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

c/ (x + 1)^4 + (x - 1)^4 = 82

<=> x^4 + 6x^2 - 40 = 0

Đặt x^2 = a

=> a^2 + 6a - 40 = 0

<=> (a - 4)(a + 10) = 0

a: =>(x^2+4x-5)(x^2+4x-21)=297

=>(x^2+4x)^2-26(x^2+4x)+105-297=0

=>x^2+4x=32 hoặc x^2+4x=-6(loại)

=>x^2+4x-32=0

=>(x+8)(x-4)=0

=>x=4 hoặc x=-8

b: =>(x^2-x-3)(x^2+x-4)=0

hay \(x\in\left\{\dfrac{1+\sqrt{13}}{2};\dfrac{1-\sqrt{13}}{2};\dfrac{-1+\sqrt{17}}{2};\dfrac{-1-\sqrt{17}}{2}\right\}\)

c: =>(x-1)(x+2)(x^2-6x-2)=0

hay \(x\in\left\{1;-2;3+\sqrt{11};3-\sqrt{11}\right\}\)

PT đã cho \(\Leftrightarrow\left(x^2-4-5x+5\right)\left(x^2-4\right)=6\left(x-1\right)^{2 }\)
\(\Leftrightarrow\left(x^2-4-5\left(x-1\right)\right)\left(x^2-4\right)=6\left(x-1\right)^2\)(*)
ĐẶt \(x^2-4=a.\)\(x-1=b\)
PT(*) có dạng \(\left(a-5b\right)a=6b^2\Leftrightarrow a^2-5ab-6b^2=0\Leftrightarrow\left(a+b\right)\left(a-6b\right)=0\)
\(\cdot a+b=0\Leftrightarrow x^2-4+x-1=0\Leftrightarrow x^2+x-5=0\)
\(\Rightarrow x_1=\frac{-1+\sqrt{21}}{2}.x_2=\frac{-1-\sqrt{21}}{2}\)
\(.a-6b=0\Leftrightarrow x^2-4-6\left(x-1\right)=0\Leftrightarrow x^2-6x+2=0\)
\(\Rightarrow x_3=3+\sqrt{7}.x_4=3-\sqrt{7}\)
THử lại: các nghiệm trên đều thỏa mãn pt 
Vậy :....
p/s : học khuya thế ==ơ

bạn còn cách nào khác giải theo sách lp9 k ????

Cái này là lớp 9 mà ?

a/

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

\(\Leftrightarrow\left[{}\begin{matrix}-5x=0\\2x^2-5x-8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{5\pm\sqrt{89}}{4}\\\end{matrix}\right.\)

b/ - Với \(x\ge3\) pt trở thành:

\(x-1+3\left(x-3\right)=6\Leftrightarrow4x=16\Rightarrow x=4\)

- Với \(x\le1\) pt trở thành:

\(1-x+3\left(3-x\right)=6\)

\(\Leftrightarrow x=1\)

- Với \(1< x< 3\) pt trở thành:

\(x-1+3\left(3-x\right)=6\)

\(\Leftrightarrow-2x=-2\Rightarrow x=1\) (loại)

c/ ĐKXĐ: \(x\ne\pm2\)

\(\left[{}\begin{matrix}\frac{x^2-6x-4}{x^2-4}=1\\\frac{x^2-6x-4}{x^2-4}=-1\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}-6x=0\\2x^2-6x-8=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=4\end{matrix}\right.\)

d/ - Với \(x\ge2\) pt trở thành:

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

\(\Leftrightarrow x^2=6\Rightarrow\left[{}\begin{matrix}x=\sqrt{6}\\x=-\sqrt{6}\left(l\right)\end{matrix}\right.\)

- Với \(x\le1\) pt trở thành:

\(1-x-2\left(2-x\right)=x^2-x-3\) làm tương tự

- Với \(1< x< 2\):

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

a.

