giúp em với mọi người ơi:<<<<<

Đặt: \(x^2-6x+9=t\left(t\ge0\right)\)

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

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

\(\Leftrightarrow t^2-15t-16=0\Leftrightarrow\left(t-16\right)\left(t+1\right)=0\Leftrightarrow t=16\left(t\ge0\right)\) 

\(\Leftrightarrow x^2-6x+9=16\Leftrightarrow\left(x-3\right)^2=16\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=4\\x-3=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-1\end{cases}}\)

Tập nghiệm của pt: \(S=\left\{7;-1\right\}\)

Đặt \(x^2-6x+9=t\)

\(\Rightarrow\)Phương trình ban đầu trở thành: \(t^2-15\left(t+1\right)=1\)

\(\Leftrightarrow t^2-15t-15=1\)\(\Leftrightarrow t^2-15t-16=0\)

\(\Leftrightarrow\left(t^2+t\right)-\left(16t+16\right)=0\)\(\Leftrightarrow t\left(t+1\right)-16\left(t+1\right)=0\)

\(\Leftrightarrow\left(t+1\right)\left(t-16\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}t+1=0\\t-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}t=-1\\t=16\end{cases}}\)

Ta thấy: \(x^2-6x+9=\left(x-3\right)^2\ge0\forall x\)

\(\Rightarrow t\ge0\)\(\Rightarrow t=16\)\(\Rightarrow x^2-6x+9=16\)

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

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

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

Vậy tập nghiệm của phương trình là: \(S=\left\{-1;7\right\}\)

dau lon dau buoi

\(ĐK:x\ge-2\)

\(\Leftrightarrow x^3+6x^2+12x+8+2\sqrt{\left(x+2\right)^3}+1-9x^2-18x-9=0\)

\(\Leftrightarrow\left(x+2\right)^3+2\sqrt{\left(x+2\right)^3}+1-\left(9x^2+18x+9\right)=0\)

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

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

ta có: ( 2 trường hợp xảy ra )

TH1: \(\left[\left(x+2\right)^3+1\right]^2=9\left(x+1\right)^2\)

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

\(\Leftrightarrow\left(x+2\right)^3-9x=8\)

\(\Leftrightarrow x^3+6x^2+12x+8-9x-8=0\)

\(\Leftrightarrow x^3+6x^2+3x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+6x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\\left[{}\begin{matrix}x=-3+\sqrt{6}\left(n\right)\\-3-\sqrt{6}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)

TH2:\(\left[{}\begin{matrix}\left(x+3\right)^3+1=0\\9\left(x+1\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)^3=-1\\\left(9x+9\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=-1\\9x=-9\end{matrix}\right.\)

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

Vậy \(S=\left\{0;-1;-3+\sqrt{6}\right\}\)

( ko bít đúng ko nha bạn ơi )

MÌNH KHÔNG BIẾT ^_^

\(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)

Đặt \(a=x^2+x\)

\(\Leftrightarrow a^2+4a=12\)

\(\Leftrightarrow a^2+4a-12=0\)

\(\Leftrightarrow a^2+6a-2a-12=0\)

\(\Leftrightarrow a\left(a+6\right)-2\left(a+6\right)=0\)

\(\Leftrightarrow\left(a+6\right)\left(a-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-6\\a=2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+x=-6\\x^2+x=2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{23}{4}=0\\x^2+2x-x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2=\frac{-23}{4}\left(loai\right)\\\left(x+2\right)\left(x-1\right)=0\end{cases}}\)

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

Vậy....

\(\left(x^2+6x+10\right)^2+\left(x+3\right)\left(3x^2+20x+36\right)=0\)

( rút gọn phá ngoặc tất cả )

\(\Leftrightarrow x^4+15x^3+85x^2+216x+208=0\)

\(\Leftrightarrow x^4+4x^3+11x^3+44x^2+41x^2+164x+52x+208=0\)

\(\Leftrightarrow x^3\left(x+4\right)+11x^2\left(x+4\right)+41x\left(x+4\right)+52\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^3+11x^2+41x+52\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^3+4x^2+7x^2+28x+13x+52\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left[x^2\left(x+4\right)+7x\left(x+4\right)+13\left(x+4\right)\right]=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+4\right)\left(x^2+7x+13\right)=0\)

\(\Leftrightarrow\left(x+4\right)^2\left(x^2+2\cdot x\cdot\frac{7}{2}+\frac{49}{4}+\frac{3}{4}\right)=0\)

\(\Leftrightarrow\left(x+4\right)^2\left[\left(x+\frac{7}{2}\right)^2+\frac{3}{4}\right]=0\)

\(\Leftrightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Vậy....

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

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

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

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

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

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

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

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

\(V...\)

\(:)\)

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

\(PT\Leftrightarrow\left(x^3+6x^2+12x+8\right)+2\sqrt{\left(x+2\right)^3}+1-9x^2-18x-9=0\\ \Leftrightarrow\left(x+2\right)^3+2\sqrt{\left(x+2\right)^3}+1-9\left(x+1\right)^2=0\\ \Leftrightarrow\left(\sqrt{\left(x+2\right)^3}+1\right)^2-9\left(x+1\right)^2=0\\ \Leftrightarrow\left[\sqrt{\left(x+2\right)^3}-3x-2\right]\left[\sqrt{\left(x+2\right)^3}+3x+4\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{\left(x+2\right)^3}=3x+2\\\sqrt{\left(x+2\right)^3}=-3x-4\end{matrix}\right.\)

\(TH_1:\sqrt{\left(x+2\right)^3}=3x+2\\ \Leftrightarrow x^3+6x^2+12x+8=9x^2+12x+4\left(x\ge-\dfrac{2}{3}\right)\\ \Leftrightarrow x^3-3x^2+4=0\\ \Leftrightarrow x^3+x^2-4x^2+4=0\\ \Leftrightarrow x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

\(TH_2:\sqrt{\left(x+2\right)^3}=-3x-4\\ \Leftrightarrow x^3+6x^2+12x+8=9x^2+24x+16\left(x\le-\dfrac{4}{3}\right)\\ \Leftrightarrow x^3-3x^2-12x-8=0\\ \Leftrightarrow x^3+x^2-4x^2-4x-8x-8=0\\ \Leftrightarrow\left(x+1\right)\left(x^2-4x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=2+2\sqrt{3}\left(ktm\right)\\x=2-2\sqrt{3}\left(tm\right)\end{matrix}\right.\)

Vậy PT có nghiệm \(S=\left\{2;2-2\sqrt{3}\right\}\)

ĐKXĐ: \(x\ge-2\)

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

Đặt \(\sqrt{x+2}=a\ge0\) pt trở thành:

\(x^3-3x.a^2+2a^3=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}=x\left(x\ge0\right)\\2\sqrt{x+2}=-x\left(x\le0\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-x-2=0\\x^2-4x-8=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=2\\x=2+2\sqrt{3}\left(loại\right)\\x=2-2\sqrt{3}\end{matrix}\right.\)