\(\dfrac{9}{7}x+\dfrac{5}{7}x=\dfrac{2}{3}\)

\(\Leftrightarrow x\left(\dfrac{9}{7}+\dfrac{5}{7}\right)=\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{14}{7}x=\dfrac{2}{3}\)

\(\Leftrightarrow2x=\dfrac{2}{3}\)

\(\Leftrightarrow x=\dfrac{2}{3}:2\)

\(\Leftrightarrow x=\dfrac{2}{3}\times\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{2}{6}\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

Ta có : \(\dfrac{x}{2}=\dfrac{1-x}{3}\)

\(\Leftrightarrow3x=2\left(1-x\right)\)

\(\Leftrightarrow3x=2-2x\)

\(\Leftrightarrow5x=2\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy ...

Ta có: \(\dfrac{x}{2}=\dfrac{1-x}{3}\)

\(\Leftrightarrow3x=2\left(1-x\right)\)

\(\Leftrightarrow3x=2-2x\)

\(\Leftrightarrow3x+2x=2\)

\(\Leftrightarrow5x=2\)

hay \(x=\dfrac{2}{5}\)

Vậy: \(x=\dfrac{2}{5}\)

= 3/4 + 1/3 = 13/12

5/4 x X = 5/8

X = 5/8 : 5/4

X = 1/2

vậy X = ...

\(\dfrac{3}{5}:\dfrac{4}{5}+\dfrac{1}{2}\times\dfrac{2}{3}=\dfrac{3}{4}+\dfrac{1}{3}+\dfrac{9}{12}+\dfrac{4}{12}=\dfrac{11}{12}\)

\(\dfrac{5}{4}\times x=\dfrac{5}{8}\)

\(x=\dfrac{5}{8}:\dfrac{5}{4}\)

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

Không có dấu "=" nên đây không được gọi là phương trình bạn nhé. Bạn cần xem lại đề.

Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{m}{2}\ne\dfrac{2}{-4}=-\dfrac{1}{2}\)

=>\(m\ne-1\)

\(\left\{{}\begin{matrix}mx+2y=1\\2x-4y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2mx+4y=2\\2x-4y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(2m+2\right)=5\\2x-4y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{2m+2}\\4y=2x-3=\dfrac{10}{2m+2}-3=\dfrac{10-6m-6}{2m+2}=\dfrac{-6m+4}{2m+2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{2m+2}\\y=\dfrac{-6m+4}{8m+8}=\dfrac{-3m+2}{4m+4}\end{matrix}\right.\)

x-3y=7/2

=>\(\dfrac{5}{2m+2}-\dfrac{3\cdot\left(-3m+2\right)}{4m+4}=\dfrac{7}{2}\)

=>\(\dfrac{10+3\left(3m-2\right)}{4m+4}=\dfrac{7}{2}\)

=>\(\dfrac{10+9m-6}{4m+4}=\dfrac{7}{2}\)

=>\(\dfrac{9m+4}{4m+4}=\dfrac{7}{2}\)

=>7(4m+4)=2(9m+4)

=>28m+28=18m+8

=>10m=-20

=>m=-2(nhận)

\(\Leftrightarrow2.5:x=2+\dfrac{2}{3}-2-\dfrac{1}{5}=\dfrac{7}{15}\)

hay \(x=\dfrac{5}{2}:\dfrac{7}{15}=\dfrac{5}{2}\cdot\dfrac{15}{7}=\dfrac{75}{14}\)

a: A=x^2-2x+1+4

=(x-1)^2+4>=4

Dấu = xảy ra khi x=1

b: =x^2-x+1/4+3/4

=(x-1/2)^2+3/4>=3/4

Dấu = xảy ra khi x=1/2

c: =2x+8-x^2-4x

=-x^2-2x+8

=-x^2-2x-1+9

=-(x^2+2x+1)+9

=-(x+1)^2+9<=9

Dấu = xảy ra khi x=-1

d: =x^2-2xy+y^2+4y^2+4y+1+2

=(x-y)^2+(2y+1)^2+2>=2

Dấu = xảy ra khi x=y và 2y+1=0

=>x=y=-1/2

\(\)đặt \(2x^2+y^2+\dfrac{28}{x}+\dfrac{1}{y}=A\)

\(=>A=2x^2+y^2-7x-y+\dfrac{28}{x}+7x+\dfrac{1}{y}+y\)

\(A=2x^2-8x+8+y^2-2y+1+x+y-9+\dfrac{28}{x}+7x+\dfrac{1}{y}+y\)

\(A=2\left(x-2\right)^2+\left(y-1\right)^2+\left(x+y\right)-9+\dfrac{28}{x}+7x+\dfrac{1}{y}+y\)

áp dụng BDT AM-GM\(=>\dfrac{28}{x}+7x+\dfrac{1}{y}+y\ge2\sqrt{28.7}+2\sqrt{1}=30\)

\(=>A\ge30+3-9=24\)

dấu"=" xảy ra<=>x=2,y=1