a)$x^2-3x+4=0\\
\Rightarrow \left(x^2-2.\frac{3}{2}x+\frac{9}{4}\right)+\frac{7}{4}=0\\
\Rightarrow \left(x-\frac{3}{2}\right)^2+\frac{7}{4}=0\\
\Rightarrow \left(x-\frac{3}{2}\right)^2=-\frac{7}{4}$ (vô lí)( DO$\left(x-\frac{3}{2}\right)^2$ luôn > hoặc = 0)
b) $x^2-5x-6=0$
$\Rightarrow x^2-6x+x-6=0\\
\Leftrightarrow x\left(x+1\right)-6\left(x+1\right)=0\\
\Leftrightarrow \left(x-6\right)\left(x+1\right)=0$
⇔\(\left[ \begin{array}{l}x-6=0\\x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\)
Vậy
c)$x^4+2x^2-3=0\\
\Rightarrow \left(x^4-x^3\right)+\left(x^3-x^2\right)+\left(3x^2-3x\right)+\left(3x-3\right)=0\\
\Rightarrow x^3\left(x-1\right)+x^2\left(x-1\right)+3x\left(x-1\right)+3\left(x-1\right)=0$
$\Rightarrow \left(x-1\right)\left(x^3+x^2+3x+3\right)=0$
Do $x^3+x^2+3x+3>0$ nên (x-1)=0
$\Rightarrow x-1=0\\
\Rightarrow x=1$
Vậy
d)$\left \{ {{2x+y=3} \atop {2x-3y=1}} \right.$
$⇔\left \{ {{2x=3-y} \atop {2x-3y=1}} \right.$
$⇔\left \{ {{2x=3-y} \atop {3-y-3y=1}} \right.$
$⇔\left \{ {{2x=3-y} \atop {-4y=1-3}} \right.$
$⇔\left \{ {{2x=3-y} \atop {-4y=-2}} \right.$
$⇔\left \{ {{2x=3-y} \atop {y=\frac{-2}{-4}=\frac{1}{2}}} \right.$
$⇔\left \{ {{2x=3-\frac{1}{2}} \atop {y=\frac{1}{2}}} \right.$
$⇔\left \{ {{2x=\frac{5}{2}} \atop {y=\frac{1}{2}}} \right.$
$⇔\left \{ {{x=\frac{5}{2}:2=\frac{5}{4}} \atop {y=\frac{1}{2}}} \right.$
Vậy $⇔\left \{ {{x=\frac{5}{4}} \atop {y=\frac{1}{2}}} \right.$
e)$\left \{ {{x-2y=-2} \atop {3-y=1}} \right.$
$⇔\left \{ {{x=-2+2y} \atop {y=3-1=2}} \right.$
$⇔\left \{ {{x=-2+2.2} \atop {y=2}} \right.$
$⇔\left \{ {{x=2} \atop {y=2}} \right.$
Vậy $⇔\left \{ {{x=2} \atop {y=2}} \right.$
f)$\sqrt{x+3}=x-2$
$\sqrt{x+3}=x-2\\
\Rightarrow
x+3=\left(x-2\right)^2\\
\Rightarrow x+3=x^2-4x+4\\
\Rightarrow x^2-4x-x+4-3=0\\
\Rightarrow x^2-5x+1=0$
$
\Rightarrow x^2-5x+1=0\\
\Rightarrow \left(x^2-2.\frac{5}{2}.x+\frac{25}{4}\right)-\frac{21}{4}=0\\
\Rightarrow \left(x-\frac{5}{2}\right)^2=\frac{21}{4}\\
x-\frac{5}{2}==\sqrt{\frac{21}{4}}=\frac{\sqrt{21}}{2}\\
\Rightarrow x=\frac{\sqrt{21}}{2}+\frac{5}{2}$\
g)
$\frac{x}{x+1}-\frac{3-2x}{x-1}=6\\
\Leftrightarrow \frac{x\left(x-1\right)-\left(3-2x\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=6\\
\Leftrightarrow \frac{x^2-x-\left(3x-2x^2+3-2x\right)}{\left(x+1\right)\left(x-1\right)}=6\\
\Leftrightarrow \frac{3x^2-2x-3}{\left(x+1\right)\left(x-1\right)}=6$
$
\Leftrightarrow 3x^2-2x-3=6\left(x+1\right)\left(x-1\right)\\
\Leftrightarrow 6\left(x^2-1\right)=3x^2-2x-3\\
\Leftrightarrow 6x^2-6=3x^2-2x-3\\
\Leftrightarrow 6x^2-6-3x^2+2x+3=0$
$6x^2-6-3x^2+2x+3=0\\
\Leftrightarrow 3x^2+2x-3=0$
Mình làm được từng này thôi! Mong bạn thông cảm
