Câu 1:

\(=\dfrac{15}{34}+\dfrac{19}{34}-1-\dfrac{15}{17}+\dfrac{1}{3}+\dfrac{3}{5}\)

\(=-\dfrac{15}{17}+\dfrac{14}{15}=\dfrac{13}{255}\)

Câu 2: 

\(=\dfrac{5^4\cdot5^4\cdot2^8}{4^4\cdot6^4\cdot3^2\cdot5}=\dfrac{5^7}{6^4\cdot3^2}\)

a: \(\dfrac{8}{18}+\dfrac{5}{3}=\dfrac{4}{9}+\dfrac{5}{3}=\dfrac{4}{9}+\dfrac{15}{9}=\dfrac{4+15}{9}=\dfrac{19}{9}\)

b: \(\dfrac{8}{24}+\dfrac{4}{48}=\dfrac{1}{3}+\dfrac{1}{12}=\dfrac{4}{12}+\dfrac{1}{12}=\dfrac{4+1}{12}=\dfrac{5}{12}\)

c: \(\dfrac{20}{15}-\dfrac{4}{45}=\dfrac{4}{3}-\dfrac{4}{45}=\dfrac{60}{45}-\dfrac{4}{45}=\dfrac{60-4}{45}=\dfrac{56}{45}\)

d: \(\dfrac{40}{32}-\dfrac{1}{2}=\dfrac{5}{4}-\dfrac{1}{2}=\dfrac{5-2}{4}=\dfrac{3}{4}\)

Lời giải:

a. ĐKXĐ: $x\geq 5$

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

$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}=4+\sqrt{x-5}$

$\Leftrightarrow 2\sqrt{x-5}=4$

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

$\Leftrightarrow x-5=4$

$\Leftrightarrow x=9$ (tm)

b. Sửa đoạn 4x-45 thành 4x-20.

ĐKXĐ: $x\geq 5$

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

$\Leftrightarrow 2\sqrt{x-5}+\frac{1}{3}\sqrt{x-5}-\frac{2}{3}\sqrt{x-5}=4$

$\Leftrightarrow \frac{5}{3}\sqrt{x-5}=4$

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

$\Leftrightarrow x-5=\frac{144}{25}=5,76$

$\Leftrightarrow x=10,76$ (tm)

Vì \(\dfrac{1}{11}>\dfrac{1}{18}>\dfrac{1}{21}>\dfrac{1}{24}>\dfrac{1}{27}>\dfrac{1}{29}\)

\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}>\dfrac{1}{4}+\dfrac{1}{11}+\dfrac{1}{18}+\dfrac{1}{21}+\dfrac{1}{24}+\dfrac{1}{27}+\dfrac{1}{29}\)\(\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}=\dfrac{1}{11}.7=\dfrac{7}{11}\)

Ta có:

\(\dfrac{7}{11}=\dfrac{7.5}{11.5}=\dfrac{35}{55};\dfrac{4}{5}=\dfrac{4.11}{5.11}=\dfrac{44}{55}\)

\(Vì\) \(\dfrac{44}{55}>\dfrac{35}{55}\)

\(\Rightarrow\dfrac{4}{5}>\dfrac{7}{11}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{11}+\dfrac{1}{18}+\dfrac{1}{21}+\dfrac{1}{24}+\dfrac{1}{27}+\dfrac{1}{29}< \dfrac{4}{5}\left(đpcm\right)\)

Ta thấy :

\(\dfrac{1}{4}+\dfrac{1}{11}< \dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{18}+\dfrac{1}{21}< \dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{6}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\dfrac{1}{24}+\dfrac{1}{27}< \dfrac{1}{24}+\dfrac{1}{24}=\dfrac{1}{12}=\dfrac{1}{3}-\dfrac{1}{4}\)

\(\dfrac{1}{29}< \dfrac{1}{20}=\dfrac{1}{4}-\dfrac{1}{5}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{11}+\dfrac{1}{18}+\dfrac{1}{21}+\dfrac{1}{24}+\dfrac{1}{27}+\dfrac{1}{29}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{11}+\dfrac{1}{18}+\dfrac{1}{21}+\dfrac{1}{24}+\dfrac{1}{27}+\dfrac{1}{29}< 1-\dfrac{1}{5}=\dfrac{4}{5}\)

\(\Rightarrow dpcm\)

Câu hỏi của Lê Khánh Nhi - Toán lớp 7 - Học toán với OnlineMath sửa n thành x cho sửa cho nó thành lũy thừa luôn

\(I=\dfrac{5}{4}+\dfrac{-1}{3}-\dfrac{5}{-24}=\dfrac{9}{8}\)

\(J=\dfrac{-19}{-9}+\dfrac{4}{-11}-\dfrac{-2}{3}=\dfrac{239}{99}\)

\(K=\dfrac{-5}{6}-\dfrac{7}{12}+\dfrac{-3}{4}=-\dfrac{13}{6}\)

\(L=\dfrac{-3}{20}+\dfrac{1}{5}-\dfrac{-5}{3}=\dfrac{103}{60}\)

\(\dfrac{45^{10}\cdot5^{20}}{75^{15}}=\dfrac{\left(3^2\cdot5\right)^{10}\cdot5^{20}}{\left(3\cdot5^2\right)^{15}}=\dfrac{3^{20}\cdot5^{10}\cdot5^{20}}{3^{15}\cdot5^{30}}=3^5=243\\ \dfrac{6^6+6^3+3^3+3^6}{-73}=\dfrac{46656+216+27+729}{-73}=-\dfrac{47628}{73}\\ \dfrac{27^7+3^{15}}{9^9-27}=\dfrac{\left(3^3\right)^7+3^{15}}{\left(3^2\right)^9-3^3}=\dfrac{3^{21}+3^{15}}{3^{18}-3^3}=\dfrac{3^{15}\left(3^6+1\right)}{3^3\left(3^{15}-1\right)}=\dfrac{3^5\cdot730}{3^{15}-1}\\ \dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)

I'm sorry, I can't help you