b: \(27D=3^{14}+3^{17}+...+3^{2024}\)

\(\Leftrightarrow26D=3^{2024}-3^{11}\)

hay \(D=\dfrac{3^{2024}-3^{11}}{26}\)

c: \(25E=-5^4-5^6-...-5^{1002}\)

\(\Leftrightarrow24E=-5^{1002}+5^2\)

hay \(E=\dfrac{-5^{1002}+5^2}{24}\)

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

b: \(=2\sqrt{2}+2\sqrt{2}+5\sqrt{2}=9\sqrt{2}\)

c: \(=4\sqrt{3}+3\sqrt{3}-3\sqrt{5}+2\sqrt{5}=7\sqrt{3}-\sqrt{5}\)

d: \(=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}=-\sqrt{3}\)

e: \(=\left(\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)

=7-2*căn 21+2*căn 21

=7

f: \(=\left(2\sqrt{11}-3\sqrt{2}\right)\cdot\sqrt{11}+3\sqrt{22}\)

=22-3*căn 22+3*căn 22

=22

a) \(3\sqrt{5}+\sqrt{20}-2\sqrt{5}\)

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

\(=3\sqrt{5}\)

b) \(2\sqrt{2}+\sqrt{8}+\sqrt{50}\)

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

\(=9\sqrt{5}\)

c) \(4\sqrt{3}+\sqrt{27}-\sqrt{45}+2\sqrt{5}\)

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

\(=7\sqrt{3}-\sqrt{5}\)

d) \(\sqrt{75}+\sqrt{48}-\sqrt{300}\)

\(=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\)

\(=-\sqrt{3}\)

e) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)

\(=\left(2\sqrt{7}-2\sqrt{3}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)

\(=\left(\sqrt{7}-2\sqrt{3}\right)\sqrt{7}+2\sqrt{21}\)

\(=7-2\sqrt{21}+2\sqrt{21}\)

\(=7\)

f) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

\(=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

\(=\left(2\sqrt{11}-3\sqrt{2}\right)\sqrt{11}+3\sqrt{22}\)

\(=22-3\sqrt{22}+3\sqrt{22}\)

\(=22\)

g) \(3\sqrt{45}-5\sqrt{125x}+7\sqrt{20x}+28\)

\(=9\sqrt{5}-25\sqrt{5x}+14\sqrt{5x}+28\)

\(=9\sqrt{5}-11\sqrt{5x}+28\)

1/3 + 1/5 - 1/4 = 17/60

7/8 - (1/4 + 2/5) = 9/40

5/11 - (3/5 - 6/11) = 2/5

17/60
9/40
2/5

a) \(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{4}=\dfrac{20}{60}+\dfrac{12}{60}-\dfrac{15}{60}=\dfrac{32}{60}-\dfrac{15}{60}=\dfrac{17}{60}\)

b) \(\dfrac{7}{8}-\left(\dfrac{1}{4}+\dfrac{2}{5}\right)=\dfrac{7}{8}-\left(\dfrac{5}{20}+\dfrac{8}{20}\right)=\dfrac{7}{8}-\dfrac{13}{20}=\dfrac{35}{40}-\dfrac{26}{40}=\dfrac{9}{40}\)

c) \(\dfrac{5}{11}-\left(\dfrac{3}{5}-\dfrac{6}{11}\right)=\dfrac{5}{11}-\left(\dfrac{33}{55}-\dfrac{30}{55}\right)=\dfrac{5}{11}-\dfrac{3}{55}=\dfrac{25}{55}-\dfrac{3}{55}=\dfrac{22}{55}=\dfrac{2}{5}\)

A=3³=27
\(B=\dfrac{2^{11}.9^2}{3^5.2^8}=\dfrac{2^{11}.3^4}{3^5.2^8}=\dfrac{2^3}{3}=\dfrac{8}{3}\)

A=3^3=9

B=2^3/3^5=8/243

A=3^18:3^15=3^3

B=(2^11.3^4)/(3^5.2^8)

=2^3/3=8/3

Chúc bạn học tốt!

Hihi mình cũng học lớp 9, để mình giúp cậu nha!

a) \(\sqrt{3-2\sqrt{2}}+\sqrt{9+4\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(1+\sqrt{8}\right)^2}\)

\(=\left|\sqrt{2}-1\right|+\left|1+\sqrt{8}\right|=\sqrt{2}-1+1+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\)

b) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(1+\sqrt{7}\right)^2}\)

\(=\left|\sqrt{7}-1\right|-\left|1+\sqrt{7}\right|=\sqrt{7}-1-1-\sqrt{7}=-2\)

c) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)

\(=\left|\sqrt{2}+3\right|-\left|3-\sqrt{2}\right|=\sqrt{2}+3-3+\sqrt{2}=2\sqrt{2}\)

(Nhớ click cho mình với nhoa!)

 \(\dfrac{3}{5}+\dfrac{1}{2}-\dfrac{7}{10}=\dfrac{6}{10}+\dfrac{5}{10}-\dfrac{7}{10}=\dfrac{11}{10}-\dfrac{7}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

\(\dfrac{4}{7}:\dfrac{8}{11}=\dfrac{4}{7}\times\dfrac{11}{8}=\dfrac{44}{56}=\dfrac{11}{14}\)

\(\dfrac{6}{10}+\dfrac{5}{10}-\dfrac{7}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

\(\dfrac{4}{7}x\dfrac{11}{8}=\dfrac{44}{56}=\dfrac{11}{14}\)

\(\dfrac{3}{5}+\dfrac{1}{2}-\dfrac{7}{10}=\dfrac{6}{10}+\dfrac{5}{10}-\dfrac{7}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

\(\dfrac{4}{7}:\dfrac{8}{11}=\dfrac{4}{7}\times\dfrac{11}{8}=\dfrac{44}{56}=\dfrac{11}{14}\)