đề bài chắc là tìm x

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tìm x đó

Câu 6: Khôg có cau nào đúng

Câu 7: C

Câu 8: B

Câu 9: B

Câu 10: D

6Không có đáp án nào đúng x=11/4

7C

8B

9B

10D

\(\dfrac{1}{\sqrt{49+20\sqrt{6}}}-\dfrac{1}{\sqrt{49-20\sqrt{6}}}+\dfrac{1}{\sqrt{7-4\sqrt{3}}}\)

\(=\dfrac{1}{\sqrt{5^2+2\cdot2\sqrt{6}\cdot5+\left(2\sqrt{6}\right)^2}}-\dfrac{1}{\sqrt{5^2-2\cdot2\sqrt{6}\cdot5+\left(2\sqrt{6}\right)^2}}+\dfrac{1}{\sqrt{2^2-2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}\)

\(=\dfrac{1}{\sqrt{\left(5+2\sqrt{6}\right)^2}}-\dfrac{1}{\sqrt{\left(5-2\sqrt{6}\right)^2}}+\dfrac{1}{\sqrt{\left(2-\sqrt{3}\right)^2}}\)

\(=\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}+\dfrac{1}{2-\sqrt{3}}\)

\(=\dfrac{5-2\sqrt{6}}{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}-\dfrac{5+2\sqrt{6}}{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}+\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)

\(=\dfrac{5-2\sqrt{6}-5-2\sqrt{6}}{1}+\dfrac{2+\sqrt{3}}{1}\)

\(=-4\sqrt{6}+2+\sqrt{3}\)

\(=\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}+\dfrac{1}{2-\sqrt{3}}\)

\(=5-2\sqrt{6}-5-2\sqrt{6}+2+\sqrt{3}\)

\(=2-4\sqrt{6}+\sqrt{3}\)

\(=\dfrac{1}{\sqrt{\sqrt{25}^2+2.\sqrt{25}.\sqrt{24}+\sqrt{24}^2}}-\dfrac{1}{\sqrt{\sqrt{25}^2-2.\sqrt{25}.\sqrt{24}+\sqrt{24}^2}}+\dfrac{1}{\sqrt{\sqrt{4}^2-2\sqrt{4}\sqrt{3}+\sqrt{3}^2}}\\ =\dfrac{1}{\sqrt{\left(\sqrt{25}+\sqrt{24}\right)^2}}-\dfrac{1}{\sqrt{\left(\sqrt{25}-\sqrt{24}\right)^2}}+\dfrac{1}{\sqrt{\left(\sqrt{4}-\sqrt{3}\right)^2}}\\ =\dfrac{1}{5+\sqrt{24}}-\dfrac{1}{5-\sqrt{24}}+\dfrac{1}{2-\sqrt{3}}\)

\(=\dfrac{5-\sqrt{24}}{25-24}-\dfrac{5+\sqrt{24}}{25-24}+\dfrac{2+\sqrt{3}}{4-3}\\ =5-\sqrt{24}-5-\sqrt{24}+2+\sqrt{3}\\ =2-4\sqrt{6}+\sqrt{3}\)

Bài 1:

a; (\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\)) x \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)

     \(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) : \(\dfrac{3}{4}\)

      \(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) x \(\dfrac{4}{3}\)

    \(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) =  \(\dfrac{1}{3}\)

      \(\dfrac{1}{4}x\) = \(\dfrac{1}{3}\) + \(\dfrac{1}{8}\)

       \(\dfrac{1}{4}\) \(x\)=  \(\dfrac{8}{24}\) + \(\dfrac{11}{24}\)

          \(\dfrac{1}{4}x=\dfrac{11}{24}\)

           \(x=\dfrac{11}{24}:\dfrac{1}{4}\)

           \(x=\dfrac{11}{24}\times4\)

           \(x=\dfrac{11}{6}\) 

b; \(\dfrac{12}{5}:x\) = \(\dfrac{14}{3}\) x \(\dfrac{4}{7}\)

     \(\dfrac{12}{5}\) : \(x\) = \(\dfrac{8}{3}\)

            \(x\) = \(\dfrac{12}{5}\) : \(\dfrac{8}{3}\)

            \(x\) = \(\dfrac{12}{5}\) x \(\dfrac{3}{8}\)

             \(x\) = \(\dfrac{9}{10}\)

c; \(x+\dfrac{2}{3}\) = 8:4 - 1

   \(x+\dfrac{2}{3}\) = 2 - 1

   \(x+\dfrac{2}{3}\) = 1

  \(x=1-\dfrac{2}{3}\)

  \(x\) = \(\dfrac{1}{3}\)

\(1,\\ a,\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow\left[{}\begin{matrix}x-4=4\\x-4=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=0\end{matrix}\right.\\ c,\Leftrightarrow2x+1=-2\Leftrightarrow x=-\dfrac{3}{2}\\ 2,\\ a,=1\\ b,=\left(\dfrac{13}{4}\right)^2=\dfrac{169}{16}\\ c,=\left(-\dfrac{7}{4}\right)^2=\dfrac{49}{16}\\ d,=\left(\dfrac{3}{7}\right)^{20}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^8=...\\ e,=\left(3\cdot5\cdot\dfrac{2}{3}\right)^2=10^2=100\)

\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{28}{49}=\dfrac{28:7}{49:7}=\dfrac{4}{9}\\ Vậy:x=\dfrac{4.9}{4}=9\\ y=\dfrac{4.21}{9}=\dfrac{28}{3}\)

\(\dfrac{x}{2}=\dfrac{3}{y}\\ \Leftrightarrow x.y=2.3=6\\ Vậy:\left[{}\begin{matrix}\left(x;y\right)=\left(1;6\right)=\left(6;1\right)\\\left(x;y\right)=\left(2;3\right)=\left(3;2\right)\end{matrix}\right.\)

\(3,\\ \dfrac{42}{54}=\dfrac{42:6}{54:6}=\dfrac{7}{9}\\ \dfrac{42}{54}=\dfrac{7}{x}=\dfrac{7}{9}\\ Vậy:x=\dfrac{7.9}{7}=9\)