Đcm học ngu k biết xài caskov

a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne-3\end{cases}}\)

b) \(P=1+\frac{x+3}{x^2+5x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)

\(\Leftrightarrow P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)

\(\Leftrightarrow P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)\)

\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow P=1+\frac{\left(x-2\right)\left(x+2\right)}{6\left(x+2\right)}\)

\(\Leftrightarrow P=1+\frac{x-2}{6}\)

\(\Leftrightarrow P=\frac{x+4}{6}\)

c) Để P = 0

\(\Leftrightarrow\frac{x+4}{6}=0\)

\(\Leftrightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Để P = 1

\(\Leftrightarrow\frac{x+4}{6}=1\)

\(\Leftrightarrow x+4=6\)

\(\Leftrightarrow x=2\)

d) Để P > 0

\(\Leftrightarrow\frac{x+4}{6}>0\)

\(\Leftrightarrow x+4>0\)(Vì 6>0)

\(\Leftrightarrow x>-4\)

a: ĐKXĐ: x∉{2;-2;3}

\(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}\cdot\frac{4x^2-8x+16}{x^2-4}\right):\frac{16}{x^2-x-6}\)

\(=\left(\frac{4x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\frac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\right):\frac{16}{\left(x-3\right)\left(x+2\right)}\)

\(=\left(\frac{4x}{x+2}-\frac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right)\cdot\frac{\left(x-3\right)\cdot\left(x+2\right)}{16}\)

\(=\frac{4x\left(x+2\right)-4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\cdot\frac{\left(x-3\right)\left(x+2\right)}{16}=\frac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)}\cdot\frac{x-3}{16}\)

\(=\frac{-16\left(x-3\right)}{16\left(x+2\right)}=\frac{-x+3}{x+2}\)

b: A<0

=>\(\frac{-x+3}{x+2}<0\)

=>\(\frac{x-3}{x+2}>0\)

=>x>3 hoặc x<-2

c: A>=5

=>\(\frac{-x+3}{x+2}-5\ge0\)

=>\(\frac{-x+3-5x-10}{x+2}\ge0\)

=>\(\frac{-6x-7}{x+2}\ge0\)

=>\(\frac{6x+7}{x+2}\le0\)

=>-2<x<=-7/6

ĐKXĐ: \(x\notin\left\{2;-2;0;3\right\}\)

Ta có: \(P=\left(\dfrac{4x}{2+x}+\dfrac{8x^2}{4-x^2}\right):\left(\dfrac{x-1}{x^2-2x}-\dfrac{2}{x}\right)\)

\(=\left(\dfrac{4x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{8x^2}{\left(x+2\right)\left(x-2\right)}\right):\left(\dfrac{x-1}{x\left(x-2\right)}-\dfrac{2\left(x-2\right)}{x\left(x-2\right)}\right)\)

\(=\dfrac{4x^2-8x-8x^2}{\left(x+2\right)\left(x-2\right)}:\dfrac{x-1-2x+4}{x\left(x-2\right)}\)

\(=\dfrac{-4x^2-8x}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x\left(x-2\right)}{-x+3}\)

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

\(=\dfrac{-4x^2}{3-x}\)

Để P<0 thì \(\dfrac{-4x^2}{3-x}< 0\)

mà \(-4x^2< 0\forall x\) thỏa mãn ĐKXĐ

nên 3-x<0

hay x>3

Kết hợp ĐKXĐ, ta được: x>3

Vậy: Để P<0 thì x>3

Lời giải:
\(\frac{4x^2-8x}{-x^2+x+6}<0\\ \Leftrightarrow \frac{4x(x-2)}{-(x^2-x-6)}<0\\ \Leftrightarrow \frac{4x(x-2)}{x^2-x-6}>0\\ \Leftrightarrow \frac{4x(x-2)}{(x+2)(x-3)}>0\)

Đến đây xảy ra 2 TH:

