a.
$x^2-11=0$
$\Leftrightarrow x^2=11$
$\Leftrightarrow x=\pm \sqrt{11}$
b. $x^2-12x+52=0$
$\Leftrightarrow (x^2-12x+36)+16=0$
$\Leftrightarrow (x-6)^2=-16< 0$ (vô lý)
Vậy pt vô nghiệm.
c.
$x^2-3x-28=0$
$\Leftrightarrow x^2+4x-7x-28=0$
$\Leftrightarrow x(x+4)-7(x+4)=0$
$\Leftrightarrow (x+4)(x-7)=0$
$\Leftrightarrow x+4=0$ hoặc $x-7=0$
$\Leftrightarrow x=-4$ hoặc $x=7$
d.
$x^2-11x+38=0$
$\Leftrightarrow (x^2-11x+5,5^2)+7,75=0$
$\Leftrightarrow (x-5,5)^2=-7,75< 0$ (vô lý)
Vậy pt vô nghiệm
e.
$6x^2+71x+175=0$
$\Leftrightarrow 6x^2+21x+50x+175=0$
$\Leftrightarrow 3x(2x+7)+25(2x+7)=0$
$\Leftrightarrow (3x+25)(2x+7)=0$
$\Leftrightarrow 3x+25=0$ hoặc $2x+7=0$
$\Leftrightarrow x=-\frac{25}{3}$ hoặc $x=-\frac{7}{2}$
f.
$x^2-(\sqrt{2}+\sqrt{8})x+4=0$
$\Leftrightarrow x^2-\sqrt{2}x-2\sqrt{2}x+4=0$
$\Leftrightarrow x(x-\sqrt{2})-2\sqrt{2}(x-\sqrt{2})=0$
$\Leftrightarrow (x-\sqrt{2})(x-2\sqrt{2})=0$
$\Leftrightarrow x-\sqrt{2}=0$ hoặc $x-2\sqrt{2}=0$
$\Leftrightarrow x=\sqrt{2}$ hoặc $x=2\sqrt{2}$
g.
$(1+\sqrt{3})x^2-(2\sqrt{3}+1)x+\sqrt{3}=0$
$\Leftrightarrow (1+\sqrt{3})x^2-(1+\sqrt{3})x-(\sqrt{3}x-\sqrt{3})=0$
$\Leftrightarrow (1+\sqrt{3})x(x-1)-\sqrt{3}(x-1)=0$
$\Leftrightarrow (x-1)[(1+\sqrt{3})x-\sqrt{3}]=0$
$\Leftrightarrow x-1=0$ hoặc $(1+\sqrt{3})x-\sqrt{3}=0$
$\Leftrightarrow x=1$ hoặc $x=\frac{3-\sqrt{3}}{2}$
a) Ta có: \(x^2-11=0\)
\(\Leftrightarrow x^2=11\)
hay \(x\in\left\{\sqrt{11};-\sqrt{11}\right\}\)
b) Ta có: \(x^2-12x+52=0\)
\(\Leftrightarrow x^2-12x+36+16=0\)
\(\Leftrightarrow\left(x-6\right)^2+16=0\)(Vô lý)
c) Ta có: \(x^2-3x-28=0\)
\(\Leftrightarrow x^2-7x+4x-28=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)
d) Ta có: \(x^2-11x+38=0\)
\(\text{Δ}=\left(-11\right)^2-4\cdot1\cdot38=121-152< 0\)
Vì Δ<0 nên phương trình vô nghiệm
e) Ta có: \(6x^2+71x+175=0\)
\(\Leftrightarrow6x^2+21x+50x+175=0\)
\(\Leftrightarrow3x\left(2x+7\right)+25\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x+7\right)\left(3x+25\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=-\dfrac{25}{3}\end{matrix}\right.\)
f) Ta có: \(x^2-\left(\sqrt{2}+\sqrt{8}\right)x+4=0\)
\(\text{Δ}=\left[-3\sqrt{2}\right]^2-4\cdot1\cdot4\)
\(=18-16=2\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{3\sqrt{2}-\sqrt{2}}{2}=\sqrt{2}\\x_2=\dfrac{3\sqrt{2}+\sqrt{2}}{2}=2\sqrt{2}\end{matrix}\right.\)