PT   0,2-(5x-10)-0,1(10x-5)=0

    <=>  0,2-5x+10-x+0,5=0

   <=>10,7-6x=0

   <=>6x=10,7

   <=>6=107/60

<=>\(\orbr{\begin{cases}0,2-\left(5x-10\right)=0\\0,1\left(10x-5\right)=0\end{cases}}\)<=>\(\orbr{\begin{cases}5x-10=0,2-0\\10x-5=0,1-0\end{cases}}\)<=>\(\orbr{\begin{cases}5x-10=0,2\\10x-5=0,1\end{cases}}\)<=>\(\orbr{\begin{cases}5x=0,2+10\\10x=0,1+5\end{cases}}\)

<=>\(\orbr{\begin{cases}5x=10,2\\10x=5,1\end{cases}}\)<=>\(\orbr{\begin{cases}x=10,2:10\\x=5,1:5\end{cases}}\)<=>\(\orbr{\begin{cases}x=1,02\\x=1,02\end{cases}}\)

VậyxE{1,02;1,02}

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

Lời giải:

1. 

PT $\Leftrightarrow (x^2+5x)^2+2(x^2+5x)-24=0$

$\Leftrightarrow t^2+2t-24=0$ (đặt $x^2+5x=t$)

$\Leftrightarrow (t-4)(t+6)=0$

$\Rightarrow t-4=0$ hoặc $t+6=0$

Nếu $t-4=0\Leftrightarrow x^2+5x-4=0$

$\Leftrightarrow x=\frac{-5\pm \sqrt{41}}{2}$

Nếu $t+6=0$

$\Leftrightarrow x^2+5x+6=0$

$\Leftrightarrow (x+2)(x+3)=0\Rightarrow x=-2$ hoặc $x=-3$

2.

PT $\Leftrightarrow (x^2-4x+1)^2+2(x^2-4x+1)-3=0$

$\Leftrightarrow t^2+2t-3=0$ (đặt $x^2-4x+1=t$)

$\Leftrightarrow (t-1)(t+3)=0$

$\Rightarrow t-1=0$ hoặc $t+3=0$

Nếu $t-1=0\Leftrightarrow x^2-4x=0\Leftrightarrow x(x-4)=0$

$\Rightarrow x=0$ hoặc $x=4$

Nếu $t+3=0\Leftrightarrow x^2-4x+4=0$

$\Leftrightarrow (x-2)^2=0\Leftrightarrow x=2$

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

1, \(x^4-19x^2-10x+8=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\\x^2-5x+2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x_1=-4\\x_2=-1\end{matrix}\right.\)

hoặc \(x^2-5x+2=0\)

\(\Rightarrow\Delta=17\left(CT:b^2-4ac\right)\)

\(\Rightarrow\left[{}\begin{matrix}x_3=\dfrac{5+\sqrt{17}}{2}\\x_4=\dfrac{5-\sqrt{17}}{2}\end{matrix}\right.\)

Vậy pt có 4 n^o là...........

Áp dụng cái này mà làm

\(a^3+b^3+c^3=\left(a+b+c\right)^3\)

\(\Leftrightarrow\left(a+b+c\right)^3-a^3-b^3-c^3=0\)

\(\Leftrightarrow3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

a) đặt \(\left(x^2+x\right)\)là \(y\)

ta có: \(3y^2-7y+4\)\(=0\)

<=>\(\left(3y-4\right)\left(y-1\right)=0\)

còn lại bạn tự xử nhé