Giải thích các bước giải:
Bài 7:
$a,\sqrt[]{3}x-\sqrt[]{27}=0$
$⇔x=\sqrt[]{9}$
$b,x\sqrt[]{3}-\sqrt[]{3}=\sqrt[]{27}-\sqrt[]{12}$
$⇔\sqrt[]{3}(x-1)=√3(√9-√4)$
$⇔x-1=3-2$
$⇔x=2$
$c,√5.x^2-√45=0$
$⇔√5.x^2=√45$
$⇔x^2=√9=3$
$⇔x=√3$
$d,\frac{x^2}{√11}-√99=0$
$⇔x^2=√9=3$
$⇔x=√3$
Bài 8:
$a,$ Đk: $x≥\frac{3}{2}$
$\sqrt[]{\frac{2x-3}{x-1} }=2$
$⇔\frac{2x-3}{x-1}=4$
$⇔2x-3=4x-4$
$⇔2x=1$
$⇔x=\frac{1}{2}$ (Loại)
$⇒$Vậy pt vô nghiệm.
$b,$ Đk: $x≥\frac{3}{2}$
$\frac{\sqrt[]{2x-3} }{\sqrt[]{x-1} }=2$
$⇔\sqrt[]{2x-3}=2.\sqrt[]{x-1}$
$⇔2x-3=4x-4
$⇔x=\frac{1}{2}$ (Loại)
$⇒$Vậy pt vô nghiệm.
$c,$ Đk: $\frac{-5}{7}$
$\frac{9x-7}{\sqrt[]{7x+5}}=\sqrt[]{7x+5}$
$⇔9x-7=(\sqrt[]{7x+5})^2=7x+5$
$⇔2x=12$
$⇔x=6$
$d,$ Đk: $x≥5$
$\sqrt[]{4x-20}+3.\sqrt[]{\frac{x-5}{9}}-\frac{1}{3}.\sqrt[]{9x-45}=4$
$⇔2.\sqrt[]{x-5}+\sqrt[]{x-5}-\sqrt[]{x-5}=4$
$⇔2.\sqrt[]{x-5}=4$
$⇔x-5=4$
$⇔x=9$
