Giải:

\(\sqrt{49x-98}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)

\(\Leftrightarrow7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8\)

ĐKXĐ: \(x-2\ge0\Leftrightarrow x\ge2\)

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

\(\Leftrightarrow7\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)

\(\Leftrightarrow2\sqrt{x-2}=8\)

\(\Leftrightarrow\sqrt{x-2}=4\)

\(\Leftrightarrow x-2=16\)

\(\Leftrightarrow x=18\) (thỏa mãn)

Vậy ...

a. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$

$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$

$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$

Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$

$\Leftrightarrow \sqrt{x}=\frac{12}{5}$

$\Leftrightarrow x=5,76$ (thỏa mãn)

d. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{49}.\sqrt{x-2}-14\sqrt{\frac{1}{49}}\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 2\sqrt{x-2}=8$

$\Leftrightarrow \sqrt{x-2}=4$

$\Leftrightarrow x=4^2+2=18$ (tm)

b. ĐKXĐ: $x^2\geq 5$

PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$

$\Leftrightarrow \sqrt{x^2-5}=0$

$\Leftrightarrow x=\pm \sqrt{5}$

\(\text{Câu 1: Sửa đề}\)

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

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

\(\text{Câu 2}:\)

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

\(\text{Câu 2}:\)

\( b)\sqrt {x + 1} - \sqrt {x - 2} = 1\left( {x \ge 2} \right)\\ \Leftrightarrow \sqrt {x + 1} = 1 + \sqrt {x - 2} \\ \Leftrightarrow x + 1 = 1 + 2\sqrt {x - 2} + x - 2\\ \Leftrightarrow - 2\sqrt {x - 2} = - 2\\ \Leftrightarrow \sqrt {x - 2} = 1\\ \Leftrightarrow x - 2 = 1\\ \Leftrightarrow x = 1 + 2 = 3\text{(thỏa mãn điều kiện)} \)

\(c)\sqrt {{x^2} + 1} + \sqrt {4{x^2} - 4x + 5}\)

\(\text{Ta có}: \sqrt {{x^2} + 1} \ge 1 \text{với mọi x}\)

\(\sqrt{x^2-4x+5}=\sqrt{\left(x-1\right)^2+4}\ge2\) \(\text{với mọi x}\)

\(\text{Vậy với mọi x thì vế trái của phương trình} \sqrt {{x^2} + 1} + \sqrt {4{x^2} - 4x + 5} \ge 3 \text{khi đó vế phải của phương trình bằng 0. Vậy phương trình vô nghiệm} \)

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

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

  \(\Leftrightarrow\) \(\sqrt{x-2}=24\)

\(\Leftrightarrow\)\(x-2=576\)\(\Leftrightarrow x=578\)

c)\(\Leftrightarrow GTTĐ\left(x-1\right)=\sqrt{2}-1\)\(TH1:x-1>0\)

\(\Rightarrow x-1=\sqrt{2}-1\)\(\Leftrightarrow x=\sqrt{2}\)

\(TH2:x-1< 0\)

\(\Rightarrow1-x=\sqrt{2}-1\)

\(\Leftrightarrow x=2+\sqrt{2}\)

d)\(TH1:x-10=0\Rightarrow x=10\)

\(TH2:\sqrt{x-4}=0\Rightarrow x=4\)

câu b) thì mik cần thêm time

Đk: x >/ 5

pt đã cho \(\Leftrightarrow7\sqrt{x-2}-14\cdot\dfrac{\sqrt{x-2}}{7}-3\sqrt{x-5}=4\)

\(\Leftrightarrow7\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-5}=4\)

\(\Leftrightarrow5\sqrt{x-2}-3\sqrt{x-5}=4\)

\(\Leftrightarrow5\sqrt{x-2}=4+3\sqrt{x-5}\)

\(\Leftrightarrow25x-50=16+9x-45+24\sqrt{x-5}\)

\(\Leftrightarrow16x-21=24\sqrt{x-5}\)

\(\Leftrightarrow\left\{{}\begin{matrix}256x^2-672x+441=576x-2880\\x\ge\dfrac{21}{16}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}256x^2-1248x+3321=0\\x\ge\dfrac{21}{16}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}256x^2-1248x+3321=0\left(vn\right)\\x\ge\dfrac{21}{16}\end{matrix}\right.\)

Kl: ptvn

a) \(\sqrt{25x+75}+3\sqrt{x-2}=2+4\sqrt{x+3}+\sqrt{9x-18}\) (ĐKXĐ : \(x\ge2\) )

\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}-4\sqrt{x+3}-3\sqrt{x-2}=2\)

\(\Leftrightarrow\sqrt{x+3}=2\)

\(\Leftrightarrow x+3=4\)

\(\Leftrightarrow x=1\) ( Thỏa mãn ĐKXĐ )

c) \(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\) (ĐKXĐ : \(x\ge-5\) )

\(\Leftrightarrow2\sqrt{x+5}+\sqrt{x+5}-\sqrt{x+5}=4\)

\(\Leftrightarrow2\sqrt{x+5}=4\)

\(\Leftrightarrow\sqrt{x+5}=2\)

\(\Leftrightarrow x+5=4\)

\(\Leftrightarrow x=-1\) ( Thỏa mãn ĐKXĐ )

Vậy.......

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