a) \(\Leftrightarrow x^2-5x-2x+10=0\)

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

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

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

Vậy \(x=5\)hoặc \(x=2\)

b) \(\Leftrightarrow\left(6x\right)^2-7^2=0\)

\(\Leftrightarrow\left(6x+7\right)\left(6x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}6x=-7\\6x=7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-7}{6}\\x=\frac{7}{6}\end{cases}}\)

Vậy \(x=\frac{-7}{6}\)hoặc \(x=\frac{7}{6}\)

a, x²-7x+10=0

<=> x²-2x-5x+10=0

<=> x.(x-2)-5.(x-2)=0

<=> (x-2).(x-5)=0

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

b, 36x²-49=0

<=> (6x)²-7²=0

<=> (6x-7).(6x+7)=0

\(\Leftrightarrow\orbr{\begin{cases}6x-7=0\\6x+7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{6}\\x=-\frac{7}{6}\end{cases}}\)

a)x²-7x+10=0

\(\Leftrightarrow x^2-2x-5x+10=0\)

\(\Rightarrow x\left(x-2\right)-5\left(x-2\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x-2\right)=0\)

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

Vậy \(x=5\) hoặc \(x=2\)

b) 36x²-49=0

\(\Leftrightarrow\left(6x\right)^2-7^2=0\)

\(\Rightarrow\left(6x+7\right)\left(6x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}6x+7=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=-7\\6x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7}{6}\\x=\dfrac{7}{6}\end{matrix}\right.\)

Vậy \(x=\dfrac{-7}{6}\) hoặc \(x=\dfrac{7}{6}\)

a) x²-7x+10=0

=> x²-2x-5x+10=0

=> x(x-2)-5(x-2)=0

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

hoặc -5(x-2)=0 -> x=2

vậy x= 0 hoặc x= 2

b) 36x²-49=0

=> (6x)²-7²=0

=> (6x-7)(6x+7)=0

=>hoặc 6x-7=0 -> 6x=7 -> x=7:6

hoặc 6x+7=0->6x=-7-> x = 6:7

vậy x=7:6 hoặc x=6:7

\(\sqrt{81x^2}-8x=\sqrt{\left(9x\right)^2}-8x=\left|9x\right|-8x=9x-8x=x\) ( vì x > 0)

\(6.\sqrt{36x^2}-36x=6.\sqrt{\left(6x\right)^2}-36x=6.\left|6x\right|-36x=6.\left(-6x\right)-36x=-36x-36x=-72x\) (vì x < 0)

Bn có thể dùng CT toán hx đc ko??/ Mk ko hỉu cái đề!

\(a,\Leftrightarrow\left(x+3\right)\left(x+3-2x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-12x+36\right)=0\\ \Leftrightarrow x\left(x-6\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

a, (x+3)² - ( 2x + 1 ).( x+3)=0              b,     x³-12x²+36x =0

=> (x+3).(x+3-2x-1)                             => x(x²-12x+36) = 0

=>(x+3).(-x+2)                                     => x(x-6)² = 0

=> x+3=0  <=> x=-3                            => x=0        <=> x=0

     -x+2=0 <=> x=-2                                 x-6= 0    <=> x=6

a) (2x - 5)² - (5 + 2x) = 0

<=> 4x² - 22x + 20 = 0 

\(\Leftrightarrow\left(2x-\dfrac{11}{2}\right)^2=\dfrac{41}{4}\)

\(\Leftrightarrow x=\dfrac{\pm\sqrt{41}+11}{4}\)

b) \(27x^3-54x^2+36x=0\)

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

\(\Leftrightarrow x=0\) (Vì \(3x^2-6x+4=3\left(x-1\right)^2+1>0\forall x\))

c) x³ + 8 - (x + 2).(x - 4) = 0

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

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

\(\Leftrightarrow x=-2\) (Vì \(x^2-3x+8=\left(x-\dfrac{3}{2}\right)^2+\dfrac{23}{4}>0\))

d) \(x^6-1=0\)

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

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

\(\Leftrightarrow x^2-1=0\) (Vì \(x^4+x^2+1>0\))

