Bài 11: Số vô tỉ. Khái niệm về căn bậc hai

a: \(\dfrac{\sqrt{81}}{\sqrt{16}}=\dfrac{9}{4}=\dfrac{36}{16}< \dfrac{81}{16}\)

b: \(\sqrt{16+25}=\sqrt{41}< 9=\sqrt{16}+\sqrt{25}\)

a)\(\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{3}{5}=\dfrac{15}{20}-\dfrac{50}{20}-\dfrac{12}{20}=-\dfrac{47}{20}\)

b) \(\sqrt{7^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}=7+\sqrt{\dfrac{1}{16}}=7+\dfrac{1}{4}=\dfrac{29}{4}\)

c) \(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}=\dfrac{1}{2}.10-\sqrt{\dfrac{1}{16}+1}=5-\sqrt{\dfrac{17}{16}}\)

b) \(\sqrt{144}-5.\sqrt{\dfrac{16}{9}}+\left|-5\dfrac{1}{3}\right|\)

=\(12-5.\dfrac{4}{3}+\dfrac{16}{3}\)

=\(12-\left(\dfrac{20}{3}-\dfrac{16}{3}\right)\)

=\(12-\dfrac{4}{3}\)

=\(\dfrac{32}{3}\)

c) \(\dfrac{21^4}{27.\left(-343\right)}+7=\dfrac{7^4.3^4}{3^3.\left(-7\right)^3}+7=-7.3+7=-21+7=-14\)

Bài 6:
a. $|x-\sqrt{3}|=3$

$\Rightarrow x-\sqrt{3}=3$ hoặc $x-\sqrt{3}=-3$

$\Leftrightarrow x=3+\sqrt{3}$ hoặc $x=-3+\sqrt{3}$

b. $|1-2x|=\sqrt{5}-1$

$\Rightarrow 1-2x=\sqrt{5}-1$ hoặc $1-2x=1-\sqrt{5}$

$\Leftrightarrow x=\frac{2-\sqrt{5}}{2}$ hoặc $x=\frac{\sqrt{5}}{2}$

Bài 7: Thiếu yêu cầu của đề. Bạn xem lại.

câu đầu:

\(2^{3n}:2^{n+2}=\left[\left(\sqrt{2}\right)^2\right]^2\\ \Leftrightarrow2^{3n-\left(n+2\right)}=2^2\\ \Leftrightarrow3n-\left(n+2\right)=2\\ \Leftrightarrow2n=4\\ \Leftrightarrow n=2\)

câu sau:

\(8^n:2^n.4=4\\ \Leftrightarrow8^n:2^n=4:4=1\\ \Leftrightarrow2^{3n}:2^n=2^0\\ \Leftrightarrow2^{3n-n}=2^0\\ \Leftrightarrow3n-n=0\\ \Leftrightarrow n=0\)

Ta có:2³ⁿ:2ⁿ⁺²=\(\left[\left(\sqrt{2}\right)^2\right]^2\)

  ⇔  2^3n-n-2=4

  ⇔  2^2n-2=2²

  ⇔  2n-2 =2

  ⇔  2n=4

  ⇔ n=2

bài giải 

PT\(\Leftrightarrow\dfrac{\left(2^n\right)^3}{2^n.4}=4\)

\(\Leftrightarrow\dfrac{2^{2n}}{4}=4\)

\(\Leftrightarrow\dfrac{4^n}{4}=4\)

\(\Rightarrow n=2\)