Gọi độ dài tấm vải thứ nhất, hai, ba lần lượt là a,b,c(m)(145>a,b,c>0)

Ta có: \(\left(1-\dfrac{1}{2}\right)a=\left(1-\dfrac{1}{3}\right)b=\left(1-\dfrac{1}{4}\right)c\)

\(\Rightarrow\dfrac{a}{2}=\dfrac{2b}{3}=\dfrac{3c}{4}\Rightarrow\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}\)

Áp dụng t/c dtsbn:

\(\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}=\dfrac{a+b+c}{2+\dfrac{3}{2}+\dfrac{4}{3}}=\dfrac{145}{\dfrac{29}{6}}=30\)

\(\Rightarrow\left\{{}\begin{matrix}a=30.2=60\\b=\dfrac{3}{2}.30=45\\c=\dfrac{4}{3}.30=40\end{matrix}\right.\)(nhận)

Vậy...

ĐKXĐ: \(x\ge0,x\ne4\)

a) \(B=\dfrac{2\sqrt{x}+2}{x-4}.\dfrac{x-4}{\sqrt{x}+2}=\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}\)

b) \(C=A\left(B-2\right)=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}.\dfrac{2\sqrt{x}+2-2\sqrt{x}-4}{\sqrt{x}+2}=\dfrac{-2}{\sqrt{x}-2}\in Z\)

\(\Rightarrow\left(\sqrt{x}-2\right)\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Kết hợp ĐKXĐ:

\(\Rightarrow x\in\left\{0;1;9;16\right\}\)

\(\left|\dfrac{x}{2}-\dfrac{2}{3}\right|-\dfrac{1}{2}=\dfrac{5}{6}\)

\(\Rightarrow\left|\dfrac{x}{2}-\dfrac{2}{3}\right|=\dfrac{5}{6}+\dfrac{1}{2}=\dfrac{4}{3}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}-\dfrac{2}{3}=\dfrac{4}{3}\\\dfrac{x}{2}-\dfrac{2}{3}=-\dfrac{4}{3}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\left(\dfrac{4}{3}+\dfrac{2}{3}\right).2=4\\x=\left(-\dfrac{4}{3}+\dfrac{2}{3}\right).2=-\dfrac{4}{3}\end{matrix}\right.\)

1C  2C  3A  4C  5A

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:

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

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

a) Ta có: \(a\perp MN,b\perp MN\Rightarrow a//b\)

b) Ta có: \(\widehat{P_1}=\widehat{P_3}=55^0\)(đối đỉnh)

Ta có: \(a//b\Rightarrow\widehat{P_3}=\widehat{Q_5}=55^0\)(đồng vị)

Ta có: \(a//b\Rightarrow\widehat{P_3}+\widehat{Q_4}=180^0\)(trong cùng phía)

\(\Rightarrow\widehat{Q_4}=180^0-55^0=125^0\)

Ta có: \(a//b\Rightarrow\widehat{Q_4}=\widehat{P_2}=125^0\)(so le trong)

\(\sqrt{16x^2\left(y^2-6y+9\right)}=\sqrt{16x^2\left(y-3\right)^2}=4\left|x\right|\left|y-3\right|=4.\left|2\right|.\left|\sqrt{5}-3\right|=8\left(3-\sqrt{5}\right)=24-8\sqrt{5}\)

\(\sqrt{\dfrac{y}{2x}}+\dfrac{y}{x}.\sqrt{\dfrac{x}{2y}}=\sqrt{\dfrac{y}{2x}}+\sqrt{\dfrac{y^2}{x^2}.\dfrac{x}{2y}}=\sqrt{\dfrac{y}{2x}}+\sqrt{\dfrac{y}{2x}}=2\sqrt{\dfrac{y}{2x}}=\dfrac{\sqrt{2xy}}{x}\)

Bài 10:

Gọi tọa độ giao điểm d1, d2 là \(D\left(x_D;y_D\right)\)

Phương trình hoành độ giao điểm của \(d_1,d_2\):

\(2x-3=-3x+7\Leftrightarrow5x=10\Leftrightarrow x=2\)

\(\Rightarrow y=2x-3=2.2-3=1\)

Vậy tọa độ giao điểm d1,d2 là: \(D\left(2;1\right)\)

Bài 11:

Gọi tọa độ giao điểm d,d' là \(A\left(x_A;y_A\right)\)

Pt hoành độ giao điểm d,d':

\(-3x+1=-x-2\Leftrightarrow2x=3\Leftrightarrow x=\dfrac{3}{2}\)

\(\Rightarrow y=-3x+1=-\dfrac{7}{2}\)

Vậy tọa độ giao điểm d,d' là: \(A\left(\dfrac{3}{2};-\dfrac{7}{2}\right)\)