\(\Leftrightarrow x=\pm1\)

\(d,x^6-1=0\\ \Leftrightarrow\left(x^2\right)^3-1^3=0\\ \Leftrightarrow\left(x^2-1\right)\left(x^4+x^2+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x^4+x^2+1=0\left(Vô.lí,vì:x^4\ge0;x^2\ge0,\forall x\in R\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\ c,\left(x^3+8\right)-\left(x+2\right)\left(x-4\right)=0\\ \Leftrightarrow\left(x^3+8\right)-\left(x^2-2x-8\right)=0\\ \Leftrightarrow x^3-x^2+2x+16=0\\ \Leftrightarrow x^3+2x^2-3x^2-6x+8x+16=0\\ \Leftrightarrow x^2\left(x+2\right)-3x\left(x+2\right)+8\left(x+2\right)=0\\ \Leftrightarrow\left(x^2-3x+8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-3x+8=0\left(Vô.lí\right)\\x+2=0\end{matrix}\right.\Leftrightarrow x=-2\)

c)(x^3+ 8) - (x + 2)(x - 4) = 0
<=> x^3 -x^2 + 2x +8 + 8 = 0
<=> x^3 -x^2 + 2x + 16 = 0
<=> (x+2)(x^2-3x+8) = 0
=> x = -2 

Đề ko có vấn đề chứ ạ ? 

\(27x^2.x+69x^2+36x=0\)

Tương đương vs pt : \(\Leftrightarrow27x^3+69x^2+36x=0\)

\(\Leftrightarrow3x\left(9x^2+23x+12\right)=0\)

TH1 : \(3x=0\Leftrightarrow x=0\)

TH2 : \(\Delta=23^2-4.12.9=529-432=97>0\)

Phương trình có 2 nghiệm phân biệt

\(x_1=\frac{-23-\sqrt{97}}{3};x_2=\frac{-23+\sqrt{97}}{3}\)

a,(x^2+9)^2-36x^2=0

   (x^2+9-6x)(x^2+9+6x=0

=>x^2+9-6x=0 hoac x^2+9+6x=0

+,x^2+9-6x=0

   x^2-6x+9=0

  ( x-3)^2    =0

  =>x-3           =0

       x              =3

+,x^2+9+6x=0

   x^2+6x+9=0

   (x+3)^2    =0

=>x+3 =0

    x      =-3

Ý dưới cũng tương tự.. 

Dấu trừ ở trước thì bạn phải đổi dấu trog ngoặc 

a) x² - 25x = 0

=> x(x - 25) = 0

=> \(\orbr{\begin{cases}x=0\\x=25\end{cases}}\)

b) (x - 3)² - 36x² = 0

=> (x - 3)² - (6x)² = 0

=> \(\left(x+6x-3\right)\left(x-6x-3\right)=0\)

=> \(\orbr{\begin{cases}7x-3=0\\-5x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{7}\\x=-\frac{3}{5}\end{cases}}\)

c) 2x(3 - x) + 2x² = 12

=> 6x - 2x² + 2x² = 12

=> 6x = 12

=> x = 2

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

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

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

=> \(\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

a. x²  - 25x = 0

\(\Leftrightarrow x\left(x-25\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-25=0\end{cases}}\)

\(\orbr{\begin{cases}x=0\\x=25\end{cases}}\)

Vậy ...

b.(x-3)² - 36x² = 0

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

\(\Leftrightarrow\orbr{\begin{cases}-5x-3=0\\7x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-3}{5}\\x=\frac{3}{7}\end{cases}}\)

Vậy...

c.2x(3-x)+2x² = 12 

<=> 6x - 2x² + 2x² = 12

<=> 6x = 12

<=> x = 2

d. x (x-2) - x + 2 =0

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}\)

Vậy...

36x - x² = 0

<=> x(36 - x) = 0

<=> x = 0 hoặc 36 - x = 0

<=> x = 0 hoặc x = 36

Vậy x = 0 hoặc x = 36

ung ho minh len 200 nha