}{\left(x-3\right)\left(x+3\right)}+\dfrac{x-4}{x-3}-\dfrac{x-1}{x+3}\right]:\left(\dfrac{x+3-1}{x+3}\right)\)

\(\left[\dfrac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right]:\dfrac{x+2}{x+3}\)

\(\dfrac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}\)

\(\dfrac{3x+6}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}=\dfrac{3}{x-3}\)

b) Để \(\left|2x+1\right|=5\)

<=> \(\left[{}\begin{matrix}2x+1=5< =>x=2\left(c\right)\\2x+1=-5< =>x=-3\left(l\right)\end{matrix}\right.\)

Thay x = 2, ta có;

B = \(\dfrac{3}{2-3}=-3\)

c) Để B = \(\dfrac{-3}{5}\)

<=> \(\dfrac{3}{x-3}=\dfrac{-3}{5}\)

<=> x - 3 = -5

<=> x = -2

d) Để B < 0

<=> \(\dfrac{3}{x-3}< 0\)

<=> x - 3 < 0

<=> x < 3

10 tháng 7 2021

a) \(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\left(x\ge0,x\ne1\right)\)

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

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

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

 

10 tháng 7 2021

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24 tháng 6 2021

`a)P=(x/(x+2)-(x^3-8)/(x^3+8)*(x^2-2x+4)/(x^2-4)):4/(x+2)`

`đk:x ne 0,x ne -2`

`P=(x/(x+2)-((x-2)(x^2+2x+4))/((x+2)(x^2-2x+4))*(x^2-2x+4)/((x-2)(x+2)))*(x+2)/4`

`=(x/(x+2)-(x^2+2x+4)/(x+2)^2)*(x+2)/4`

`=(x^2+2x-x^2-2x-4)/(x+2)^2*(x+2)/4`

`=-4/(x+2)^2*(x+2)/4`

`=-1/(x+2)`

`b)P<0`

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

Vì `-1<0`

`<=>x+2>0`

`<=>x> -2`

`c)P=1/x+1(x ne 0)`

`<=>-1/(x+2)=1/x+1`

`<=>1/x+1+1/(x+2)=0``

`<=>x+2+x(x+2)+x=0`

`<=>x^2+4x+2=0`

`<=>` \(\left[ \begin{array}{l}x=\sqrt2-2\\x=-\sqrt2-2\end{array} \right.\) 

`d)|2x-1|=3`

`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-2\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=2(l)\\x=-1(tm)\end{array} \right.\) 

`x=-1=>P=-1/(-1+2)=-1`

`e)P=-1/(x+2)` thì nhỏ nhất cái gì nhỉ?

24 tháng 6 2021

a) đk: \(x\ne-2;2\)

 \(P=\left[\dfrac{x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}.\dfrac{x^2-2x+4}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{4}{x+2}\)

\(\left[\dfrac{x}{x+2}-\dfrac{x^2+2x+4}{\left(x+2\right)^2}\right].\dfrac{x+2}{4}\)

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

b) Để P < 0

<=> \(\dfrac{-1}{x+2}< 0\)

<=> x +2 > 0

<=> x > -2 ( x khác 2)

c) Để P= \(\dfrac{1}{x}+1\)

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

<=> \(\dfrac{1}{x}+\dfrac{1}{x+2}+1=0\)

<=> \(\dfrac{x+2+x+x\left(x+2\right)}{x\left(x+2\right)}=0\)

<=> x2 + 4x + 2 = 0

<=> (x+2)2 = 2

<=> \(\left[{}\begin{matrix}x=\sqrt{2}-2\left(c\right)\\x=-\sqrt{2}-2\left(c\right)\end{matrix}\right.\)

d) Để \(\left|2x-1\right|=3\)

<=> \(\left[{}\begin{matrix}2x-1=3< =>x=2\left(l\right)\\2x-1=-3< =>x=-1\left(c\right)\end{matrix}\right.\)

Thay x = -1, ta có:

P = \(\dfrac{-1}{-1+2}=-1\)

 

a: \(E=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)

\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)

\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)

b: |x-3|=2

=>x-3=2 hoặc x-3=-2

=>x=5(nhận) hoặc x=1(loại)

Khi x=5 thì \(E=\dfrac{5^2}{5-1}=\dfrac{25}{4}\)

c: Để E=1/2 thì \(\dfrac{x^2}{x-1}=\dfrac{1}{2}\)

\(\Leftrightarrow2x^2-x+1=0\)

hay \(x\in\varnothing\)

 

19 tháng 5 2022

f) \(A=\dfrac{x^2}{x-1}=\dfrac{x^2-x+x-1+1}{x-1}=\dfrac{x\left(x-1\right)+x-1+1}{x-1}=x+1+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+2\ge2\sqrt{\left(x-1\right).\dfrac{1}{x-1}}+2=4\)\(A=4\Leftrightarrow x=2\)

