4 If you aren't alert, you computer will be infected with viruses

5 If we use much pesticide on vegetables, they will become poisonous

9 Because Nam needs to improve his social skills, he attends the activity

11 How about using public buses instead of motorbike 

12 They enjoy protecting the environment unpolluted 

15 Mr Brown was disappointed that his son didn't write to him

a: ta có: \(5\cdot8^{12}\cdot9^5\cdot\left(-3\right)^2\cdot\left(-16\right)^5\)

\(=5\cdot\left(2^3\right)^{12}\cdot\left(3^2\right)^5\cdot3^2\cdot\left(-1\right)\cdot2^{20}\)

\(=-5\cdot3^{12}\cdot2^{56}\)

Ta có: \(7\cdot49^3\cdot81^5\cdot\left(-7\right)^{10}\cdot\left(-8\right)^5\)

\(=-7\cdot\left(7^2\right)^3\cdot\left(3^4\right)^5\cdot7^{10}\cdot\left(2^3\right)^5\)

\(=-7\cdot7^6\cdot3^{20}\cdot7^{10}\cdot2^{15}=-7^{17}\cdot2^{15}\cdot3^{20}\)

Ta có: \(B=\frac{5\cdot8^{12}\cdot9^5\cdot\left(-3\right)^2\cdot\left(-16\right)^5}{7\cdot49^3\cdot81^5\cdot\left(-7\right)^{10}\cdot\left(-8\right)^5}\)

\(=\frac{-5\cdot3^{12}\cdot2^{56}}{-7^{17}\cdot2^{15}\cdot3^{20}}=\frac{5\cdot2^{41}}{7^{17}\cdot3^8}\)

b: \(13\cdot4^7\cdot9^{15}-9^7\cdot\left(-16\right)^4\)

\(=13\cdot2^{14}\cdot3^{20}-3^{14}\cdot2^{16}\)

\(=3^{14}\cdot2^{14}\left(13\cdot3^6-2^2\right)\)

\(15\cdot3^{12}\cdot2^8+3\cdot27^5\cdot\left(-8\right)^{10}\)

\(=3\cdot5\cdot3^{12}\cdot2^8+3\cdot\left(3^3\right)^5\cdot\left(2^3\right)^{10}=3^{13}\cdot5\cdot2^8+3^{16}\cdot2^{30}\)

\(=3^{13}\cdot2^8\left(5+3^3\cdot2^{22}\right)\)
Ta có: \(C=\frac{13\cdot4^7\cdot9^{15}-9^7\cdot\left(-16\right)^4}{15\cdot3^{12}\cdot2^8+3\cdot27^5\cdot\left(-8\right)^{10}}\)

\(=\frac{3^{14}\cdot2^{14}\left(13\cdot3^6-2^2\right)}{3^{13}\cdot2^8\left(5+3^3\cdot2^{22}\right)}=\frac{3\cdot2^6\cdot\left(13\cdot3^6-2^2\right)}{5+3^3\cdot2^{22}}\)

\(49,=\dfrac{38}{5}-\left(\dfrac{19}{7}+\dfrac{28}{5}\right)\)

\(=\dfrac{38}{5}-\dfrac{19}{7}-\dfrac{28}{5}\)

\(=\left(\dfrac{38}{5}-\dfrac{28}{5}\right)-\dfrac{19}{7}\)

\(=2-\dfrac{19}{7}=-\dfrac{5}{7}\)

\(50,=\dfrac{25}{81}.\dfrac{15}{22}=\dfrac{125}{594}\)

Câu 50 tus sửa đề 

\(-\dfrac{1}{9}.\dfrac{15}{22}:\dfrac{-25}{9}=-\dfrac{1}{9}.\dfrac{15}{22}.\dfrac{-9}{25}=\dfrac{15}{22}.\dfrac{1}{25}=\dfrac{3}{110}\)

em chụp gần hơn đc ko,chữ bé quá

Giúp em với ạ !!!! Câu 6 câu 7 câu 9 ạ !!!

Chụp thế này mà teo tức:(

...=-15-9=-24

...=4+7=11

...=-12-13=-25

...=13+10=23

...=-4-7=-11

...=-9+13=4

...=12+0=12

...=0-7=-7

...=0+3=3

...=15-17=-2

...=-6

...=-23

...=15-15=0

-15-(+9) = -24

4-(-7) = 11

-12+(-13) = -25

-(-13)-(-10) = 23

-4-(+7) = -11

(-9)-(-13) = 4

-(-12)+0 = 12

0-(+7) = -7

0-(-3) = 3

15-17 = -2

-18+12 = -6

-5-18 = -23

-(-15)+(-15) = 0

câu 5: 

x=3,6

y=6,4

câu 6: chụp lại đề

câu 7:

a)ĐKXĐ: \(x\ge0\)

\(3\sqrt{x}=\sqrt{12}\\ \Rightarrow9x=12\\ \Rightarrow x=\dfrac{4}{3}\)

b) ĐKXĐ: \(x\ge6\)

\(\sqrt{x-6}=3\\ \Rightarrow x-6=9\\ \Rightarrow x=15\)

Câu 5: 

Áp dụng định lý Pi-ta-go ta có:

\(AB^2+AC^2=BC^2\\ \Rightarrow BC=\sqrt{6^2+8^2}\\ \Rightarrow BC=10\)

Áp dụng HTL ta có: \(x.BC=AB^2\Rightarrow x.10=6^2\Rightarrow x=3,6\)

Áp dụng HTL ta có: \(x.BC=AC^2\Rightarrow x.10=8^2\Rightarrow x=6,4\)