\(2.|5x-3|-x=\)\(7\)

\(\Leftrightarrow\) \(2.|5x-3|=7+x\)

\(\Leftrightarrow\) \(|5x-3|=\frac{7+x}{2}\)

\(\Leftrightarrow\)\(5x-3=\pm\left(\frac{7+x}{2}\right)\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}5x-3=\frac{7+x}{2}\\5x-3=\frac{-7-x}{2}\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}\frac{2.\left(5x-3\right)}{2}=\frac{7+x}{2}\\\frac{2.\left(5x-3\right)}{2}=\frac{-7-x}{2}\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}\frac{10x-6}{2}=\frac{7+x}{2}\\\frac{10x-6}{2}=\frac{-7-x}{2}\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}10x-6=7+x\\10x-6=-7-x\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}10x-x=7+6\\10x+x=-7+6\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}9x=13\\11x=-1\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{13}{9}\\x=\frac{-1}{11}\end{cases}}\)

Vậy x = \(\frac{13}{9}\) , x = \(\frac{-1}{11}\)

\(a,\frac{x-1}{12}=\frac{75}{x-1}\)

\(\Rightarrow\left(x-1\right)\left(x-1\right)=12\cdot75\)

\(\Rightarrow\left(x-1\right)^2=900\)

\(\Rightarrow x-1=\pm30\)

\(\Rightarrow\orbr{\begin{cases}x=31\\x=-29\end{cases}}\)

\(b,\left|12-|4x+1|\right|=5\)

\(\Rightarrow\orbr{\begin{cases}12-\left|4x+1\right|=5\\12-\left|4x+1\right|=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\left|4x+1\right|=7\\\left|4x+1\right|=17\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}4x+1=7\text{ hoặc}4x+1=-7\\4x+1=17\text{ hoặc}4x+1=-17\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}or\text{ }x=-2\\x=4or\text{ }x=-\frac{9}{2}\end{cases}}\)

x - 1/12 = 75/x - 1

<=> (x - 1).(x - 1) = 12.75

<=> x² - 2x + 1 = 900

<=> x² - 2x + 1 - 900 = 0

<=> x² - 2x + 1 - 899 = 0

<=> (x + 29)(x - 31) = 0

        x + 29 = 0 hoặc x - 31 = 0

        x = 0 - 29          x = 0 + 31

        x = -29              x = 31

=> x = -29 hoặc x = 31

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)

\(\Rightarrow\frac{x+z}{2+4}=\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\) mà x + z = 18

\(\Rightarrow\frac{18}{6}=\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)

\(\Rightarrow3=\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)

\(\Rightarrow\hept{\begin{cases}x=3\cdot2=6\\y=3\cdot3=9\\z=3\cdot4=12\end{cases}}\)

#)Giải :

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{x}{2}=\frac{z}{4}=\frac{x+z}{2+4}=\frac{18}{6}=3\)

\(\hept{\begin{cases}\frac{x}{2}=3\\\frac{y}{3}=3\\\frac{z}{4}=3\end{cases}\Rightarrow\hept{\begin{cases}x=6\\y=9\\z=12\end{cases}}}\)

Vậy x = 6; y = 9; z = 12

#)Giải :

a)Theo đầu bài, ta có : \(n=a^2+b^2\)

\(\Rightarrow2n=2a^2+2b^2\Rightarrow2n=a^2+2ab+b^2+a^2-2ab+b^2=\left(a+b\right)^2+\left(a-b\right)^2\)

\(\Rightarrowđpcm\)

b)Theo đầu bài, ta có : \(2n=a^2+b^2\)

\(\Rightarrow n=\frac{a^2}{2}+\frac{b^2}{2}\Rightarrow\left(\frac{a^2}{4}+2.\frac{a}{2}.\frac{b}{2}+\frac{b^2}{4}\right)+\left(\frac{a^2}{4}+2.\frac{a}{2}.\frac{b}{2}+\frac{b^2}{4}\right)=\frac{\left(a+b\right)^2}{2}+\frac{\left(a-b\right)^2}{2}\)

\(\Rightarrowđpcm\)

\(=>\left(m-2\right)^3=-3=>m-2=-\sqrt[3]{3}\)

\(=>m=-\sqrt[3]{3}+2\)

bạn có thể giải đơn giản đc ko mk lớp 8 à

\(A=\frac{5x+4}{3x-1}>0\)

\(\Leftrightarrow\hept{\begin{cases}5x+4>0\\3x-1>0\end{cases}}\) hoặc \(\hept{\begin{cases}5x+4< 0\\3x-1< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}5x>-4\\3x>1\end{cases}}\) hoặc \(\hept{\begin{cases}5x< -4\\3x< 1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x>-\frac{4}{5}\\x>\frac{1}{3}\end{cases}}\) hoặc \(\hept{\begin{cases}x< -\frac{4}{5}\\x< \frac{1}{3}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x>\frac{1}{3}\\x< -\frac{4}{5}\end{cases}}\)

b, tương tự nhưng xét trái dấu

Để mình giải câu b) cho \(A< 0\Leftrightarrow\frac{5x+4}{3x-1}< 0\)

\(\Leftrightarrow\hept{\begin{cases}5x+4>0\\3x-1< 0\end{cases}}\)hoặc \(\hept{\begin{cases}5x+4< 0\\3x-1>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}5x>-4\\3x< 1\end{cases}}\)hoặc \(\hept{\begin{cases}5x< -4\\3x>1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x>\frac{-4}{5}\\x< \frac{1}{3}\end{cases}}\left(TM\right)\)hoặc \(\Leftrightarrow\hept{\begin{cases}x< \frac{-4}{5}\\x>\frac{1}{3}\end{cases}}\left(L\right)\)

Vậy \(A< 0\Leftrightarrow\frac{-4}{5}< x< \frac{1}{3}\)

#)Giải :

a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)

b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)