\(\left\{{}\begin{matrix}x^4+y^4=34\\y=2-x\end{matrix}\right.\)

\(\Rightarrow x^4+\left(x-2\right)^4=34\)

Đặt \(x-1=t\)

\(\Rightarrow\left(t+1\right)^4+\left(t-1\right)^4=34\)

\(\Leftrightarrow t^4+6t^2-16=0\Rightarrow\left[{}\begin{matrix}t^2=2\\t^2=-8\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}t=\sqrt{2}\Rightarrow x=\sqrt{2}+1\Rightarrow y=1-\sqrt{2}\\t=-\sqrt{2}\Rightarrow x=1-\sqrt{2}\Rightarrow y=1+\sqrt{2}\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}xy^2-x^2y+6x-y^2-y-6=0\\x^2y-xy^2+6y-x^2-x-6=0\end{matrix}\right.\) (1)

Lần lượt cộng 2 vế và trừ 2 vế ta được:

\(\left\{{}\begin{matrix}-x^2-y^2+5x+5y-12=0\\2xy\left(y-x\right)+7\left(x-y\right)+\left(x-y\right)\left(x+y\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-5\left(x+y\right)+12=0\\\left(y-x\right)\left(2xy-x-y-7\right)=0\end{matrix}\right.\)

Th1: \(\left\{{}\begin{matrix}x=y\\x^2+y^2-5\left(x+y\right)+12=0\end{matrix}\right.\)

\(\Rightarrow2x^2-10x+12=0\Rightarrow...\)

TH2: \(\left\{{}\begin{matrix}2xy-\left(x+y\right)-7=0\\x^2+y^2-5\left(x+y\right)+12=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2xy-\left(x+y\right)-7=0\\\left(x+y\right)^2-2xy-5\left(x+y\right)+12=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2v-u-7=0\\u^2-2v-5u+12=0\end{matrix}\right.\)

\(\Rightarrow u^2-6u+5=0\)

\(\Leftrightarrow...\)

a) \(\frac{1}{x+2}+\frac{2}{x+3}=\frac{6}{x+4}\)

ĐKXĐ \(x\ne-2,-3,-4\)

=> \(\frac{1}{x+2}+\frac{2}{x+3}-\frac{6}{x+4}=0\)

=> \(\frac{3x+7}{\left(x+2\right)\left(x+3\right)}-\frac{6}{x+4}=0\)

=> \(\frac{\left(3x+7\right)\left(x+4\right)-6\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x+4\right)}=0\)

=> (3x + 7)(x + 4) - 6(x² + 5x + 6) = 0

=> 3x² + 19x + 28 - 6x² - 30x - 36 = 0

=> -3x² - 11x - 8 = 0

=> -3x² - 3x - 8x - 8 = 0

=> -3x(x + 1) - 8(x + 1) = 0

=> (x + 1)(-3x - 8) = 0

=> \(\orbr{\begin{cases}x=-1\\x=-\frac{8}{3}\end{cases}}\)

Vậy ...

b) Thiếu dữ liệu cuả đề 

c) \(\frac{6x+22}{x+2}-\frac{2x+7}{x+3}=\frac{x+4}{x^2+5x+6}\)

ĐKXĐ \(x\ne-2;-3\)

=> \(\frac{\left(6x+22\right)\left(x+3\right)-\left(x+2\right)\left(2x+7\right)}{\left(x+2\right)\left(x+3\right)}=\frac{x+4}{\left(x+2\right)\left(x+3\right)}\)

=> \(6x^2+40x+66-x\left(2x+7\right)-2\left(2x+7\right)=x+4\)

=> \(6x^2+40x+66-2x^2-7x-4x-14=x+4\)

=> 4x² + 29x + 52 = x + 4

=> 4x² + 29x + 52 - x - 4 = 0

=> 4x² + 28x + 48 = 0

=> 4(x² + 7x + 12) = 0

=> x² + 7x +12 = 0

=> x² + 3x + 4x + 12 = 0

=> x(x + 3) + 4(x + 3) = 0

=> (x + 3)(x + 4) = 0

=> \(\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\) 

Mà \(x\ne-2,-3\)nên x = -3 loại

Vậy x = -4

a, ĐK: \(x\ge1\)

Đặt \(\sqrt{5x-1}=a;\sqrt{x-1}=b\left(a,b\ge0\right)\)

\(pt\Leftrightarrow\left(a+b\right)\left(\dfrac{a^2+b^2}{2}-ab\right)=a^2-b^2\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2=2\left(a-b\right)\left(a+b\right)\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)\left(a-b-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=b+2\end{matrix}\right.\)

TH1: \(a=b\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}\Leftrightarrow x=0\left(l\right)\)

TH2: \(a=b+2\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}+2\)

\(\Leftrightarrow5x-1=x-1+4+4\sqrt{x-1}\)

\(\Leftrightarrow4x-4-4\sqrt{x-1}=0\)

\(\Leftrightarrow4x-4-4\sqrt{x-1}+1=1\)

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

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

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

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