Question

Giải PT :

a) \(x^2-11=0\)

\(b,x^2-2\sqrt{13}x+13=0\)

c, \(\sqrt{x^2-10x+25}=7-2x\)

Answer 1

a) \(x^2-11=0\)

<=> \(x^2-\sqrt{11}=0\)

<=> \(\left(x-\sqrt{11}\right)\left(x+\sqrt{11}\right)=0\)

<=> \(\left[{}\begin{matrix}x-\sqrt{11}=0\\x+\sqrt{11}=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{11}\\x=-\sqrt{11}\end{matrix}\right.\) => x = \(\pm\sqrt{11}\) Vậy S ={ \(\pm\sqrt{11}\)}

b) \(x^2-2\sqrt{13}x+13=0\)

\(\Leftrightarrow\left(x-\sqrt{13}\right)^2=0\)

=> x = \(\sqrt{13}\)

Vậy S = {\(\sqrt{13}\) }

\(c\)) \(\sqrt{x^2-10x+25}=7-2x\)

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=7-2x\)

\(\Leftrightarrow\left|x-5\right|=7-2x\)

=> Có 2 TH xảy ra

* Khi x - 5 \(\ge0\Leftrightarrow x\ge5\) Ta có PT :

x - 5 = 7 - 2x

<=> 3x = 12

=> x= 4 (KTM)

* Khi x - 5 < 0 => x < 5

Ta có pT

-x + 5 = 7-2x

<=> x = 2 (TM)

Vậy S = { 2 }

Answer 2

\(a\text{)} x^2-11=0\\ x^2=11\\ x=\pm\sqrt{11}\)

\(b\text{)}\:x^2-2\sqrt{13x}+13=0\\ \left(x-\sqrt{13}\right)^2=0\\ x-\sqrt{13}=0\\ x=\sqrt{13}\)

\(c\text{)}\:\sqrt{x^2-10x+25}=7-2x\\ \left|x-5\right|=7-2x\\ \Rightarrow\left[{}\begin{matrix}x-5=7-2x\left(với\:x\ge5\right)\\5-x=7-2x\left(với\:x< 5\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

Answer 3

a)easy

b)chuyển vế bình phương x=căn 13

c)pt \(x^2-10x+25=(x-5)^2\)

Answer 4

a, \(x^2-11=0\)

\(\Rightarrow x^2=11\Rightarrow x=\pm\sqrt{11}\)

Vậy........

b, \(x^2-2\sqrt{13}x+13=0\)

\(\Rightarrow x^2-\sqrt{13}x-\sqrt{13}x+13=0\)

\(\Rightarrow x\left(x-\sqrt{13}\right)-\sqrt{13}\left(x-\sqrt{13}\right)=0\)

\(\Rightarrow\left(x-\sqrt{13}\right)^2=0\)

\(\Rightarrow x=\sqrt{13}\)

Vậy...........

c, \(\sqrt{x^2-10x+25}=7-2x\)

\(\Rightarrow\sqrt{x^2-5x-5x+25}=7-2x\)

\(\Rightarrow\sqrt{\left(x-5\right)^2}=7-2x\)

\(\Rightarrow x-5=7-2x\Rightarrow3x=12\Rightarrow x=4\)

Chúc bạn học tốt!!!