Giải bài 41 trang 67

Đề bài

Tìm x, biết:

a) $\frac{1}{2}\sqrt{x}-\frac{3}{2}\sqrt{9x}+24\sqrt{\frac{x}{64}}=-17$ với $x\ge 0$

b) $\sqrt{\frac{x}{5}}=4$ với $x\ge 0$

c) $\sqrt{25x^2}=10$

d) $\sqrt{{\left(2x-1\right)}^2}=3$

e) $2-\sqrt[3]{35-x}=0$

Lời giải chi tiết

a) $\frac{1}{2}\sqrt{x}-\frac{3}{2}\sqrt{9x}+24\sqrt{\frac{x}{64}}=-17$

$\begin{array}{c}\frac{1}{2}\sqrt{x}-\frac{3}{2}\sqrt{9x}+24\sqrt{\frac{x}{64}}=-17\\ \frac{1}{2}\sqrt{x}-\frac{9}{2}\sqrt{x}+3\sqrt{x}=-17\\ \sqrt{x}\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\\ \sqrt{x}\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\\ -\sqrt{x}=-17\\ \sqrt{x}=17\\ x=289\left(tm\right)\end{array}$

Vậy $x=289$.

b) $\sqrt{\frac{x}{5}}=4$

$\begin{array}{c}\sqrt{\frac{x}{5}}=4\\ \frac{x}{5}=16\\ x=80\left(tm\right)\end{array}$

Vậy $x=80$.

c) $\sqrt{25x^2}=10$

$\begin{array}{c}\sqrt{25x^2}=10\\ 25x^2=100\\ x^2=4\end{array}$

$x=2$ hoặc $x=-2$

Vậy $x=2$;$x=-2$

d) $\sqrt{{\left(2x-1\right)}^2}=3$

${\left(2x-1\right)}^2=9$

$2x-1=3$ hoặc $2x-1=-3$

$2x=4$ hoặc $2x=-2$

$x=2$ hoặc $x=-1$

Vậy $x=2$;$x=-1$

e) $2-\sqrt[3]{35-x}=0$

$\begin{array}{c}\sqrt[3]{35-x}=2\\ 5-x=8\\ x=-3\end{array}$

Vậy $x=-3.$