Toán Tìm x: $\text{√x=27}$ $x^2=7$ $\text{5x-x}$ 07/09/2021 By Rylee Tìm x: $\text{√x=27}$ $x^2=7$ $\text{5x-x}$
(Phương trình số 3 chưa đủ) $\sqrt{x}=27$ $\Leftrightarrow x=27^2=729$ $x^2=7$ $\Leftrightarrow x^2= (\pm \sqrt{7})^2$ $\Leftrightarrow x=\pm \sqrt{7}$ $5x-x=0$ $\Leftrightarrow 4x=0$ $\Leftrightarrow x=0$ Trả lời
$\sqrt[]{x}$ = $27^{}$ ⇔ ($\sqrt[]{x}$)² = ($27^{}$ )² ⇔ x = 729 $x^{2}$ = 7 ⇔\(\left[ \begin{array}{l}x^2 =\sqrt[]{7} \\x^2 =-\sqrt[]{7} \end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=\sqrt[]{7}\\x=- \sqrt[]{7}\end{array} \right.\) 5x – x = 0 ⇔ 4x = 0 ⇔ x = 0 Trả lời
(Phương trình số 3 chưa đủ)
$\sqrt{x}=27$
$\Leftrightarrow x=27^2=729$
$x^2=7$
$\Leftrightarrow x^2= (\pm \sqrt{7})^2$
$\Leftrightarrow x=\pm \sqrt{7}$
$5x-x=0$
$\Leftrightarrow 4x=0$
$\Leftrightarrow x=0$
$\sqrt[]{x}$ = $27^{}$
⇔ ($\sqrt[]{x}$)² = ($27^{}$ )²
⇔ x = 729
$x^{2}$ = 7
⇔\(\left[ \begin{array}{l}x^2 =\sqrt[]{7} \\x^2 =-\sqrt[]{7} \end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\sqrt[]{7}\\x=- \sqrt[]{7}\end{array} \right.\)
5x – x = 0
⇔ 4x = 0
⇔ x = 0