27 . 39 + 27 . 63

2,8 - 1,5 \(\sqrt{x}\) = 3,5

1,5\(\sqrt{x}\) = 2,8 - 3,5

1,5\(\sqrt{x}\) = - 0,7

\(\sqrt{x}\) = ( - 0,7 ) : 1,5

\(\sqrt{x}\) = -7/15

x = 49/225

\(\sqrt{\frac{98}{8}}\)=7/2=3,5

quên, bấm nút tl, làm tip

\(\sqrt{x}\)= (2,8 - 3,5)/1,5 = -0,7/1,5

vô nghĩa vì căn bậc 2 k âm, pt vn

c

\(\sqrt{x}+9=4\)

\(\sqrt{x}=4-9=-5\)

Vì \(\sqrt{x}\) ≥ 0

⇒ Không tồn tại x

\(\begin{array}{l}a)\frac{x}{{ - 3}} = \frac{7}{{0,75}}\\ \Rightarrow x.0,75 = ( - 3).7\\ \Rightarrow x = \frac{{( - 3).7}}{{0,75}} =  - 28\end{array}\)

Vậy x = 28

\(\begin{array}{l}b) - 0,52:x = \sqrt {1,96} :( - 1,5)\\ - 0,52:x = 1,4:( - 1,5)\\ x = \dfrac{(-0,52).(-1,5)}{1,4}\\x = \frac{39}{{70}}\end{array}\)

Vậy x = \(\frac{39}{{70}}\)

\(\begin{array}{l}c)x:\sqrt 5  = \sqrt 5 :x\\ \Leftrightarrow \frac{x}{{\sqrt 5 }} = \frac{{\sqrt 5 }}{x}\\ \Rightarrow x.x = \sqrt 5 .\sqrt 5 \\ \Leftrightarrow {x^2} = 5\\ \Leftrightarrow \left[ {_{x =  - \sqrt 5 }^{x = \sqrt 5 }} \right.\end{array}\)

Vậy x \( \in \{ \sqrt 5 ; - \sqrt 5 \} \)

Chú ý:

Nếu \({x^2} = a(a > 0)\) thì x = \(\sqrt a \) hoặc x = -\(\sqrt a \)

a: \(\dfrac{x}{-3}=\dfrac{7}{0.75}=\dfrac{28}{3}\)

=>\(x=\dfrac{28\left(-3\right)}{3}=-28\)

b: \(-\dfrac{0.52}{x}=\dfrac{\sqrt{1.96}}{-1.5}=\dfrac{1.4}{-1.5}\)

=>\(x=0.52\cdot\dfrac{1.5}{1.4}=\dfrac{39}{70}\)

c: \(\dfrac{x}{\sqrt{5}}=\dfrac{\sqrt{5}}{x}\)

=>\(x^2=5\)

=>\(x=\pm\sqrt{5}\)

\(5x^2+7,1=\text{√}49\)

\(\Rightarrow5x^2+7,1=7\)

\(\Rightarrow5x^2=7-7,1=-0,1\)

\(\Rightarrow x^2=\left(-0,1\right):5=\left(-0,02\right)\)

\(\Rightarrow x\in\varnothing\)

\(5x^2+7,1=\sqrt{49}\)

\(\Rightarrow5x^2+7,1=7\)

\(\Rightarrow5x^2=-0,1\)

\(\Rightarrow x^2=-0,1:5\Rightarrow x^2=-0,02\Rightarrow x=-\sqrt{0,02}\) hoặc \(x=\sqrt{0,02}\)

Vậy x=\(\sqrt{0,02}\)hoặc \(x=-\sqrt{0,02}\)