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: \(x.BC=AC^2\Rightarrow x.10=8^2\Rightarrow x=6,4\)

Question 2: David has volunteered for 2 years

Question 3: I think collecting stamps is interesting

mình làm câu 2, 3 rồi nhé

https://hoc24.vn/cau-hoi/anh-chi-giup-em-tieng-anh-7-cau-1-va-2-voi-a-chieu-e-thi-aem-cam-on-anh-chi-rat-nhieu-a.3054793966970

câu 1: Lan had a high fever, so she stayed home from school yesterday

\(d,=\dfrac{3y}{5x\left(x-y\right)}\\ e,=\dfrac{5x\left(x+2\right)\left(2-x\right)}{4\left(x-2\right)\left(x+2\right)}=\dfrac{-5x}{4}\\ f,=\dfrac{3\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)\left(6-x\right)}=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\\ g,=\dfrac{3xy\left(x-3y\right)\left(x+3y\right)}{2x^2y^2\left(x-3y\right)}=\dfrac{3\left(x+3y\right)}{2xy}\\ h,=\dfrac{45x^2y\left(x-y\right)\left(x+y\right)}{10xy\left(y-x\right)}=\dfrac{-9x\left(x+y\right)}{2}\\ i,=\dfrac{12\left(a-b\right)\left(a+b\right)\left(a^2+ab+b^2\right)}{3\left(a+b\right)\left(a-b\right)^2}=\dfrac{4\left(a^2+ab+b^2\right)}{a-b}\)

e: \(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=-\dfrac{5}{2}\)

Bài 2:

\(\dfrac{1}{2}:\dfrac{5}{4}=x:\dfrac{10}{3}\Leftrightarrow\dfrac{1}{2}.\dfrac{4}{5}=\dfrac{3}{10}x\Leftrightarrow\dfrac{3}{10}x=\dfrac{2}{5}\Leftrightarrow x=\dfrac{2}{5}:\dfrac{3}{10}=\dfrac{4}{3}\)

Bài 3:

Áp dụng t/c dtsbn:

\(\dfrac{x}{4}=\dfrac{y}{12}=\dfrac{x+y}{4+12}=\dfrac{48}{16}=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.4=12\\y=3.12=36\end{matrix}\right.\)

Gọi xy là tiếp tuyến tại A của (O)

=>góc xAC=góc ABC

xy//DE

=>góc xAE=góc AED

=>góc AED=góc ABC

Xét ΔAED và ΔABC có

góc AED=góc ABC

góc EAD chung

=>ΔAED đồng dạng với ΔABC

=>AE/AB=AD/AC

=>AE*AC=AB*AD

\(a,P=\dfrac{3\sqrt{a}-3}{\sqrt{a}\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\left(a\ge0;a\ne1\right)\\ P=\dfrac{3\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}=\dfrac{3\left(\sqrt{a}+1\right)}{\sqrt{a}}\\ b,a=4\Leftrightarrow\sqrt{a}=2\\ \Leftrightarrow P=\dfrac{3\left(2+1\right)}{2}=\dfrac{9}{2}\)

Câu 27: D

Câu 29: D

Câu 30: B