\(-x^4+4x^2-5x^2+20=0\\\Rightarrow -(x^4-4x^2)-(5x^2-20)=0\\\Rightarrow-x^2(x^2-4)-5(x^2-4)=0\\\Rightarrow(x^2-4)(-x^2-5)=0\\\Rightarrow-(x-2)(x+2)(x^2+5)=0\\\Rightarrow(2-x)(x+2)=0(vì.x^2+5>0\forall x)\)

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

1) x² - 7x =  0

=> x(x - 7) = 0

=> \(\orbr{\begin{cases}x=0\\x=7\end{cases}}\)

2) -3x² + 5x = 0

=> x(-3x + 5) = 0

=> \(\orbr{\begin{cases}x=0\\-3x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}\)

3) x² - 19x - 20 = 0

=> x² - 20x + x - 20 = 0

=> x(x - 20) + (x - 20) = 0

=> (x + 1)(x - 20) = 0

=> \(\orbr{\begin{cases}x+1=0\\x-20=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=20\end{cases}}\)

4) x² - 5x - 24 = 0

=> x² - 8x + 3x - 24 = 0

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

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

=> \(\orbr{\begin{cases}x+3=0\\x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)

1) x² - 7x = 0

<=> x( x - 7 ) = 0

<=> \(\orbr{\begin{cases}x=0\\x=7\end{cases}}\)

2) -3x² + 5x = 0

<=> x( -3x + 5 ) = 0

<=> \(\orbr{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}\)

3) x² - 19x - 20 = 0

<=> x² + x - 20x - 20 = 0

<=> x( x + 1 ) - 20( x + 1 ) = 0

<=> ( x - 20 )( x + 1 ) = 0

<=> \(\orbr{\begin{cases}x=20\\x=-1\end{cases}}\)

4) x² - 5x - 24 = 0

<=> x² + 3x - 8x - 24 = 0

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

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

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

Chử đề sai câu 1:

1, x(x-2)+x-2=0

a) 2x²-4x-x+2=0

=> 2x(x-2)-(x-2)=0

=> (2x-1)(x-2)=0

=> \(\left[{}\begin{matrix}2x-1=0\\x-2=0\end{matrix}\right.\)

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

b) 3x²-12x+5x-20=0

=> 3x(x-4)+5.(x-4)=0

=> (x-4)(3x+5)=0

=> \(\left[{}\begin{matrix}x-4=0\\3x+5=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=4\\x=-\dfrac{5}{3}\end{matrix}\right.\)

c)x³+2x²-x²-2x+2x+4=0

=> x²(x+2)-x(x+2)+2(x+2)=0

=>(x²-x+2)(x+2)=0

=> x=-2( vi x²-x+2>0)

d) x³-x²-4x²+4x+4x-4=0

=> x²(x-1)-4x(x-1)+4(x-1)=0

=>(x-1)(x²-4x+4)=0

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

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

2x²-5x+2=0

⇔2x²-x-4x+2=0

⇔x(2x-1)-2(2x-1)=0

⇔(x-2)(2x-1)=0

⇔\(\left[{}\begin{matrix}x-2=0\\2x-1=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=2\\2x=1\Leftrightarrow x=\dfrac{1}{2}\end{matrix}\right.\)

sậy S=\(\left\{2;\dfrac{1}{2}\right\}\)

x³+x²+4=0

⇔x³+2x²-x²-2x+2x+4=0

⇔(x³+2x²)-(x²+2x)+(2x+4)=0

⇔x²(x+2)-x(x+2)+2(x+2)=0

⇔(x+2)(x²-x+2)=0

⇔x+2=0 và x²-x+2=0

⇔x=-2 và \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\)(vô lý)

vậy S={-2}

l) (x + 9) . (x² – 25) = 0  

<=> (x + 9) . (x – 5) . (x + 5) = 0   

<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy S = \(\left\{-9,5,-5\right\}\)

      e) |x - 4 |< 7         

<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)

Vậy S = \(\left\{11;-3\right\}\)

I,(x+9).(x^2-25)=0

tương đương:x+9=0

                       x^2-25=0

tương đương : x=-9

                       x=5

e,\(\left|x-4\right|\)=7

tương đương x-4=4

                       x-4=-4

tương đương :x=0

                        x=-8

Bài 1: 

l) Ta có: \(\left(x+9\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{-9;5;-5\right\}\)

e) Ta có: |x-4|<7

mà \(\left|x-4\right|\ge0\forall x\)

nên \(\left|x-4\right|\in\left\{0;1;2;3;4;5;6\right\}\)

\(\Leftrightarrow x-4\in\left\{0;1;-1;2;-2;3;-3;4;-4;5;-5;6;-6\right\}\)

hay \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)

Vậy: \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)

f) Ta có: \(40< 31+\left|x\right|< 47\)

\(\Leftrightarrow\left|x\right|+31\in\left\{41;42;43;44;45;46\right\}\)

\(\Leftrightarrow\left|x\right|\in\left\{10;11;12;13;14;15\right\}\)

hay \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)

Vậy: \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)

g) Ta có: \(\left|x+3\right|\le2\)

\(\Leftrightarrow\left|x+3\right|\in\left\{0;1;2\right\}\)

\(\Leftrightarrow x+3\in\left\{0;1;-1;2;-2\right\}\)

hay \(x\in\left\{-3;-2;-4;-1;-5\right\}\)

Vậy: \(x\in\left\{-3;-2;-4;-1;-5\right\}\)

a) Ta có: \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2=\frac{1}{4}\)

\(\Rightarrow x-\frac{1}{2}=\frac{1}{2}\Rightarrow x=1\)

hoặc \(x-\frac{1}{2}=-\frac{1}{2}\Rightarrow x=-\frac{1}{2}+\frac{1}{2}=0\)

Vậy \(x\in\left\{1;0\right\}\)

b)\(20\%.x+\frac{2}{5}x=x-4\)

\(\Rightarrow\frac{1}{5}x+\frac{2}{5}x-x=-4\)

\(\Rightarrow\frac{-2}{5}x=-4\)

\(\Rightarrow x=-4:\left(-\frac{2}{5}\right)=10\)

Vậy x=10

c) 3(x+1)+4(x+3)-2(x+16) = 18

=> 3x+3+4x+12-2x-32 =18

=> (3x+4x-2x)+(3+12-32)=18

=> 5x - 17 = 18

=> 5x = 18 + 17 = 35

=> x = 35 : 5 = 7

Vậy x = 7

sao3x+3 rồi 4x +12 là sao mình hk hiểu