\(a,\Leftrightarrow x^2-2x-x^2+5x=6\\ \Leftrightarrow3x=6\\ \Leftrightarrow x=2\)

\(b,\Leftrightarrow x^2-6x+9-x+9=0\\ \Leftrightarrow x^2-7x+18=0\\ \Leftrightarrow\left(x^2-7x+\dfrac{49}{4}\right)+\dfrac{23}{4}=0\\ \Leftrightarrow\left(x-\dfrac{7}{2}\right)^2+\dfrac{23}{4}=0\left(vôlí\right)\)

\(a,\Leftrightarrow25x^2-70x+49-25x^2=32\\ \Leftrightarrow-70x=-17\Leftrightarrow x=\dfrac{17}{70}\\ b,\Leftrightarrow x^2-6x+9+x^2+2x+1-5=0\\ \Leftrightarrow2x^2-4x+5=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+3=0\\ \Leftrightarrow2\left(x-1\right)^2=-3\Leftrightarrow\left(x-1\right)^2=-\dfrac{3}{2}\left(\text{vô lí}\right)\\ \Leftrightarrow x\in\varnothing\)

\(\Leftrightarrow-\dfrac{2}{5}\left(4x-3\right)^2=-\dfrac{5}{18}\)

\(\Leftrightarrow\left(4x-3\right)^2=\dfrac{25}{36}\)

\(\Leftrightarrow4x-3\in\left\{\dfrac{5}{6};-\dfrac{5}{6}\right\}\)

hay \(x\in\left\{\dfrac{23}{24};\dfrac{13}{24}\right\}\)

Đơn thức là: x/(3y), x, -2/3, xy-2, 0

Đơn thức là \(-5xy^2;x;-\dfrac{2}{3};0\)

=>x-3=0

hay x=3

\(\left(x-3\right)\left(2x^2+3\right)=0\\ \Rightarrow x-3=0\left(vì.2x^2+3>0\right)\\ \Rightarrow x=3\)

2:

a: =>x-1=0 hoặc 3x+1=0

=>x=1 hoặc x=-1/3

b: =>x-5=0 hoặc 7-x=0

=>x=5 hoặc x=7

c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)

d: =>x=0 hoặc x^2-1=0

=>\(x\in\left\{0;1;-1\right\}\)

Bạn tách ra từng câu thoi nhe .

\(x^2-4x+3=0\\ \Rightarrow\left(x^2-3x\right)-\left(x-3\right)=0\\ \Rightarrow x\left(x-3\right)-\left(x-3\right)=0\\ \Rightarrow\left(x-1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

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

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

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

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

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

Ta có: \(\left(1-x\right)^2+\left(x-x^2\right)+3=0\)

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

\(\Leftrightarrow4-x=0\)

hay x=4

Vậy: S={4}

$⇔x^2-2x+1+x-x^2+3=0$

$⇔-x=-4$

$⇔x=4$

Vậy phương trình đã cho có tập nghiệm S={4}

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

\(< =>\left(x^2-9\right)^2-\left[3\left(x-3\right)\right]^2=0\)

\(< =>\left(x^2-9\right)^2-\left(3x-9\right)^2=0\)

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

\(< =>\left(x^2+3x-18\right)\left(x^2-3x\right)=0\)

\(=>\left[{}\begin{matrix}x^2+3x-18=0\\x^2-3x=0\end{matrix}\right.< =>\left[{}\begin{matrix}\left(x+6\right)\left(x-3\right)=0\\x\left(x-3\right)=0\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=-6\\x=3\\x=0\end{matrix}\right.\)

Nhanh đấy tốc độ đấy =))