đây là bài lớp 10 chứ nhỉ

ta có \(AC=20\times2=40\text{ hải lí}\), \(AB=15\times2=30\text{ hải lí}\)

áp dụng định lý cosin ta có :

\(BC=\sqrt{AB^2+AC^2-2AB.AC\text{c}osA}=\sqrt{40^2+30^2-2\times30\times40\times cos60^o}\simeq36.06\text{ hải lí}\)

a)pt<=> \(-5\sqrt{x-2}=-2\)<=>\(5\sqrt{x-2}=2\)<=>\(25\left(x-2\right)=4\)(binh phuong 2 ve) <=> \(x-2=\frac{4}{25}\Rightarrow x=\frac{54}{25}\)vay \(x=\frac{54}{25}\)

b)pt<=> \(3\sqrt{x-1}=5-x\Leftrightarrow\sqrt{x-1}=\frac{5-x}{3}\Leftrightarrow x-1=\frac{25-10x+x^2}{9}\)\(\Leftrightarrow\frac{9x-9}{9}=\frac{25-10x+x^2}{9}\Leftrightarrow9x-9=25-10x+x^2\Leftrightarrow x^2-19x+34=0\)\(\Leftrightarrow x\left(x-2\right)-17\left(x-2\right)=0\Leftrightarrow\left(x-17\right)\left(x-2\right)=0\Rightarrow\orbr{\begin{cases}x-17=0\Rightarrow x=17\\x-2=0\Rightarrow x=2\end{cases}}\)vay \(x=17\)hoac \(x=2\)

c)pt \(\Leftrightarrow4x^2-12x+9=1\)(binh phuong 2 ve) \(\Leftrightarrow4x^2-12x+8=0\Leftrightarrow4x\left(x-2\right)-4\left(x-2\right)=0\Leftrightarrow4\left(x-1\right)\left(x-2\right)=0\Rightarrow\orbr{\begin{cases}x-1=0\Rightarrow x=1\\x-2=0\Rightarrow x=2\end{cases}}\)vay x=1 hoac x = 2

mik can cau b, cau a mik t giai dc r

bài đâu

\(\orbr{\frac{1}{1-\sqrt{x}}-\frac{1}{\sqrt{x}}]}\div\orbr{\begin{cases}\\\end{cases}(2\sqrt{x}-1)(\frac{1}{1-\sqrt{x}}+\frac{\sqrt{x}}{1-\sqrt{x}+x})]}\)

sori mng em bị lag xíu

so do chan chan co dang xxxx5, chac vay