Chọn B

B

bị sang chấn tâm lí cái câu hỏi cụa bn ghê , tách `5->6` câu th bn:)

:v què tay hmu :<

từ bài 2 trở đi là lú rồi=)))

c: Ta có: \(x^3+3x^2+3x-7=0\)

\(\Leftrightarrow x+1=2\)

hay x=1

b: Ta có: \(x\left(x-3\right)-4x+12=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

a: \(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)

hay x=1/2

b: \(\Leftrightarrow\left(3x-1\right)^2=0\)

hay x=1/3

c: \(\Leftrightarrow\left(x+4\right)^2=0\)

hay x=-4

a) ⇒ \(\left(x-\dfrac{1}{2}\right)^2\)= 0

⇒ \(x-\dfrac{1}{2}=0\)

⇒ \(x=\dfrac{1}{2}\)

b) ⇒ \(\left(3x-1\right)^2=0\)

⇒ \(3x-1=0\)

⇒ \(3x=1\)

⇒ \(x=\dfrac{1}{3}\)

c) ⇒ \(\left(x+4\right)^2=0\)

⇒ \(x+4=0\)

⇒ \(x=-4\)

\(a,\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ b,\Rightarrow x^2=-1\left(vô.lí\right)\Rightarrow x\in\varnothing\\ c,\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\\ d,\Rightarrow x^2=3\Rightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)

a) \(\Rightarrow\left(x-3\right)\left(x+3\right)=0\)

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

b) \(x^2+1=0\)

\(\Rightarrow x^2=-1\left(VLý.do.x^2\ge0\forall x\right)\)

Vậy \(S=\varnothing\)

c) \(\Rightarrow x=\pm\sqrt{2}\)

d) \(\Rightarrow x^2=3\Rightarrow x=\pm\sqrt{3}\)

a ) x³ - 9x=0

<=> x (x² - 3 )= 0

<=> x(x+3)(x-3)

<=> x=0

     hoặc x=0-3=-3

      hoặc x=0+3=3

ĐK \(x\ge\frac{4}{7}\)

PT <=> \(x^2+6x-1+2=2\sqrt{\left(7x-4\right)\left(x^2-x+3\right)}\)

<=> \(\left(\sqrt{x^2-x+3}-\sqrt{7x-4}\right)^2+2=0\) vô nghiệm do VT>0 với mọi \(x\ge\frac{4}{7}\)

Vậy PT vô nghiệm