\(\sqrt{\frac{98}{8}}\)=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

a, 4x²=15-(-21)

=36

x²=36:4

x²=4

x²=2²

x=2

b. 5x²+7,1=\(\sqrt{49}\)

\(\Rightarrow\)5x²+7,1=7

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

^{\(\Rightarrow\)}5x² =14,1

\(\Rightarrow\)x² =\(\dfrac{14,1}{5}\)

\(\Rightarrow\)x =\(\sqrt{\dfrac{14,1}{5}}\)

cho mk 1 tick đúng và câu tiếp thao sẽ hiện ra

27 . 39 + 27 . 63

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

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

=>\(x^2=5\)

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

ĐKXĐ: \(x\ge2\)

Từ pt đã cho suy ra:

\(7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8\)

⇒ \(2\sqrt{x-2}=8\) ⇒ \(x=18\)

Bài này chắc tác giả đánh sai tử thức của phân thức cuối cùng, biểu thức B phải là  \(B=\frac{1}{C}\)  trong đó \(C=\left(\frac{\sqrt{x}+3}{x+\sqrt{x}+1}-\frac{\sqrt{x}-3}{x\sqrt{x}-1}\right)\cdot\frac{x\sqrt{x}-\sqrt{x}+x^2-1}{\sqrt{x}}\)
Ta có \(C=\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}-\frac{\left(\sqrt{x}-3\right)}{x\sqrt{x}-1}\right)\cdot\frac{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{x+\sqrt{x}}{x\sqrt{x}-1}\cdot\frac{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}=\left(\sqrt{x}+1\right)^2\)

Thành thử ta được \(C=\left(\sqrt{x}+1\right)^2\)

 Ta có \(x=98+20\sqrt{6}=\left(5\sqrt{2}+4\sqrt{3}\right)^2\to\sqrt{x}=5\sqrt{2}+4\sqrt{3}\to\)

hay \(C=\left(\sqrt{x}+1\right)^2=x+2\sqrt{x}+1=99+20\sqrt{6}+2\left(5\sqrt{2}+4\sqrt{3}\right)\)

\(=99+20\sqrt{6}+10\sqrt{2}+8\sqrt{3}\to B=\frac{1}{C}=\frac{1}{99+20\sqrt{6}+10\sqrt{2}+8\sqrt{3}}\)

\(\text{Câu 1: Sửa đề}\)

\( a)M = \left( {1 - \dfrac{{4\sqrt x }}{{x - 1}} + \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{x - 2\sqrt x }}{{x - 1}}\\ M = \left[ {1 - \dfrac{{4\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} + \dfrac{1}{{\sqrt x - 1}}} \right].\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x - 2\sqrt x }}\\ M = \left[ {1 + \dfrac{{ - 4\sqrt x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right].\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x - 2\sqrt x }}\\ M = \left[ {1 + \dfrac{{ - 3\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right].\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x - 2\sqrt x }}\\ M = \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) - 3\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x - 2\sqrt x }}\\ M = \sqrt x \left( {\sqrt x - 3} \right).\dfrac{1}{{x - 2\sqrt x }}\\ M = \dfrac{{x - 3\sqrt x }}{{x - 2\sqrt x }} \)

\( b)M = \dfrac{1}{2} \Rightarrow \dfrac{{x - 3\sqrt x }}{{x - 2\sqrt x }} = \dfrac{1}{2}\\ \Leftrightarrow 2\left( {x - 3\sqrt x } \right) = x - 2\sqrt x \\ \Leftrightarrow 2x - 6\sqrt x = x - 2\sqrt x \\ \Leftrightarrow - 4\sqrt x = - x\\ \Leftrightarrow 16x = {x^2}\\ \Leftrightarrow 16x - {x^2} = 0\\ \Leftrightarrow x\left( {16 - x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ 16 - x = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 16 \end{array} \right. \)

\(\text{Câu 2}:\)

\( a)\sqrt {49x - 98} - 14\sqrt {\dfrac{{x - 2}}{{49}}} = 3\sqrt {x - 2} + 8\left( {x \ge 2} \right)\\ \Leftrightarrow 7\sqrt {x - 2} - 3\sqrt {x - 2} = 8 + 14\sqrt {\dfrac{{x - 2}}{{49}}} \\ \Leftrightarrow 4\sqrt {x - 2} = 8 + 14\sqrt {\dfrac{{x - 2}}{{49}}} \\ \Leftrightarrow 4\sqrt {x - 2} = 8 + 14\dfrac{{\sqrt {x - 2} }}{7}\\ \Leftrightarrow 4\sqrt {x - 2} = 8 + 2\sqrt {x - 2} \\ \Leftrightarrow 4\sqrt {x - 2} - 2\sqrt {x - 2} = 8\\ \Leftrightarrow 2\sqrt {x - 2} = 8\\ \Leftrightarrow \sqrt {x - 2} = 4\\ \Leftrightarrow x - 2 = 16\\ \Leftrightarrow x = 16 + 2 = 18 \text{(thỏa mãn điều kiện)} \)