a) \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3x-24}\) \(ĐK:x\ne8\)

\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3\left(x-8\right)}\)

\(\Leftrightarrow\frac{3.3}{6.\left(x-8\right)}+\frac{6.\left(3x-20\right)}{6\left(x-8\right)}-\frac{2\left(3x-102\right)}{6\left(x-8\right)}=\frac{-1}{8}\)

\(\Leftrightarrow\frac{9+18x-120-6x+204}{6\left(x-8\right)}=\frac{-1}{8}\)

\(\Leftrightarrow\frac{12x+93}{6\left(x-8\right)}=\frac{-1}{8}\)

\(\Leftrightarrow8\left(12x+93\right)=-6\left(x-8\right)\)

\(\Leftrightarrow96x+744=-6x+48\)

\(\Leftrightarrow102x=-696\)

\(\Leftrightarrow x=\frac{-116}{17}\) (nhận)

Vậy .....

b) \(\frac{1}{3-x}+\frac{14}{x^2-9}=\frac{x-4}{3+x}+\frac{7}{3+x}\) \(ĐK:x\ne\pm3\)

\(\Leftrightarrow\frac{1}{3-x}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{x-4}{3+x}+\frac{7}{3+x}\)

\(\Leftrightarrow-\frac{3+x}{\left(x-3\right)\left(3+x\right)}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)

\(\Leftrightarrow\frac{-3-x+14}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)

\(\Leftrightarrow-3-x+14=x^2-3x-4x+12+7x-21\)

\(\Leftrightarrow x=-5\) (nhận)

Vậy ....

Lời giải:
a) ĐKXĐ: $x\neq \pm 3; x\neq 0$

\(A=\frac{3-x}{x+3}.\frac{(x+3)^2}{(x-3)(x+3)}.\frac{x+3}{3x^2}\)

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

b)

Với $x=-\frac{1}{2}\Rightarrow A=-\frac{-\frac{1}{2}+3}{3(\frac{-1}{2})^2}=\frac{-10}{3}$

c)

Để $A< 0\Leftrightarrow -\frac{x+3}{3x^2}< 0$

$\Rightarrow x+3>0\Rightarrow x>-3$

Vậy $x>-3; x\neq 3; x\neq 0$

Lời giải:

a) ĐK: $x\neq 8$

PT \(\Leftrightarrow \frac{3}{2(x-8)}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3(x-8)}\)

\(\Leftrightarrow \frac{36}{24(x-8)}+\frac{24(3x-20)}{24(x-8)}+\frac{3(x-8)}{24(x-8)}=\frac{8(13x-102)}{24(x-8)}\)

\(\Rightarrow 36+24(3x-20)+3(x-8)=8(13x-102)\)

\(\Leftrightarrow x=12\) (t/m)

b)

ĐK: $x\neq \pm 2$

PT \(\Leftrightarrow \frac{(x-1)(x-2)}{(x+2)(x-2)}-\frac{x(x+2)}{(x-2)(x+2)}=\frac{5x-2}{(2-x)(x+2)}=\frac{2-5x}{(x-2)(x+2)}\)

\(\Rightarrow (x-1)(x-2)-x(x+2)=2-5x\)

$\Leftrightarrow 0=0$

Vậy PT có nghiệm $x\in\mathbb{R}$ và $x\neq \pm 2$

a/ \(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x+4\right)^2+2\left(x^2+5x+4\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+10\right)=0\)

b/ ĐKXĐ; ...

\(\Leftrightarrow\frac{x^2}{x^2+4x+4}+12x+5=3x^2+6x+2\)

\(\Leftrightarrow\frac{x^2+\left(12x+5\right)\left(x^2+4x+4\right)}{x^2+4x+4}=3x^2+6x+2\)

\(\Leftrightarrow\frac{\left(4x+10\right)\left(3x^2+6x+2\right)}{x^2+4x+4}=3x^2+6x+2\)

\(\Leftrightarrow\left[{}\begin{matrix}3x^2+6x+2=0\\\frac{4x+10}{x^2+4x+4}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3x^2+6x+2=0\\x^2=6\end{matrix}\right.\)

a) Ta có :

(x² + 3x + 2)(x² + 7x + 12) = 24

⇔ ( x + 1 ) ( x + 2 ) (x + 3 ) ( x + 4 ) = 24

⇔ ( x + 1 ) ( x + 4 ) ( x + 2 ) ( x + 3 ) - 24 = 0

⇔ ( x² + 5x + 4 ) ( x² + 5x + 6 ) - 24 = 0

Đặt t = x² + 5x + 4, ta có :

t ( t + 2 ) - 24 = 0

⇔ t² + 2t +1 - 25 = 0

⇔ ( t + 1 )² - 5² = 0

⇔ ( t - 4 ) ( t + 6 ) = 0

\(\Leftrightarrow\left[{}\begin{matrix}t-4=0\\t+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=4\\t=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+4=4\\x^2+5x+4=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+5x=0\\x^2+5x+10=0\end{matrix}\right.\)

Sau đó tìm x bạn tự làm nha

Ý b) là - 3 à !?