-Vậy \(A_{min}=4\)

21 tháng 1 2021

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21 tháng 1 2021

Bổ sung phần c và d luôn:

c, C = \(\dfrac{2}{5}\)

\(\Leftrightarrow\) \(\dfrac{x^2-1}{2x^2+3}\) = \(\dfrac{2}{5}\)

\(\Leftrightarrow\) 5(x2 - 1) = 2(2x2 + 3)

\(\Leftrightarrow\) 5x2 - 5 = 4x2 + 6

\(\Leftrightarrow\) x2 = 11

\(\Leftrightarrow\) x2 - 11 = 0

\(\Leftrightarrow\) (x - \(\sqrt{11}\))(x + \(\sqrt{11}\)) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-\sqrt{11}=0\\x+\sqrt{11}=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\sqrt{11}\left(TM\right)\\x=-\sqrt{11}\left(TM\right)\end{matrix}\right.\)

d, Ta có: \(\dfrac{x^2-1}{2x^2+3}\) = \(\dfrac{x^2+\dfrac{3}{2}-\dfrac{5}{2}}{2\left(x^2+\dfrac{3}{2}\right)}\) = \(\dfrac{1}{2}\) - \(\dfrac{5}{4\left(x^2+\dfrac{3}{2}\right)}\)

C nguyên \(\Leftrightarrow\) \(\dfrac{5}{4\left(x^2+\dfrac{3}{2}\right)}\) nguyên \(\Leftrightarrow\) 5 \(⋮\) 4(x2 + \(\dfrac{3}{2}\))

\(\Leftrightarrow\) 4(x2 + \(\dfrac{3}{2}\)\(\in\) Ư(5)

Xét các TH:

4(x2 + \(\dfrac{3}{2}\)) = 5 \(\Leftrightarrow\) x2 = \(\dfrac{-1}{4}\) \(\Leftrightarrow\) x2 + \(\dfrac{1}{4}\) = 0 (Vô nghiệm)

4(x2 + \(\dfrac{3}{2}\)) = -5 \(\Leftrightarrow\) x2 = \(\dfrac{-11}{4}\) \(\Leftrightarrow\) x2 + \(\dfrac{11}{4}\) = 0 (Vô nghiệm)

4(x2 + \(\dfrac{3}{2}\)) = 1 \(\Leftrightarrow\) x2 = \(\dfrac{-5}{4}\) \(\Leftrightarrow\) x2 + \(\dfrac{5}{4}\) = 0 (Vô nghiệm)

4(x2 + \(\dfrac{3}{2}\)) = -1 \(\Leftrightarrow\) x2 = \(\dfrac{-7}{4}\) \(\Leftrightarrow\) x2 + \(\dfrac{7}{4}\) = 0 (Vô nghiệm)

Vậy không có giá trị nào của x \(\in\) Z thỏa mãn C \(\in\) Z

Chúc bn học tốt! (Ko bt đề sai hay ko nữa :v)

10 tháng 1 2021

a) đặt mẫu chứng là x-2

a: \(A=\dfrac{-\left(x+2\right)^2-2x\left(x-2\right)-4x^2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x-3\right)}{\left(x-3\right)^2}\)

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

\(=\dfrac{-7x^2-4}{\left(x+2\right)}\cdot\dfrac{-1}{x-3}=\dfrac{7x^2+4}{\left(x+2\right)\left(x-3\right)}\)

b: Khi x=1/3 thì \(A=\dfrac{7\cdot\dfrac{1}{9}+4}{\left(\dfrac{1}{3}-2\right)\left(\dfrac{1}{3}-3\right)}=\dfrac{43}{40}\)

a) Ta có: \(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)

\(=\dfrac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{11x-3}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{2x^2-6x+x^2+4x+3+11x-3}{\left(x-3\right)\left(x+3\right)}\)

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

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

b)

ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)

Ta có: P=AB

\(=\dfrac{3x}{x-3}\cdot\dfrac{x-3}{x+1}\)

\(=\dfrac{3x}{x+1}\)

Để \(P=\dfrac{9}{2}\) thì \(\dfrac{3x}{x+1}=\dfrac{9}{2}\)

\(\Leftrightarrow9\left(x+1\right)=6x\)

\(\Leftrightarrow9x-6x=-9\)

\(\Leftrightarrow3x=-9\)

hay x=-3(loại)

Vậy: Không có giá trị nào của x để \(P=\dfrac{9}{2}\)

a: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{\sqrt{x}-1+1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{5}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}}=\dfrac{x-1}{\sqrt{x}}+\dfrac{5}{\sqrt{x}}=\dfrac{x+4}{\sqrt{x}}\)

b: Để A=5 thì \(x+4=5\sqrt{x}\)

=>x=1(loại) hoặc x=16(nhận)