câu a tự quy đồng cùng  mẫu rồi làm thôi :"))

b) \(\left[x.\left(x-1\right)\right].\left[\left(x-2\right).\left(x+1\right)\right]=24\)

\(\Leftrightarrow\left(x^2-x\right).\left(x^2-x-2\right)=24\)

Đặt \(x^2-x=k\), ta có:

\(k.\left(k-2\right)=24\)

\(\Leftrightarrow k^2-2k+1=25\)

\(\Leftrightarrow\left(k-1\right)^2=5^2\Leftrightarrow\orbr{\begin{cases}k-1=5\\k-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}k=6\\k=-4\end{cases}}}\)

\(k=6\Rightarrow x^2-x=6\Rightarrow x^2-x-6=0\)

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

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

\(k=-4\Rightarrow x^2-x+4=0\Rightarrow x^2-x+\frac{1}{4}+\frac{15}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=-\frac{15}{4}\left(\text{loại}\right)\)

c)\(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow x^4+2x^3+2x^2+4x+3x^2-12=0\)

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}\text{vì }x^2+x+6>0\left(\text{tự c/m}\right)}\)

p/s: bn tự kết luận nha :))

a/ Đặt (x^2 - 5x) = a thì ta có

a^2 + 10a + 24 = 0

<=> (a + 4)(a + 6) = 0

Làm nốt

b/ (x - 4)(x - 5)(x - 6)(x - 7) = 1680

<=> (x - 4)(x - 7)(x - 5)(x - 6) = 1680

<=> (x^2 - 11x + 28)(x^2 - 11x + 30) = 1680

Đặt x^2 - 11x + 28 = a thì ta có

a(a + 2) = 1680

<=> (a - 40)(a + 42) = 0

Làm nốt

pt phải có dấu bằng giữa hai vế chứ bạn hik như thiếu đề ạk bn ktra lại xem

bằng 0 nha cậu. Mình ghi thiếu

a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)

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

\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)

Suy ra: \(9-3x+10x-2=4\)

\(\Leftrightarrow7x+7=4\)

\(\Leftrightarrow7x=-3\)

hay \(x=-\dfrac{3}{7}\)

Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)

(x - 1)^3 - x(x + 1)^2 = 5x(2 - x) - 11(x + 2)

<=> -5x^2 + 2x - 1 = -5x² - x - 22

<=> 2x - 1 = -5x² - x - 22 + 5x²

<=> 2x - 1 = -x - 22

<=> 2x - 1 + x = -22

<=> 3x - 1 = -22

<=> 3x = -22 + 1

<=> 3x = -21

<=> x = -7

Vậy: phương trình có nghiệm duy nhất là: S = {-7}

a) ta có : \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)

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

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

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

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

\(\Leftrightarrow\left(x\left(x-2\right)-3\left(x+2\right)\right)\left(x\left(x-1\right)-4\left(x-1\right)\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x-1\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\\x-4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\\x=4\end{matrix}\right.\) vậy \(x=1;x=2;x=3;x=4\)

b) ta có : \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)

\(\Leftrightarrow\left(x^2+x+1\right)^2+\left(x^2+x+1\right)-12=0\)

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

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

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

ta có : \(x^2+x+5>0\forall x\)

\(\Rightarrow pt\Leftrightarrow x^2+x-2=0\Leftrightarrow x^2-x+2x-2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\) vậy \(x=1;x=-2\)

giải giùm mấy câu kia luôn nha bạn ơi

a/ Đặt \(x^2+5x=t\)

\(\Rightarrow t^2-2t-24=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-4\end{matrix}\right.\)

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

b/ Đặt \(x^2-x=t\)

\(\Leftrightarrow t^2-2=t\Leftrightarrow t^2-t-2=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-x=-1\\x^2-x=2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\left(vn\right)\\x^2-x-2=0\end{matrix}\right.\) (casio)

c/ \(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)

Đặt \(x^2+x=t\)

\(\Rightarrow t\left(t-2\right)-24=0\Rightarrow t^2-2t-24=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-4\end{matrix}\right.\)

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