Question

tìm x

\(\sqrt{4\left(1-x^2\right)}-6=0\)

Answer 1

ĐK:....

\(\sqrt{4\left(1-x^2\right)}-6=0\)

\(\Leftrightarrow\sqrt{4\left(1-x^2\right)}=6\)

\(\Leftrightarrow4\left(1-x^2\right)=36\)

\(\Leftrightarrow4x^2+32=0\)

\(\Leftrightarrow x^2+8=0\)( vô lí )

Vậy \(x\in\varnothing\)

Answer 2

\(\sqrt{4\left(1-x^2\right)}-6=0\) ( 1)

ĐKXĐ : .......

(1) \(\Leftrightarrow\sqrt{4\left(1-x^2\right)}=6\)

\(\Leftrightarrow4\left(1-x^2\right)=36\)

\(\Leftrightarrow4-4x^2=36\)

\(\Leftrightarrow-4x^2=40\)

\(\Leftrightarrow x^2=-10\left(VL\right)\)

Vậy phương trình vô nghiệm

Answer 3

tìm x

$\sqrt{4\left(1-x^2\right)}-6=0$

Ta có :

Điều kiện : \(x\ne\pm1\)

\(\Rightarrow\sqrt{4\left(1-x^2\right)}=6\)

\(\Leftrightarrow4\left(1-x^2\right)=36\)

\(\Leftrightarrow1-x^2=9\)

\(\Leftrightarrow x^2=-8\)(Vo lí vì \(x^2\ge0\))

Vậy ko có giá trị x thỏa mãn