Câu 1:

Ta có 2x - y = 8 => 2x - y + 9 = 17

Mà 3x + y = 17 => 2x - y + 9 = 3x + y

<=> 9 - y = x + y <=> 9 = x + 2y <=> x = 9 - 2y

Mà 2x - y = 8 => 18 - 4y - y = 8 => 18 - 5y = 8 => y = 2 => x = 5

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

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

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:

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

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

24.

\(M=\left|1-\sqrt{3}\right|+1-\sqrt{3}=\sqrt{3}-1+1-\sqrt{3}=0\)

Đáp án A

9.

\(\sqrt{0,4.90\left(2-x\right)^2}=\sqrt{36\left(2-x\right)^2}=6\left|2-x\right|=6\left(x-2\right)=6x-12\)

Đáp án D

h=1,4+tan39o.400 ≈325(m)