TH1: $4x(x-2)>0$ và $(x+2)(x-3)>0$

$4x(x-2)>0\Leftrightarrow x> 2$ hoặc $x<0(1)$

$(x+2)(x-3)>0\Leftrightarrow x> 3$ hoặc $x<-2(2)$

Từ $(1); (2)\Rightarrow x>3$ hoặc $x<-2$

TH2: $4x(x-2)<0$ và $(x+2)(x-3)<0$

$4x(x-2)<0\Leftrightarrow 0< x< 2(3)$

$(x+2)(x-3)<0\Leftrightarrow -2< x< 3(4)$

Từ $(3); (4)\Rightarrow 0< x< 2$

Vậy $x>3$ hoặc $x< -2$ hoặc $0< x< 2$

Lời giải:
a. Để biểu thức xác định thì:

$x^2-x-6\geq 0$

$\Leftrightarrow (x+2)(x-3)\geq 0$

$\Leftrightarrow x\geq 3$ hoặc $x\leq -2$

b. Để biểu thức xác định thì:

$4x-x^2-5\geq 0$

$\Leftrightarrow x^2-4x+5\leq 0$

$\Leftrightarrow (x-2)^2+1\leq 0$

$\Leftrightarrow (x-2)^2\leq -1< 0$ (vô lý)

Vậy không tồn tại $x$ để bt xác định

c. Để biểu thức xác định thì:

$x^2-8x+15>0$

$\Leftrightarrow (x-3)(x-5)>0$

$\Leftrightarrow x>5$ hoặc $x< 3$

a) ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le-2\end{matrix}\right.\)

b) ĐKXĐ: \(x\in\varnothing\)

c) ĐKXĐ: \(\left[{}\begin{matrix}x>5\\x< 3\end{matrix}\right.\)

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(a.P=1+\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)=\dfrac{x+3}{x+2}:\left(\dfrac{2}{x-2}-\dfrac{3}{x^2-4}-\dfrac{1}{x+2}\right)=\dfrac{x+3}{x+2}.\dfrac{\left(x+2\right)\left(x-2\right)}{2x+4-3-x+2}=\left(x+3\right).\dfrac{x-2}{x+3}=x-2\left(x\ne\pm2;x\ne-3\right)\)

\(b.P=0\Leftrightarrow x-2=0\Leftrightarrow x=2\left(KTM\right)\)

\(P=1\Leftrightarrow x-2=1\Leftrightarrow x=3\left(TM\right)\)

\(c.P>0\Leftrightarrow x-2>0\Leftrightarrow x>2\)

ĐKXĐ: x<>1/2; x<>-1/2

\(\frac{1+8x}{4+8x}-\frac{4x}{12x-6}+\frac{\frac{32}{3}x^2}{4-16x^2}=0\)

=>\(\frac{8x+1}{4\left(2x+1\right)}-\frac{4x}{6\left(2x-1\right)}-\frac{32x^2}{3\cdot4\left(2x-1\right)\left(2x+1\right)}=0\)

=>\(\frac{8x+1}{4\left(2x+1\right)}-\frac{2x}{3\left(2x-1\right)}-\frac{32x^2}{12\cdot\left(2x-1\right)\left(2x+1\right)}=0\)

=>\(\frac{3\left(8x+1\right)\left(2x-1\right)}{12\left(2x+1\right)\left(2x-1\right)}-\frac{8x\left(2x+1\right)}{12\left(2x+1\right)\left(2x-1\right)}-\frac{32x^2}{12\left(2x+1\right)\left(2x-1\right)}=0\)

=>3(8x+1)(2x-1)-8x(2x+1)-32x^2=0

=>\(3\left(16x^2-8x+2x-1\right)-16x^2-8x-32x^2=0\)

=>\(48x^2-18x-3-48x^2-8x=0\)

=>-26x=3

=>x=-3/26(nhận)

dễ tự làm