Question

[Hà Mai Thùy]

giúp mik với ! mik đang cần gấp!!!

Answer 1

[Akai Haruma]

5a.

$\frac{1}{2}-\sqrt{x}=0$

$\Rightarrow \sqrt{x}=\frac{1}{2}$
$\Rightarrow x=(\frac{1}{2})^2=\frac{1}{4}$

5b.

$\frac{5}{11}\sqrt{x}-\frac{1}{6}=\frac{1}{3}$

$\Rightarrow \frac{5}{11}\sqrt{x}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$

$\Rightarrow \sqrt{x}=\frac{1}{2}: \frac{5}{11}=\frac{11}{10}$

$\Rightarrow x=(\frac{11}{10})^2=\frac{121}{100}$

5c.

$-\frac{4}{3}\sqrt{x}+\frac{8}{5}=\frac{1}{3}+\frac{2}{3}=1$

$\Rightarrow -\frac{4}{3}\sqrt{x}=1-\frac{8}{5}=\frac{-3}{5}$

$\Rightarrow \frac{4}{3}\sqrt{x}=\frac{3}{5}$

$\Rightarrow \sqrt{x}=\frac{3}{5}: \frac{4}{3}=\frac{9}{20}$
$\Rightarrow x=(\frac{9}{20})^2=\frac{81}{400}$

Answer 2

[Akai Haruma]

5d.

$x-6\sqrt{x}=0$

$\Rightarrow \sqrt{x}(\sqrt{x}-6)=0$

$\Rightarrow \sqrt{x}=0$ hoặc $\sqrt{x}-6=0$

$\Rightarrow \sqrt{x}=0$ hoặc $\sqrt{x}=6$

$\Rightarrow x=0$ hoặc $x=36$

5e.

$1-3x^2=7$

$3x^2=1-7=-6$

$x^2=-2<0$ (vô lý)

Do đđ không tồn tại $x$ thỏa mãn đề.

5f.

$7x^2-4=1$

$7x^2=1+4=5$

$x^2=\frac{5}{7}=(\sqrt{\frac{5}{7}})^2=(-\sqrt{\frac{5}{7}})^2$

$\Rightarrow x=\pm \sqrt{\frac{5}{7}}$