Ns thật thì k phải dễ mà là quá dễ :))

@Tuấn Anh Phan Nguyễn thừa sức lm đc nhé bn!

a, Theo tính chất của tỉ lệ thuận ta có:

\(\dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}=\dfrac{x_1}{-\dfrac{3}{4}}=\dfrac{2}{\dfrac{1}{7}}\)

\(\Rightarrow x_1=\left(-\dfrac{3}{4}\cdot2\right):\dfrac{1}{7}=-\dfrac{3}{2}\cdot7=-\dfrac{21}{2}\)

Vậy..............................

b, Theo t/c của tỉ lệ thuận ta có:

\(\dfrac{x_1}{x_2}=\dfrac{y_1}{y2}\) hay \(\dfrac{x_1}{-4}=\dfrac{y_1}{3}\)

Áp dụng t/c của dãy tỉ số = nhau ta có:

\(\dfrac{x_1}{-4}=\dfrac{y_1}{3}=\dfrac{y_1-x_1}{3-\left(-4\right)}=-\dfrac{2}{7}\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=-\dfrac{2}{7}\cdot\left(-4\right)=\dfrac{8}{7}\\y_1=-\dfrac{2}{7}\cdot3=-\dfrac{6}{7}\end{matrix}\right.\)

Vậy.............

Xử câu b trc =)

b) \(3\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)

\(\Leftrightarrow\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{9}:3=-\dfrac{1}{27}\)

\(\Leftrightarrow\left(3x-\dfrac{1}{2}\right)^3=\left(-\dfrac{1}{3}\right)^3\)

\(\Leftrightarrow3x-\dfrac{1}{2}=-\dfrac{1}{3}\)

\(\Leftrightarrow3x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\)

\(\Leftrightarrow x=\dfrac{1}{6}:3=\dfrac{1}{18}\)

a, Vì \(-\dfrac{1}{2}< 0\) => \(-\dfrac{1}{2}\) là số nhỏ nhất trong 3 số

Ta có: \(\dfrac{4}{9}:\dfrac{3}{7}=\dfrac{4\cdot7}{3\cdot9}=\dfrac{28}{27}>1\)

\(\Rightarrow\dfrac{4}{9}>\dfrac{3}{7}>-\dfrac{1}{2}\)

e, \(\dfrac{3^{10}+1}{3^9+1}\) và \(\dfrac{3^9+1}{3^8+1}\)

Ta có: \(A=\dfrac{3^{10}+1}{3^9+1}>1>\dfrac{3^{10}+1+2}{3^9+1+2}=\dfrac{3^{10}+3}{3^9+3}=\dfrac{3\left(3^9+1\right)}{3\left(3^8+1\right)}\)

\(=\dfrac{3^9+1}{3^8+1}=B\)

\(\Rightarrow A>B\)

\(\left[\left(x+\dfrac{1}{3}\right)+\dfrac{1}{6}\right]-\dfrac{1}{2}:2=\dfrac{3}{4}\)

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

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}+\dfrac{1}{4}=1\)

\(\Rightarrow x=1-\dfrac{1}{2}=\dfrac{1}{2}\)

Vậy......

a, \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)

=> \(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)

=> \(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

b, Dễ ==> tự lm

c,( \(2\dfrac{4}{5}x-50\)) \(:\dfrac{2}{3}=51\)

=> \(2\dfrac{4}{5}x-50=51\cdot\dfrac{2}{3}=34\)

=> \(\dfrac{14}{5}x=34+50=84\)

=> \(x=84:\dfrac{14}{5}=30\)

d, \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\-2x=-\dfrac{2}{3}\Rightarrow x=\dfrac{1}{3}\end{matrix}\right.\)

e, \(\dfrac{2}{3}-\dfrac{1}{2}x=\dfrac{5}{12}\)

=> \(-\dfrac{1}{2}x=\dfrac{5}{12}-\dfrac{2}{3}=-\dfrac{1}{4}\)

=> \(x=-\dfrac{1}{4}:\left(-\dfrac{1}{2}\right)=\dfrac{1}{2}\)