x(x+8)=20

=>\(x^2+8x-20=0\)

=>(x+10)(x-2)=0

=>\(\left[{}\begin{matrix}x=-10\\x=2\end{matrix}\right.\)

\(x\)(\(x+8\)) = 20

\(x^2\) + 8\(x\)  = 20

\(x^2\) + 8\(x\) - 20 = 0

(\(x^2\) + 10\(x\)) - (2\(x\) + 10) = 0

\(x\)(\(x+10\)) - 2(\(x+10\)) = 0

  (\(x+10\))(\(x-2\)) = 0

  \(\left[{}\begin{matrix}x+10=0\\x-2=0\end{matrix}\right.\)

   \(\left[{}\begin{matrix}x=-10\\x=2\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-10; 2}

a: Oz là phân giác của góc xOy

=>\(\widehat{xOz}=\widehat{yOz}=\dfrac{\widehat{xOy}}{2}=\dfrac{80^0}{2}=40^0\)

gọi oz là tia PG của oy 😤 đề sai r

\(4^2+5=y^2+y\\ =>16+5=y^2+y\\ =>21=y^2+y\\ =>y^2+y-21=0\\ =>\left(y^2+\dfrac{1}{2}y+\dfrac{1}{2}y+\dfrac{1}{4}\right)-\dfrac{85}{4}=0\\ =>\left[y\left(y+\dfrac{1}{2}\right)+\dfrac{1}{2}\left(y+\dfrac{1}{2}\right)\right]=\dfrac{85}{4}\\ =>\left(y+\dfrac{1}{2}\right)\left(y+\dfrac{1}{2}\right)=\dfrac{85}{4}\\ =>\left(y+\dfrac{1}{2}\right)^2=\left(\dfrac{\sqrt{85}}{2}\right)^2\\ TH1:y+\dfrac{1}{2}=\dfrac{\sqrt{85}}{2}\\ =>y=\dfrac{\sqrt{85}-1}{2}\\ TH2:y+\dfrac{1}{2}=-\dfrac{\sqrt{85}}{2}\\ =>y=\dfrac{-\sqrt{85}-1}{2}\)

\(100-x:15=10,5\\ x:15=100-10,5\\ x:15=89,5\\ x=15\times89,5\\ x=1342,5\)

Vậy: ... 

100 - x : 15 = 10,5

x : 15 = 100 - 10,5

x : 15 = 89,5

x = 89,5 × 15

x = 1342,5

                                     ~~~ the end~~~

\(\left(\dfrac{3}{4}\times x\right):\dfrac{1}{2}=\dfrac{4}{5}\\ \dfrac{3}{4}\times x=\dfrac{4}{5}\times\dfrac{1}{2}\\ \dfrac{3}{4}\times x=\dfrac{4\times1}{5\times2}\\ \dfrac{3}{4}\times x=\dfrac{2}{5}\\ x=\dfrac{2}{5}:\dfrac{3}{4}\\ x=\dfrac{2}{5}\times\dfrac{4}{3}\\ x=\dfrac{2\times4}{5\times3}\\ x=\dfrac{8}{15}\)

Vậy: ...

\(15\times8-\left(17-30+83\right)-144:6\\ =15\times8-\left[\left(17+83\right)-30\right]-24\\ =120-\left(100-30\right)-24\\ =120-70-24\\ =50-24\\ =26\)

\(12+\left(x\times3+5\right):5=16\\ \left(x\times3+5\right):5=16-12\\ \left(x\times3+5\right):5=4\\ x\times3+5=4\times5\\ x\times3+5=20\\ x\times3=20-5\\ x\times3=15\\ x=15:3\\ x=5\)

Vậy: .. 

\(\left(\dfrac{-4}{5}+\dfrac{-3}{2}:x\right)\cdot\dfrac{1}{3}-\dfrac{1}{2}=\dfrac{1}{2}\\ \left(\dfrac{-4}{5}+\dfrac{-3}{2}:x\right)\cdot\dfrac{1}{3}=\dfrac{1}{2}+\dfrac{1}{2}\\ \left(\dfrac{-4}{5}+\dfrac{-3}{2}\right):x\cdot\dfrac{1}{3}=1\\ \dfrac{-4}{5}+\dfrac{-3}{2}:x=1:\dfrac{1}{3}\\ -\dfrac{4}{5}+\dfrac{-3}{2}x=3\\ \dfrac{3}{2}x=\dfrac{-4}{5}-3\\ \dfrac{3}{2}x=-\dfrac{19}{5}\\ x=\dfrac{-19}{5}:\dfrac{3}{2}\\x =-\dfrac{38}{15}\)

Số ml nước chị dùng để tưới rau là:

`500:2=250(ml)`

Số ml nước còn lại là:

`500-250=250(ml)`

Số ml nước chị dùng để đổ vào bình hoa là:

`1/5 xx 250 = 50(ml)`

Lượng nước còn lại là:

`250-50=200(ml)`

ĐS: ...

số nước cj An tưới ra là: 500 : 2 = 250 ml

số nước cj An đổ vào bình hoa là: 250 x 1/5 = 50 ml

đáp số: lượng nước đổ vào bình hoa là 50 ml nước

\(a,7\cdot7\cdot7\cdot7\\ =7^1\cdot7^1\cdot7^1\cdot7^1\\ =7^{1+1+1+1}=7^4\\ b,7\cdot35\cdot7\cdot25\\ =7\cdot\left(5\cdot7\right)\cdot7\cdot\left(5^2\right)\\ =7\cdot5\cdot7\cdot7\cdot5^2\\ =\left(7\cdot7\cdot7\right)\cdot\left(5\cdot5^2\right)\\ =\left(7^1\cdot7^1\cdot7^1\right)\cdot\left(5^1\cdot5^2\right)\\ =7^{1+1+1}\cdot5^{1+2}\\ =7^3\cdot5^3\\ c,2\cdot3\cdot8\cdot12\cdot24\\ =2\cdot3\cdot2^3\cdot\left(2^2\cdot3\right)\cdot\left(2^3\cdot3\right)\\ =\left(2\cdot2^3\cdot2^2\cdot2^3\right)\cdot\left(3\cdot3\cdot3\right)\\ =\left(2^1\cdot2^3\cdot2^2\cdot2^3\right)\cdot\left(3^1\cdot3^1\cdot3^1\right)\\ =2^{1+3+2+3}\cdot3^{1+1+1}\\ =2^9\cdot3^3\\ d,x\cdot x\cdot y\cdot y\cdot x\cdot y\cdot x\\ =\left(x\cdot x\cdot x\cdot x\right)\cdot\left(y\cdot y\cdot y\right)\\ =\left(x^1\cdot x^1\cdot x^1\cdot x^1\right)\cdot\left(y^1\cdot y^1\cdot y^1\right)\\ =x^{1+1+1+1}\cdot y^{1+1+1}\\ =x^4\cdot y^3\)