I think school uniforms have many benefits for students. Firstly, wearing uniforms helps students not distinguish rich and poor. Rich or poor students are required to wear the same clothes to school, and poor students will not be inferior to their peers. On Monday, students wear uniforms look very beautiful and simple. The uniforms are beautifully designed so students can wear comfortably . On Tuesday, students will not take time to select clothes to school. This will help students concentrate on learning instead of thinking how I would wear, what to wear to school. Finally, the uniforms will help students save money on shopping. They will not spend too much money to buy clothes because they always have to wear uniforms to school. In conclusion, wearing uniforms is very beneficial to students.

mình viết là đề students should wear uniform when they are at school

Bạn hình như sai đề ? Nếu đề bạn đúng thì mình không biết làm ..

Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{m}{3}< >-\dfrac{1}{m}\)

=>\(m^2\ne-3\)(luôn đúng)

Ta có: \(\left\{{}\begin{matrix}mx-y=2\\3x+my=3m\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=mx-2\\3x+m\left(mx-2\right)=3m\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=mx-2\\3x+m^2x-2m=3m\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=mx-2\\x\left(m^2+3\right)=5m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5m}{m^2+3}\\y=m\cdot\dfrac{5m}{m^2+3}-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5m}{m^2+3}\\y=\dfrac{5m^2-2m^2-6}{m^2+3}=\dfrac{3m^2-6}{m^2+3}\end{matrix}\right.\)

\(\left(x+y\right)\cdot\left(m^2+3\right)+8=0\)

=>\(\dfrac{5m+3m^2-6}{m^2+3}\cdot\left(m^2+3\right)+8=0\)

=>\(3m^2+5m-6+8=0\)

=>\(3m^2+5m+2=0\)

=>(m+1)(3m+2)=0

=>\(\left[{}\begin{matrix}m=-1\\m=-\dfrac{2}{3}\end{matrix}\right.\)

Bài 2:

Thay x=-1 vào (P), ta được:

\(y=\left(-1\right)^2=1\)

Thay x=2 vào (P), ta được:

\(y=2^2=4\)

vậy: A(-1;1); B(2;4)

Gọi (d): y=ax+b(a\(\ne\)0) là phương trình đường thẳng AB

Thay x=-1 và y=1 vào (d), ta được:

\(a\cdot\left(-1\right)+b=1\)

=>-a+b=1(1)

Thay x=2 và y=4 vào (d), ta được:

\(2\cdot a+b=4\)

=>2a+b=4(2)

Từ (1) và (2) ta có hệ phương trình: \(\left\{{}\begin{matrix}-a+b=1\\2a+b=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-3a=-3\\-a+b=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=1\\b=a+1=1+1=2\end{matrix}\right.\)

Vậy: phương trình AB là y=x+2

\(ĐK:x\ne\dfrac{1}{2};x\ne1;x\ne\dfrac{3}{2};x\ne2;x\ne\dfrac{5}{2}\\ PT\Leftrightarrow\dfrac{1}{\left(2x-1\right)\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(3x-2\right)}+\dfrac{1}{\left(3x-2\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(5x-2\right)}=\dfrac{4}{21}\\ \Leftrightarrow2\left[\dfrac{\dfrac{1}{2}}{\left(x-\dfrac{1}{2}\right)\left(x-1\right)}+\dfrac{\dfrac{1}{2}}{\left(x-1\right)\left(x-\dfrac{3}{2}\right)}+\dfrac{\dfrac{1}{2}}{\left(x-\dfrac{3}{2}\right)\left(x-2\right)}+\dfrac{\dfrac{1}{2}}{\left(x-2\right)\left(x-\dfrac{5}{2}\right)}\right]=\dfrac{4}{21}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-\dfrac{1}{2}}+\dfrac{1}{x-\dfrac{3}{2}}-\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-\dfrac{3}{2}}+\dfrac{1}{x-\dfrac{5}{2}}-\dfrac{1}{x-2}=\dfrac{2}{21}\\ \Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-\dfrac{5}{2}}=\dfrac{2}{21}\\ \Leftrightarrow\dfrac{x-\dfrac{5}{2}-x+1}{\left(x-1\right)\left(x-\dfrac{5}{2}\right)}=\dfrac{2}{21}\\ \Leftrightarrow\dfrac{-\dfrac{3}{2}}{x^2-\dfrac{7}{2}x+\dfrac{5}{2}}=\dfrac{2}{21}\\ \Leftrightarrow x^2-\dfrac{7}{2}x+\dfrac{5}{2}=-\dfrac{63}{4}\\ \Leftrightarrow4x^2-14x+10=-63\\ \Leftrightarrow4x^2-14x+73=0\\ \Leftrightarrow x\in\varnothing\)

  tìm x, y, z biết 

Giải

\(\dfrac{x}{y}=\dfrac{3}{4}\\ =>\dfrac{x}{3}=\dfrac{y}{4}\\ \dfrac{y}{z}=\dfrac{3}{5}\\ =>\dfrac{y}{3}=\dfrac{z}{5}\\\dfrac{x}{3}=\dfrac{y}{4}\\ =>\dfrac{x}{3.3}=\dfrac{y}{4.3}\\ =>\dfrac{x}{9}=\dfrac{y}{12}\\ \dfrac{y}{3}=\dfrac{z}{5}\\ =>\dfrac{y}{3.4}=\dfrac{z}{4.5}\\ =>\dfrac{y}{12}=\dfrac{z}{20}\\ =>\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)

Áp dụng tính chất dãy tỉ số = 

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}=\dfrac{2x-3y+z}{2.9-3.12+20}=\dfrac{6}{2}=3\\ =>\left\{{}\begin{matrix}x=3.9=27\\y=3.12=36\\z=3.20=60\end{matrix}\right.\)

  tìm x, y, z biết 

\(2x=3y\\ =>\dfrac{x}{3}=\dfrac{y}{2}\\ 4y=5z\\ =>\dfrac{y}{5}=\dfrac{z}{4}\\ \dfrac{x}{3}=\dfrac{y}{2}\\ =>\dfrac{x}{3.5}=\dfrac{y}{2.5}\\ =>\dfrac{x}{15}=\dfrac{y}{10}\\ \dfrac{y}{5}=\dfrac{z}{4}\\ =>\dfrac{y}{5.2}=\dfrac{z}{4.2}\\ =>\dfrac{y}{10}=\dfrac{z}{8}\\ =>\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}=\dfrac{x+y+z}{15+10+8}=\dfrac{11}{33}=\dfrac{1}{3}\\ =>\left\{{}\begin{matrix}x=\dfrac{1}{3}.15=5\\y=\dfrac{1}{3}.10=\dfrac{10}{3}\\z=\dfrac{1}{3}.8=\dfrac{8}{3}\end{matrix}\right.\)

  tìm x, y, z biết

\(a,\dfrac{3^{10}.11+9^5.5}{27^3.2^4}.x=-9\\ =>\dfrac{3^{10}.11+\left(3^2\right)^5.5}{\left(3^3\right)^3.2^4}.x=-9\\ =>\dfrac{3^{10}.\left(11+5\right)}{3^9.2^4}.x=-9\\ =>\dfrac{3^{10}.16}{3^9.2^4}.x=-9\\ =>\dfrac{3^{10}.2^4}{3^9.2^4}.x=-9\\ =>3^1.x=-9\\ =>x=-9:3\\ =>x=-3\)