\(y\left(y-5\right)\left(y-10\right)\left(y-15\right)< 0\)y(y-5)(y-10)(y-15)<0

\(\left(y^2-15y\right)\left(y^2-15y+50\right)< 0\)(y^2-15y)(y^2-15y+50)

\(\left(z\right)\left(z+50\right)< 0\)

\(-50< z< 0\Rightarrow\hept{\begin{cases}y^2-15y< 0\Rightarrow0< y< 15\\y^2-15>-50dungvoi.\forall y\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}y>0\\y< 15\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2-5>0\Leftrightarrow\orbr{\begin{cases}x>5\\x< -5\end{cases}}\\x^2-5< 15\Rightarrow-10< x< 10\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2-5>0\Rightarrow x< -5hoac.x>5\\x^2-5< 10\Rightarrow-10< x< 10\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}-10< x< -5\\5< x< 10\end{cases}}\)

Để đẳng thức trên xảy ra thì phải có ít nhất 1 số âm hoặc 3 số âm

TH1:có 1 số âm

=>x²-20 < 0 <x²-15

=>15 < x² <20

=> x²=16 

=> x = +-4

TH2:có 3 số âm

=> x²-10 < 0 <x²-5

=> 5 < x² <10

=> x² =9

=>x=+-3. Vậy x=3;x=-3;x=4hoặc x=-4

Chắc lun đó bạn ạ.Chúc bạn học giỏi nha!

a) \(\left(x+2\right)^2-9=0\)

\(\Rightarrow\left(x+2\right)^2=9\)

\(\Rightarrow\left(x+2\right)^2=3^2\)

\(\Rightarrow x+2=3\)

\(\Rightarrow x=3-2=1\)

a) ( x + 2 )² = 9

=> ( x + 2 ) ² = 9

=> ( x + 2 )² = 3²

=> x + 2 = + 3

=> \(\orbr{\begin{cases}x+2=-3\\x+2=3\end{cases}}\)

=> \(\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)

Vậy x = -1; 5

b) ( x + 2 )² - x² + 4 = 0

=> ( x + 2 )² - ( x² - 4 ) = 0

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

=> ( x + 2 ) ( x + 2 -  x + 2 ) = 0

=> ( x + 2 ) . 4 = 0

=> x + 2 = 0 

=> x = - 2

Vậy x = - 2 

c)  5 ( 2x - 3 )² - 5 ( x + 1 )² - 15( x + 4 ) ( x - 4 )  = - 10

=> 5 ( 4x² - 12x + 9 ) - 5 ( x² + 2x + 1 ) - 15 ( x² - 4² ) = - 10

=> 20x² - 60x + 45 - 5x² - 10x - 5 - 15x² + 240 = -10

=> - 70x + 280 = - 10

=> - 70x = - 290

=> x = \(\frac{29}{7}\)

Vậy x = \(\frac{29}{7}\)

d)  x ( x + 5 ) ( x - 5 ) - ( x + 2 ) ( x² - 2x + 4 ) = 3

=> x ( x² - 25 ) - ( x³ - 8 ) = 3

=> x³ - 25x - x³ + 8 = 3

=> - 25x + 8 = 3

=> - 25x = -5

=> x = \(\frac{1}{5}\)

Vậy x = \(\frac{1}{5}\)

Lại là 1 bạn cùng quê hương >.<

H t bận lắm gần đi học rồi hướng dẫn cách làm thôi nha

\(x^2-20< x^2-15< x^2-10< x^2-5\)

Để tích trên lẻ thì có 1 số lẻ các thừa số là âm

\(\Rightarrow\)có 1 số hoặc 3 số là âm

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-20< 0\\x^2-15>0\\x^2-10>0\\x^2-5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-20< 0\\x^2-15< 0\\x^2-10< 0\\x^2-5>0\end{matrix}\right.\end{matrix}\right.\)

Giải ra nhé :))

\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-5-x-1\right)\left(2x-5+x+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\3x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy \(S=\left\{6;\dfrac{4}{3}\right\}\)

\(10,\left(x+3\right)^2-x^2=45\)

\(\Leftrightarrow x^2+6x+9-x^2-45=0\\ \Leftrightarrow6x=36\\ \Leftrightarrow x=6\)

Vậy \(S=\left\{6\right\}\)

\(11,\left(5x-4\right)^2-49x^2=0\\ \Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\\ \Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\\ \Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(S=\left\{-2;\dfrac{1}{3}\right\}\)

\(12,16\left(x-1\right)^2-25=0\\ \Leftrightarrow4^2\left(x-1\right)^2-5^2=0\\ \Leftrightarrow\left[4\left(x-1\right)\right]^2-5^2=0\\ \Leftrightarrow\left(4x-4\right)^2-5^2=0\\ \Leftrightarrow\left(4x-4-5\right)\left(4x-4+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-9=0\\4x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{1}{4};\dfrac{9}{4}\right\}\)

1: Ta có: \(4x^2-36=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

2: Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=11\)

\(\Leftrightarrow x^2-2x+1+4x-x^2=11\)

\(\Leftrightarrow2x=10\)

hay x=5

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

`<=>[(x=2/3),(x=-5):}`

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

`<=>(5x-1)(x-3)(x+3)=0`

`<=>[(x=1/5),(x=3),(x=-3):}`

`12)x(x-3)+3(x-3)=0`

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

`<=>[(x=3),(x=-3):}`

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

`<=>(x+5)(x-1)=0`

`<=>[(x=-5),(x=1):}`

\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)