bài nâng cao à bạn !!!!!!!!!

\(pt\Leftrightarrow\left\{{}\begin{matrix}62x+12y=\frac{14.6}{a}-59\\6x-4y=\frac{0.2}{a}-3\end{matrix}\right.\)

Giải pt thu được \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{19}{50a}-\frac{17}{10}=\frac{38-170a}{100a}\\y=\frac{47}{100a}-\frac{21}{20}=\frac{47-105a}{100a}\end{matrix}\right.\)

\(\Rightarrow\frac{x}{y}=\frac{38-170a}{47-105a}\)

Bài 3: 

a: \(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\\dfrac{3}{4}x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

b: \(\Leftrightarrow\left\{{}\begin{matrix}3x+2>0\\\dfrac{2}{3}x-5< 0\end{matrix}\right.\Leftrightarrow-\dfrac{2}{3}< x< \dfrac{15}{2}\)

c: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x+2=0\\\dfrac{2}{5}x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{3}{4}=-2\\\dfrac{2}{5}x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{8}{3}\\x=6:\dfrac{2}{5}=15\end{matrix}\right.\)

|x-1/3|+4/5=14/5

|x-1/3|=2

=>x-1/3=2 hoặc x-1/3=-2

=>x=7/3 hoặc x=-5/3

vậy x=7/3 hoặc x=-5/3

tk mk nha

-1,6666.....

mai bn đăng lại nhé mk lm cho h đi ngủ

x thuộc Z

\(\left(x-2\right):\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)

\(\left(x-2\right):\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\left(x-2\right):\frac{2}{9}=\frac{16}{9}\)

\(x-2=\frac{32}{91}\)

\(x=\frac{32}{91}+2\)

\(x=\frac{212}{91}\)

1) \(x^2+\frac{1}{x^2}+16y^2+\frac{1}{y^2}=10\)

\(\Leftrightarrow\left(x^2+2\cdot x\cdot\frac{1}{x}+\frac{1}{x^2}\right)+\left(16y^2+2\cdot4y\cdot\frac{1}{y}+\frac{1}{y^2}\right)=0\)

\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2+\left(4y+\frac{1}{y}\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x+\frac{1}{x}=0\\4y+\frac{1}{y}=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x^2+1=0\\4y^2+1=0\end{cases}}\) ( vô lí )

Phương trình vô nghiệm

Câu 1 giống bạn kia:

Câu 2:Sửa đề nhé, tại thấy a,b thuộc N

\(M=\frac{b}{7\left(a+b\right)}\) ( đkxđ:\(a\ne-b\))

\(\Rightarrow\frac{1}{M}=\frac{7a}{b}+7\ge7\)\(\)(Vì \(a,b\in N\Rightarrow a,b\ge0\))

\(\Rightarrow M\le7\)

\(\Rightarrow M\)đạt GTLN là 7 khi \(\text{a=0}\) và  \(b\ne0\)

\(\left|x+\frac{1}{3}\right|+\frac{4}{5}=\left|-3,2+\frac{2}{5}\right|+\left(27-\frac{3}{5}\right)\left(27-\frac{3^2}{6}\right)...\left(27-\frac{3^5}{9}\right)...\left(27-\frac{3^{2010}}{2014}\right)\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}+\left(27-\frac{3^2}{6}\right)\left(27-\frac{3^3}{7}\right)...\left(27-27\right)...\left(27-\frac{3^{2010}}{2014}\right)\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|=2\)

\(\Rightarrow\hept{\begin{cases}x+\frac{1}{3}=2\\x+\frac{1}{3}=-2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=-\frac{7}{3}\end{cases}}}\)

bạn ơi, có một chỗ chưa chuẩn .bạn kiểm tra lại giú mình. chỗ vế trái bạn thiếu \(\left(27-\frac{3}{5}\right)\). bạn bổ sung vào cho đúng nhé. dù sao vẫn cảm ơn bạn.

\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)

\(\Rightarrow\left|x-\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)

\(\Rightarrow\left|x-\frac{1}{3}\right|=2\)

\(\Rightarrow x-\frac{1}{3}=\hept{\begin{cases}2\\-2\end{cases}}\)

\(\Rightarrow x=\hept{\begin{cases}\frac{7}{3}\\\frac{-5}{3}\end{cases}}\)