Có: \(\dfrac{\left(-3\right)^{10}x15^5}{25^3x\left(-9\right)^7}=\dfrac{3^{10}.\left(3.5\right)^5x}{-\left(3^2\right)^7\left(5^2\right)^3x}\)

\(=\dfrac{3^{15}.5^5x}{-3^{14}.5^6x}\)\(=\dfrac{3^{14}.5^5\left(3x\right)}{3^{14}.5^5\left(-5x\right)}=\dfrac{3x}{-5x}=-\dfrac{3}{5}\)

Vậy...

`63.(-25)+25.(-37)`

`=(-63).25+25.(-37)`

`=25.[(-63)+(-37)]`

`=25.(-100)`

`=-2500`

\(63\times\left(-25\right)+25\times\left(-37\right)\\ =63\times\left(-1\right)\times25+25\times\left(-37\right)\\ =25\times\left(-63-37\right)\\ =25\times\left(-100\right)\\ =-2500\)

\(5^{2021}.5^2:5^{2020}-75:25+5.2^3-2022^0=5^{2021+2-2020}-3+5.8-1=5^3-3+40-1=125-3+40-1=161\)

161

10 C

11 A

12 D

13 B

14 D

15 D

16 B

17 A

18 C

19 D

20 B

21 B

22 C

23 A

24 A

25 C

23.17-23.14

=23.(17-14)

=23.3 = 69

23.(17-14)=23.3=69

Bài 5: 

a: BC=10cm

b: HA=4,8cm

HB=3,6(cm)

HC=6,4(cm)

Bài 6:

\(x^3=6+3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\\ \Leftrightarrow x^3=6+3x\sqrt[3]{1}\\ \Leftrightarrow x^3-3x=6\\ y^3=34+3\sqrt[3]{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}\left(\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}\right)\\ \Leftrightarrow y^3=34+3y\sqrt[3]{1}\\ \Leftrightarrow y^3-3y=34\\ \Leftrightarrow P=x^3-3x+y^3-3y+1980=6+34+1980=2020\)

gfrưerrrrrrrrrrr

\(3x=4z\Rightarrow\dfrac{x}{4}=\dfrac{z}{3}\); \(\dfrac{x}{5}=\dfrac{y}{6}\)

\(\Rightarrow\dfrac{x}{20}=\dfrac{z}{15}=\dfrac{y}{24}\)

\(\Rightarrow\dfrac{x}{20}=\dfrac{z}{15}=\dfrac{y}{24}=\dfrac{x-y+z}{20-24+15}=\dfrac{121}{11}=11\)

\(\Rightarrow x=20.11=220;z=15.11=165;y=264\)

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