Đề mik ko hiểu lắm

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

\(\Leftrightarrow6x^2\left(x-4\right)+2x\left(x-4\right)=0\)

\(\Leftrightarrow2x\left(x-4\right)\left(3x+1\right)=0\)

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

Ta có : (2x - 3)² - (x + 4²)² = 0

=> (2x - 3)² = (x + 4²)² 

=> 2x - 3 = (x + 16)

=> 2x - x = 16 + 3

=> x = 19

Bài này là Bình phương của 1 tổng với bp của 1 hiệu , giải nhưvậy đúng ko bạn ?

sorry mk thiếu 1 trường hợp : 

2x - 3 = -x - 4

=> 2x + x = -4 + 3

=> 3x = -1

=> x = -1/3

b) 6x(x - 5) + 5x - 25 = 0

=> 6x(x - 5) + 5(x - 5) = 0

=> (6x + 5)(x - 5) = 0

\(\Leftrightarrow\orbr{\begin{cases}6x+5=0\\x-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}6x=-5\\x=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{6}\\x=5\end{cases}}\)

a)      \(2\left(x+5\right)-x^2-5x=0\)

  \(\Leftrightarrow2x+10-x^2-5x=0\)

 \(\Leftrightarrow-x^2-3x+10=0\)

\(\Leftrightarrow x^2+3x-10=0\)

 \(\Leftrightarrow x^2-2x+5x-10=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)

b) \(x^3-6x^2+12x-8=0\)

\(\Leftrightarrow\left(x^3-8\right)-\left(6x^2-12x\right)=0\)

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

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

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

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

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

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

c)\(16x^2-9\left(x+1\right)^2=0\)

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

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

\(\Leftrightarrow\left(x-1\right)\left(7x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\7x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)

d) \(x^3+x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

e)\(x^2-2x-3=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)

Cảm ơn bạn nha

a)  ( 3x - 1 ) ( 2x + 7 )  - ( x + 1 ) ( 6x + 5 ) = 16 

<=> 6x² + 21x - 2x - 7 - ( 6x² - 5x + 6x - 5) = 16

<=> 6x² + 21x - 2x - 7 - ( 6x² + x - 5 )        = 16 

<=> 6x²+ 21x - 2x - 7 - 6x² -x + 5              = 16 

<=> 18x - 2                                             = 16 

<=>  18x                                                 = 18 

=>        x                                                 = 1

Vậy....  

1) \(x^4-6x^3-x^2+54x-72=0\)

\(\Leftrightarrow x^3\left(x-2\right)-4x^2\left(x-2\right)-9x\left(x-2\right)+36\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-4x^2-9x+36\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-4\right)-9\left(x-4\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2-9\right)=0\)

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

Tự làm nốt...

2) \(x^4-5x^2+4=0\)

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

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

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

Tự làm nốt...

\(x^4-2x^3-6x^2+8x+8=0\)

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

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

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

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

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left[\left(x-1\right)^2-\left(\sqrt{3}\right)^2\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{3}\right)=0\)

...

\(2x^4-13x^3+20x^2-3x-2=0\)

\(\Leftrightarrow2x^3\left(x-2\right)-9x^2\left(x-2\right)+2x\left(x-2\right)+\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x^3-9x^2+2x+1\right)=0\)

Bí

\(2x^3-9x^2+2x+1\)

\(=2x^3-x^2-8x^2+4x-2x+1\)

\(=x^2\left(2x-1\right)-4x\left(2x-1\right)-\left(2x-1\right)\)

\(=\left(2x-1\right)\left(x^2-4x-1\right)\)

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

\(=\left(2x-1\right)\left[\left(x-2\right)^2-5\right]\)

.......

Bài 1

Em xem lại đề nhé

a. Ta có VP=\(x^4-y^4=\left(x^2\right)^2-\left(y^2\right)^2\)

\(=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)

\(=\left(x+y\right)\left(x^3+xy^2-x^2y-y^3\right)\)

\(=VT\)

b. 

1.\(\left(x-3\right)\left(x-2\right)-\left(x+10\right)\left(x-5\right)=0\)

\(\Leftrightarrow x^2-5x+6-\left(x^2+5x-50\right)=0\)

\(\Leftrightarrow-10x=-56\Rightarrow x=\frac{56}{10}\)

2.\(\left(2x-1\right)\left(3-x\right)+\left(x-2\right)\left(x+3\right)=\left(1-x\right)\left(x-2\right)\)

\(=-2x^2+7x-3+x^2+x-6=-x^2+3x-2\)

\(\Leftrightarrow5x=7\Leftrightarrow x=\frac{7}{5}\)

Em cảm ơn chị ạ! :))

\(2,\\ a,\sqrt{4x-4}+\sqrt{9x-9}-\sqrt{25x-25}=7\left(x\ge1\right)\\ \Leftrightarrow2\sqrt{x-1}+3\sqrt{x-1}-5\sqrt{x-1}=7\\ \Leftrightarrow0\sqrt{x-1}=7\Leftrightarrow x\in\varnothing\\ b,\sqrt{2x^2-3}=4\left(x\le-\dfrac{\sqrt{6}}{2};\dfrac{\sqrt{6}}{2}\le x\right)\\ \Leftrightarrow2x^2-3=16\\ \Leftrightarrow x^2=\dfrac{19}{2}\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{19}{2}}\left(tm\right)\\x=-\sqrt{\dfrac{19}{2}}\left(tm\right)\end{matrix}\right.\)

\(1,\\ A=\sqrt{5+4x}+\sqrt{7-3x}\\ ĐKXĐ:\left\{{}\begin{matrix}5+4x\ge0\\7-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{5}{4}\\x\le\dfrac{7}{3}\end{matrix}\right.\)

Bài 2:

a) \(\sqrt{4x-4}+\sqrt{9x-9}-\sqrt{25x-25}=7\left(đk:x\ge1\right)\)

\(\Leftrightarrow2\sqrt{x-1}+3\sqrt{x-2}-5\sqrt{x-1}=7\)

\(\Leftrightarrow0=7\left(VLý\right)\)

Vậy \(S=\varnothing\)

b) \(\sqrt{2x^2-3}=4\left(đk:-\sqrt{\dfrac{3}{2}}\ge x\ge\sqrt{\dfrac{3}{2}}\right)\)

\(\Leftrightarrow2x^2-3=16\)

\(\Leftrightarrow2x^2=19\Leftrightarrow x^2=\dfrac{19}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{19}{2}}\left(tm\right)\\x=-\sqrt{\dfrac{19}{2}}\left(tm\right)\end{matrix}\right.\)