Bài 1:
a,\(|x-3|+|2-x|=0\)
b,\(\left(2-\dfrac{3}{4}x\right).\left(x+1\right)=0\)
bài 2:
a,A=\(\dfrac{\dfrac{-6}{7}+\dfrac{6}{13}-\dfrac{6}{29}}{\dfrac{9}{7}-\dfrac{9}{13}+\dfrac{9}{29}}\)
b,B=\(\dfrac{\dfrac{2}{15}-\dfrac{2}{21}+\dfrac{2}{39}}{0,25-\dfrac{5}{28}+\dfrac{5}{52}}\)
c,C=\(\dfrac{50-\dfrac{4}{15}+\dfrac{2}{15}-\dfrac{2}{17}}{100-\dfrac{8}{13}+\dfrac{4}{15}-\dfrac{4}{17}}:\dfrac{1+\dfrac{2}{21}-\dfrac{5}{121}}{\dfrac{65}{121}-\dfrac{26}{71}-13}\)
1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)
\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)
\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)
\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)
Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)
Bài 2:
a: \(=\dfrac{-6\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{29}\right)}{9\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{29}\right)}=\dfrac{-6}{9}=\dfrac{-2}{3}\)
b: \(=\dfrac{\dfrac{2}{15}-\dfrac{2}{21}+\dfrac{2}{39}}{\dfrac{10}{40}-\dfrac{10}{56}+\dfrac{10}{104}}\)
\(=\dfrac{\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{13}\right)}{\dfrac{10}{8}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{2}{3}:\dfrac{5}{4}=\dfrac{2}{3}\cdot\dfrac{4}{5}=\dfrac{8}{15}\)
c: \(=\dfrac{2\left(25-\dfrac{2}{13}+\dfrac{1}{15}-\dfrac{1}{17}\right)}{4\left(25-\dfrac{2}{13}+\dfrac{1}{15}-\dfrac{1}{17}\right)}:\dfrac{1+\dfrac{2}{21}-\dfrac{5}{121}}{13\left(\dfrac{5}{121}-\dfrac{2}{21}-1\right)}\)
=2/4:(-1)/13=2/4x(-13)=-13/2