\(a)(x^2 + 3x + 2)(x^2 + 7x + 12) = 24\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)

\(\Leftrightarrow\left(x+1\right)\left(x+4\right)\left(x+3\right)\left(x+2\right)=24\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)

Đặt: \(x^2+5x+5=y\) ta được: \(\left(y-1\right)\left(y+1\right)=24\)

\(\Leftrightarrow y^2=25\Leftrightarrow\left[{}\begin{matrix}y=-5\\y=5\end{matrix}\right.\)

Với: \(y=-5\Rightarrow x^2+5x+5=-5\Leftrightarrow x^2+5x+10=0\)

Có: \(\Delta=-15< 0\) vô nghiệm.

Với: \(y=5\Rightarrow x^2+5x+5=5\Leftrightarrow\left(x+5\right)x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy pt có tập \(n_0S=\left\{0;-5\right\}\)

\(b,\frac{x^2}{\left(x+2\right)^2}=3x^2-6x-3\left(Đkxđ:x\ne-2\right)\)

\(\Leftrightarrow x^2=\left(x+2\right)^2\left(3x^2-6x-3\right)\)

\(\Leftrightarrow\left(x^2+4x+4\right)\left(3x^2-6x-3\right)-x^2=0\)

\(\Leftrightarrow\left(x^2+4x+4\right)\left(3x^2+6x+1\right)+\left(x^2+4x+4\right)\left(-12x-5\right)-x^2=0\)

\(\Leftrightarrow\left(x^2+4x+4\right)\left(3x^2+6x+2\right)-\left(12x^3+48x^2+48x+5x^2+20x+20+x^2\right)=0\)

\(\Leftrightarrow\left(x^2+4x+4\right)\left(3x^2+6x+2\right)-2\left(2x+5\right)\left(3x^2+6x+2\right)=0\)

\(\Leftrightarrow\left(3x^2+6x+2\right)\left(x^2-6\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\pm\sqrt{6}\\x=-1\pm\frac{1}{\sqrt{3}}\end{matrix}\right.\)

Vậy ..............

bạn giúp mình giải 3 câu này nhé

a) (x-1)x(x+1)(x+2) = 24

<=> [(x-1)(x+2)][x(x+1) = 24

<=> (x^2+x-2)(x^2+x) = 24     (1)

Đặt t=x^2+x-1 = (x+1/2)^2 - 5/4    (*)

(1) trở thành (t-1)(t+1) = 24

<=> t^2 - 1 - 24 = 0

<=> t^2 - 25 = 0

<=> t^2 = 25

<=> t=5 hoặc t=-5

Mà t >= -5/4 ( từ *) => t = (x+1/2)^2-5/4 = 5

<=> (x+1/2)^2 = 25/4

Đến đây dễ r`

c) x^4 + 3x^3 + 4x^2 + 3x + 1 = 0

<=> x^4 + x^3 + 2x^3 + 2x^2 + 2x^2 + 2x + x + 1 = 0

<=> (x+1)(x^3 + 2x^2 + 2x + 1) = 0

<=> (x +1)(x^3 + x^2 + x^2 + x + x + 1) = 0

<=> (x+1)^2.(x^2+x+1) = 0

Mà x^2+x+1 = (x+1/2)^2 + 3/4 > 0

Nên x+1=0 <=> x=-1

Vậy ...

b, cộng 1 vào 4 phân thức đầu,trừ 4 ở pt cuối ,rồi đặt đc NTC (x+100)

c)Ta có: \(x^4+3x^3+4x^2+3x+1=0\)

\(\Leftrightarrow x\left(x^3+2x^2+2x+1\right)+1\left(x^3+2x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+2x^2+2x+1\right)=0\)

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

Ta có: \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\) nên vô nghiệm

Suy ra x + 1 =0 hay x = -1

X=0 hoặc -1

a) \(x^3-2x^2-5x+6=0\)

\(x^3-x^2-x^2+x-6x+6=0\)

\(x^2\left(x-1\right)-x\left(x-1\right)-6\left(x-1\right)=0\)

\(\left(x-1\right)\left(x^2-x-6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x^2-x-6=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x^2-2x+3x-6=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\\left(x+3\right)\left(x-2\right)=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=\left\{2;-3\right\}\end{cases}}\)

\(a,x^3-2x^2-5x+6=0\)

\(\Leftrightarrow\left(x^3-x^2\right)-\left(x^2-x\right)-\left(6x-6\right)=0\)

\(\Leftrightarrow x^2\left(x-1\right)-x\left(x-1\right)-6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x^2-3x\right)+\left(2x-6\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x\left(x-3\right)+2\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x-3\right)=0\)

\(\Leftrightarrow x-1=0\left(h\right)x+2=0\left(h\right)x-3=0\)

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

Vậy \(x\in\left\{-2;1;3\right\}\)

P/S: (h) là hoặc nhé

\(b,\left|5-3x\right|=3x-5\)

*Nếu \(x\ge\frac{5}{3}\)thì

\(3x-5=3x-5\)Luôn đúng \(\forall x\ge\frac{5}{3}\)

*Nếu \(x< \frac{5}{3}\)thì

\(5-3x=3x-5\)

\(\Leftrightarrow6x=10\)

\(\Leftrightarrow x=\frac{5}{3}\)(loại vì ko thỏa mãn khoảng đag xét)

Vậy \(x\ge\frac{5}{3}\)

Cách khác : dùng tính chất của trị tuyệt đối

\(\left|5-3x\right|=3x-5\)

Vì \(VT\ge0\Rightarrow VP\ge0\)

                  \(\Leftrightarrow3x-5\ge0\)

                   \(\Leftrightarrow x\ge\frac{5}{3}\)

Vậy ...........