\(\Leftrightarrow\left(x^2-6x+9\right)^2-1-15\left(x^2-6x+10\right)=0\)

\(\Leftrightarrow\left(x^2-6x+8\right)\left(x^2-6x+10\right)-15\left(x^2-6x+10\right)=0\)

\(\Leftrightarrow\left(x^2-6x+10\right)\left(x^2-6x-7\right)=0\)

\(\Leftrightarrow\left(x^2-6x+10\right)\left(x^2+x-7x-7\right)=0\)

\(\Leftrightarrow\left(x^2-6x+10\right)\left(x+1\right)\left(x-7\right)=0\)

\(Vi:x^2-6x+10=0\Leftrightarrow\left(x-3\right)^2+1>0,\forall x\)

\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

\(hay:x-7=0\Leftrightarrow x=7\)

\(V...\)

\(:)\)

giúp em với mọi người ơi:<<<<<

\(\left(x^2-6x+9\right)+15\left(x^2-6x+10\right)=1\)

\(\Leftrightarrow\left(x-3\right)^2+15\left[\left(x-3\right)^2+1\right]=1\)

\(\Leftrightarrow16\left(x-3\right)^2+15=1\)

\(\Leftrightarrow16\left(x-3\right)^2=-14\)

=> Phương trình vô nghiệm 

\(\left(x^2-6x+9\right)-15\left(x^2-6x+10\right)=1\)

Đặt : \(x^2-6x+9=\left(x-3\right)^2=t\) thay vào pt ta được :

\(t^2-15\left(t+1\right)=1\)

\(\Leftrightarrow t^2-15t-16=0\)

\(\Leftrightarrow\left(t+1\right)\left(t-16\right)=0\)

\(\Leftrightarrow t=\left\{{}\begin{matrix}16\\-1\end{matrix}\right.\)

với : \(t=-1\) thì \(\left(x-3\right)^2=-1\)

\(\Rightarrow ptvonghiem\)

Với : \(t=16\) thì \(\left(x-3\right)^2=16\)

\(\Leftrightarrow x\in\left\{{}\begin{matrix}7\\-1\end{matrix}\right.\)

\(vay...\)

thôi chết rồi em nhầm đáng lẽ là : \(\left(x^2-6x+9\right)^2\) á

\(\left(x+3\right)^2\left(x^2+6x+1\right)=9\)

\(\Leftrightarrow\left(x^2+6x+9\right)\left(x^2+6x+1\right)=9\)

Đặt: \(x^2+6x+5=t\)thì:

\(\left(1\right)\Leftrightarrow\left(t-4\right)\left(t+4\right)=9\)

\(\Leftrightarrow t^2-25=0\)

\(\Leftrightarrow\left(t-5\right)\left(t+5\right)=0\)

\(\Leftrightarrow\left(x^2+6x\right)\left(x^2+6x+10\right)=0\)

\(\Leftrightarrow x\left(x+6\right)=0\left(x^2+6x+10=\left(x+3\right)^2+1>0\right)\)

.... bạn tự giả tiếp

Chúc bạn hc tốt :D

b: Đặt \(x^2+5x+4=a\)

\(\Leftrightarrow a=5\sqrt{a+24}\)

\(\Leftrightarrow a^2=25a+600\)

\(\Leftrightarrow a^2-25a-600=0\)

\(\Leftrightarrow\left(a-40\right)\left(a+15\right)=0\)

\(\Leftrightarrow a=-15\)

hay S=∅

\(\left|x^2-9\right|=\left|-7\right|\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-9=7\\x^2-9=-7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=16\\x^2=2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\pm4\\x=\pm\sqrt{2}\end{cases}}\)

x² + 6x - 16 > 2x - 7

<=> x² + 6x - 2x > -7 + 16

<=> x² + 4x > 9

<=> x² + 4x + 4 > 9 + 4

<=> ( x + 2 )² > 13

<=> ( x + 2 )² > \(\left(\pm\sqrt{13}\right)^2\)

<=> \(\orbr{\begin{cases}x+2>\sqrt{13}\\x+2>-\sqrt{13}\end{cases}\Rightarrow}\orbr{\begin{cases}x>\sqrt{13}-2\\x>-2-\sqrt{13}\end{cases}}\)

1: \(\Leftrightarrow6\left(3x-1\right)+3\left(6x-2\right)=4\left(1-3x\right)\)

=>18x-6+18x-6=4-12x

=>36x-12=4-12x

=>48x=16

hay x=1/3

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

=>(2x-1)(3x-4)=0

=>x=1/2 hoặc x=4/3

\(3\left(x^2-6x+10\right)^2+2\left(x^2-6x\right)-65=0\\ \Leftrightarrow3\left(\left(x-3\right)^2+1\right)^2+2\left(x^2-6x+9\right)-83=0\\ \Leftrightarrow3\left(\left(x-3\right)^4+2.\left(x-3\right)^2+1\right)+2\left(x-3\right)^2-83=0\\ \Leftrightarrow3\left(x-3\right)^4+6\left(x-3\right)^2+3+2\left(x-3\right)^2-83=0\\ \Leftrightarrow3\left(x-3\right)^4+8\left(x-3\right)^2-80=0\\ \Leftrightarrow\left(x-3\right)^2\left(3.\left(x-3\right)^2+8\right)=80\\ \Leftrightarrow\left(x-3\right)^2.\left(3x^2-18x+27+8\right)=80\)

Phá ra rồi giải tiếp !!