`M=sqrt{(3a-1)^2}+2a-3`

`=|3a-1|+2a-3`

`=3a-1+2a-3(do \ a>=1/3)`

`=5a-4`

`N=sqrt{(4-a)^2}-a+5`

`=|4-a|-a+5`

`=a-4-a+5(do \ a>4)`

`=1`

`I=sqrt{(3-2a)^2}+2-7`

`=|3-2a|-5`

`=3-2a-5(do \ a<3/2)`

`=-2-2a`

`K=(a^2-9)/4*sqrt{4/(a-2)^2}`

`=(a^2-9)/4*|2/(a-2)|`

`=(a^2-9)/(2|a-2|)`

Nếu `3>a>2=>|a-2|=a-2`

`=>K=(a^2-9)/(2(a-2))`

Nếu `a<2=>|a-2|=2-a`

`=>K=(a^2-9)/(2(2-a))`

\(M=\left|3a-1\right|+2a-3\)

Mà \(a-\dfrac{1}{3}\ge0\)

\(\Rightarrow M=3a-1+2a-3=5a-4\)

\(N=\left|4-a\right|-a+5\)

Mà \(4-a< 0\)

\(\Rightarrow N=a-4-a+5=1\)

\(I=\left|3-2a\right|-5\)

Mà \(a-\dfrac{3}{2}< 0\)

\(\Rightarrow I=3-2a-5=-2a-2\)

K, Ta có : \(a-3< 0\)

\(\Rightarrow K=\dfrac{2\left(a^2-9\right)}{4\left|a-2\right|}=\dfrac{\left(a-3\right)\left(a+3\right)}{\left|2a-4\right|}\)
 

a: ĐKXĐ: a<>0; a<>1; a<>-1

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

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

b: Khi a=1/2 thì K=(1/4-1):1/2=-3/4*2=-3/2

Bài 1 :

Để \(\dfrac{x^3+x^2-x-1}{x^3+2x-3}=0\) thì \(x^3+x^2-x-1=0\)

\(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

Vậy,.........

a Để đây là hàm số bậc nhất thì |k-3|<>1

hay \(k\notin\left\{4;2\right\}\)

b: Để đây là hàm số bậc nhất thì k^2-4=0 và k-2<>0

=>k=-2

c: Để đây là hàm số bậc nhất thì \(\dfrac{\sqrt{3-k}}{k+2}< >0\)

=>k<=3 và k<>-2

d: Để đây là hàm số bậc nhất thì k>0; k<>4

a) 1/(b+2)*(sqr(a)+c)=5;

b) sqr(k)+sqr(k+1)<>sqr(k+2);

c) (8*x-7)>1;

d) sqr(b)-4*a*c>=0;

đ) (1/n)*(1/(n+1))*(1/(n+2))<0.01;

e) (a-3)*(a+5)=0

Bài 2:

a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)

b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)

\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)

\(=x^4-22x^3+108x^2-45x\)

c: \(=12x^5-18x^4+30x^3-24x^2\)

d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)

\(ĐKXĐ:x\ge0,x\ne1\)

\(K=\left[\dfrac{x+3\sqrt{x}+2}{x+\sqrt{x}-2}-\dfrac{x+\sqrt{x}}{x-1}\right]:\left[\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\right]\)

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

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

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

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

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

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

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

\(K=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)

b.

Ta có: \(24+\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=24+\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.2\sqrt{5}.3+9}}}\)

\(=24+\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=24+\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=24+\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}=24+\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

\(=24+\sqrt{\sqrt{5}-\sqrt{5}+1}=24+1=25\)

Thay \(x=25\) vào \(K\) ta được:

\(K=\dfrac{\sqrt{x}+1}{2\sqrt{x}}=\dfrac{\sqrt{25}+1}{2.\sqrt{25}}=\dfrac{6}{10}=\dfrac{3}{5}\)

c.

Ta có: \(\dfrac{1}{K}-\dfrac{\sqrt{x}+1}{8}\ge1\)

\(\Rightarrow\dfrac{1}{K}-\dfrac{\sqrt{x}+1}{8}-1\ge0\)

\(\Rightarrow\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{8}-1\ge0\)

\(\Rightarrow\dfrac{16\sqrt{x}}{8\sqrt{x}+8}-\dfrac{x+2\sqrt{x}+1}{8\sqrt{x}+8}-\dfrac{8\sqrt{x}+8}{8\sqrt{x}+8}\ge0\)

\(\Rightarrow\dfrac{16\sqrt{x}-x-2\sqrt{x}-1-8\sqrt{x}-8}{8\sqrt{x}+8}\ge0\)

\(\Rightarrow\dfrac{6\sqrt{x}-x-9}{8\sqrt{x}+8}\ge0\)

\(\Rightarrow\dfrac{-\left(\sqrt{x}-3\right)^2}{8\sqrt{x}+8}\ge0\)

Ta có: \(\left\{{}\begin{matrix}-\left(\sqrt{x}-3\right)^2\le0\\8\sqrt{x}+8\ge0\end{matrix}\right.\)

⇒ Không có \(x\) thỏa mãn