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Ta có :

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

HT

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

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

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

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

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

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

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

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

a, Hoành độ giao điểm (P) ; (d) thỏa mãn pt 

\(x^2=2x-m\Leftrightarrow x^2-2x+m=0\)

Để pt có 2 nghiệm pb khi \(\Delta'=1-m>0\Leftrightarrow m< 1\)

Vậy với m < 1 thì (P) cắt (d) tại 2 điểm pb 

b, Theo Vi et \(\hept{\begin{cases}x_1+x_2=2\\x_1x_2=m\end{cases}}\)

Ta có : \(\frac{1}{x_1^2}+\frac{1}{x_2^2}=2\Leftrightarrow\frac{x_1^2+x_2^2}{x_1^2x_2^2}=2\)

\(\Leftrightarrow\frac{\left(x_1+x_2\right)^2-2x_1x_2}{\left(x_1x_2\right)^2}=2\)Thay vào ta có : 

\(\Leftrightarrow\frac{4-2m}{m^2}=2\Leftrightarrow4-2m=2m^2\Leftrightarrow2m^2+2m-4=0\)

mà a + b + c = 0 => 2 + 2 - 4 = 0 

vậy pt có 2 nghiệm 

\(m_1=1\left(ktm\right);m_2=-2\left(tm\right)\)

one cộng one bằng two

two cộng one bằng three ok

Áp dụng BĐT Cô-si cho hai số dương x và y, ta có: \(\sqrt{xy}\le\frac{x+y}{2}\)

\(\Leftrightarrow\sqrt{xy}+y\le\frac{x+y}{2}+y=\frac{x+y+2y}{2}=\frac{x+3y}{2}\)

\(\Leftrightarrow\frac{x+3y}{\sqrt{xy}+y}\ge\frac{x+3y}{\frac{x+3y}{2}}=2\)

Dấu "=" xảy ra khi \(x=y\)

Vậy GTNN của P là 2 khi \(x=y\)