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22 tháng 12 2021

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21 tháng 12 2021

Ta có :

\(tanx=2=x=90^0\)

HT

21 tháng 12 2021

đổi 24p= 2/5 giờ, 18p=3/10 giờ

gọi độ dài quãng đường AB là x (km) (x>0)

thời gian xe dự định đi từ A đến B là: x/50 ( giờ) 

độ dài quãng đường đầu xe đi được là : 50. 2/5= 20 (km/h)

=> độ dài quãng đường còn lại là x-20(km/h) 

thời gian đi đoạn đường xấu lúc sau là : (x-20)/40 ( giờ) 

ta có phương trình: 2/5 + (x-20)/40= x/50 + 3/10

<=> ...<=> x =80( km)(thỏa mãn0

vậy quãng đường AB dài 80km

21 tháng 12 2021

x2 + 2x + 1 = (x + 2). căn bậc 2 của x2 + 1

Vậy ta có:

(x2 + 2x + 1) . (x2 + 1) = (x + 2).(x2 + 1)

(x2 + 2x + 1) . (x2 + 1) = x.(x2 + 1) + 2.(x2 + 1)

(x2 + 2x + 1) . (x2 + 1) = x3 + x + 2x2 + 2

= x2.(x2 + 1) + 2x.(x2 + 1) + x2 + 1 = x3 + x + 2x2 + 2

= x4 + x2 + 2x3 + 2x + x2 + 1 = x3 + x + 2x2 + 2

= x4 + 2x2 + 2x + 2x3 + 1 = x3 + x + 2x2 + 2

= x4 + 2x + 2x3 + 1 = x3 + x + 2

DD
21 tháng 12 2021

\(x^2+2x+1=\left(x+2\right)\sqrt{x^2+1}\)

\(\Leftrightarrow x^2-3=\left(x+2\right)\left(\sqrt{x^2+1}-2\right)\)

\(\Leftrightarrow x^2-3=\left(x+2\right)\frac{x^2+1-4}{\sqrt{x^2+1}+2}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-3=0\\1=\frac{x+2}{\sqrt{x^2+1}+2}\end{cases}}\)

\(\Leftrightarrow x^2-3=0\)

\(\Leftrightarrow x=\pm\sqrt{3}\)