\(\sqrt{-2x^2+6}=x-1\left(đk:\sqrt{3}\ge x\ge1\right)\)

\(\Leftrightarrow-2x^2+6=x^2-2x+1\)

\(\Leftrightarrow3x^2-2x-5=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\x=\dfrac{5}{3}\left(tm\right)\end{matrix}\right.\)

đội tuyển toán tự làm đi m 

\(1.\sqrt{16-8x+x^2}=4-x\)

\(\sqrt{\left(4-x\right)^2}=4-x\)

\(4-x-4+x=0\)

= 0 phương trình vô nghiệm.

\(2.\sqrt{4x^2-12x+9}=2x-3\)

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

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

= 0 phương trình vô nghiệm.

a: Ta có: \(\sqrt{16-8x+x^2}=4-x\)

\(\Leftrightarrow\left|4-x\right|=4-x\)

hay \(x\le4\)

b: Ta có: \(\sqrt{4x^2-12x+9}=2x-3\)

\(\Leftrightarrow\left|2x-3\right|=2x-3\)

hay \(x\ge\dfrac{3}{2}\)

a/ \(\sqrt{16-8x+x^2}=4-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\\sqrt{\left(4-x\right)^2}=4-x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\\left|4-x\right|=4-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le4\\\left[{}\begin{matrix}4-x=4-x\left(loại\right)\\4-x=x-4\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x=4\)

Vậy...

b/ \(\sqrt{4x^2-12x+9}=2x-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\\sqrt{\left(2x-3\right)^2}=2x-3\end{matrix}\right.\)

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

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy...

\(\left(x^2-9\right)^2=12x+1\)

\(\Leftrightarrow x^4-18x^2+81=12x+1\)

\(\Leftrightarrow x^4-18x^2+81-12x-1=0\)

\(\Leftrightarrow x^4-2x^3+2x^3-4x^2-14x^2+28x-40x+80=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+2x^2-14x-40\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2+6x+10\right)=0\)

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

(x^2-9)^2=12x-1 

<=>x^4-18x^2-12x+80=0 

<=>x^4-2x^3+2x^3-4x^2-14x^2+28x-40x+80... 

<=>(x-2)(x^3+2x^2-14x-40)=0 

<=>(x-2)(x-4)(x^2+6x+10)=0 

Ta thấy x^2+6x+10=(x+3)^2+1>0 

=>x=2 hhoặc x=4

(x^2-9)^2=12x-1 

<=>x^4-18x^2-12x+80=0 

<=>x^4-2x^3+2x^3-4x^2-14x^2+28x-40x+80... 

<=>(x-2)(x^3+2x^2-14x-40)=0 

<=>(x-2)(x-4)(x^2+6x+10)=0 

Ta thấy x^2+6x+10=(x+3)^2+1>0 

=>x=2 hhoặc x=4

\(\left(x^2+2x\right)^2-6x^2+12x+9=0\Leftrightarrow x^4+4x^3+4x^2-6x^2+12x+9=0\\ \Leftrightarrow x^4+4x^3-2x^2+12x+9=0\Leftrightarrow x^2+4x-2+\frac{12}{x}+\frac{9}{x^2}=0\\ \Leftrightarrow\left(x^2+\frac{9}{x^2}\right)+4\left(x+\frac{3}{x}\right)-2=0\)

Đặt \(k=x+\frac{3}{x}\Rightarrow x^2+\frac{9}{x^2}=k^2-6\)

Ta đc \(k^2-6+4k-2=0\Leftrightarrow k^2+4k-8=0\)

\(\left(x^2+2x\right)^2\)\(-6x^2\)\(+12x+9\)=0

⇔\(\left(x^2\right)^2\)\(+2.2x.x^2\)+\(2x^2\)-6x²+12x+9=0

⇔ x⁴+ 4x³+2x²-6x²+12x+9=0

⇔ x²+4x³-4x² +12x=-9

⇔x²+ 4x(x-x+3)=-9

⇔x²+12x=-9

⇔x(x+12)=-9

⇔ {x=-9 hoặc x+12=-9}

⇔ {x=-9 hoặc x=-21}

S={-9;-21}

1) \(9x^4+8x^2-1=0\)

\(\Leftrightarrow9x^4+9x^2-x^2-1=0\)

\(\Leftrightarrow9x^2\left(x^2+1\right)-\left(x^2+1\right)=0\)

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

\(\Rightarrow9x^2-1=0\)

\(\Leftrightarrow x=\dfrac{\pm1}{3}\)

Vậy...

2)  \(\Delta=\left(m-1\right)^2-4\left(-m^2+m-1\right)\) \(=5m^2-6m+5\)

Có: \(5m^2-6m+5=5\left(m^2-\dfrac{6}{5}m+\dfrac{9}{25}\right)+\dfrac{16}{5}\)

\(=5\left(m-\dfrac{3}{5}\right)^2+\dfrac{16}{5}\ge\dfrac{16}{5}>0\forall m\in R\)

\(\Rightarrow\Delta>0\forall m\in R\)

Vậy: PT luôn có 2 nghiệm phân biệt với mọi m.

vô nghiệm nha bn!

2.lxl-12x-x=-3-9
2.lxl-13x=-12
2x-13x=-12;x>=0
2.(-x)-13x=-12;x<0
x=12/11;x>=0
x=4/5;x<0