\(3^{x+4}=9^{2x-1}\)

\(\Rightarrow3^{x+4}=3^{4x-2}\)

\(\Rightarrow x+4=4x-2\)

\(\Rightarrow3x=6\Rightarrow x=2\)

3x+4=92x−13x+4=92x−1

⇒3x+4=34x−2⇒3x+4=34x−2

Bài làm:

a) \(\left|\frac{1}{2}x-\frac{5}{2}\right|-1=-\frac{1}{2}\)

\(\Leftrightarrow\left|\frac{1}{2}x-\frac{5}{2}\right|=\frac{1}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{5}{2}=\frac{1}{2}\\\frac{1}{2}x-\frac{5}{2}=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

+ Nếu x = 6

\(\left|12-\frac{1}{3}y\right|=\frac{5}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}12-\frac{1}{3}y=\frac{5}{6}\\12-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{67}{6}\\\frac{1}{3}y=\frac{77}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{67}{2}\\y=\frac{77}{2}\end{cases}}\)

+ Nếu x = 4

\(\left|8-\frac{1}{3}y\right|=\frac{5}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}8-\frac{1}{3}y=\frac{5}{6}\\8-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{43}{6}\\\frac{1}{3}y=\frac{53}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{43}{2}\\y=\frac{53}{2}\end{cases}}\)

Vậy ta có 4 cặp số (x;y) thỏa mãn: \(\left(6;\frac{67}{2}\right);\left(6;\frac{77}{2}\right);\left(4;\frac{43}{2}\right);\left(4;\frac{53}{2}\right)\)

b) \(\frac{3}{2}x-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{5}{3}\)

\(\Leftrightarrow\frac{3}{2}x-\frac{1}{2}x+\frac{1}{3}=\frac{5}{3}\)

\(\Leftrightarrow x=\frac{4}{3}\)

Thay vào ta được:

\(\frac{2.\frac{4}{3}+y}{\frac{4}{3}-2y}=\frac{5}{4}\)

\(\Leftrightarrow\frac{32}{3}+4y=\frac{20}{3}-10y\)

\(\Leftrightarrow14y=-4\)

\(\Rightarrow y=-\frac{2}{7}\)

Vậy ta có 1 cặp số (x;y) thỏa mãn: \(\left(\frac{4}{3};-\frac{2}{7}\right)\)

áp dụng các hằng đẳng thức thôi mà :)

a)\(x^2-2x+1=25\)

=>\(\left(x-1\right)^2=25\)

=>\(\orbr{\begin{cases}x-1=-5\\x-1=5\end{cases}}\)

b)\(3\left(x-1\right)^2-3x\left(x-5\right)=1\)

=>\(3\left[\left(x-1\right)^2-x\left(x-5\right)\right]=1\)

=>\(3\left(x^2-2x+1-x^2+5x\right)=1\)

=>\(3\left(3x+1\right)=1\)

=>\(3x+1=\frac{1}{3}\)

=>\(3x=\frac{-2}{3}\)

=>\(x=\frac{-2}{9}\)

c)\(\left(5-2x\right)^2-16=0\)

=>\(\left(5-2x\right)^2-4^2=0\)

=>\(\left(5-2x-4\right)\left(5-2x+4\right)=0\)

=>\(\orbr{\begin{cases}5-2x-4=0\\5-2x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{9}{2}\end{cases}}}\)

\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)

\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)

\(\Leftrightarrow-3x=-\frac{13}{4}\)

\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)

\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)

\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)

\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)

\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)

\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)

\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)

\(\Leftrightarrow x=-\frac{6}{11}\)

d,e,f Tương tự

a; \(\dfrac{1}{2}x\) + \(\dfrac{5}{2}\) = \(\dfrac{7}{2}\)\(x\) - \(\dfrac{3}{4}\)

     \(\dfrac{3}{4}\) + \(\dfrac{5}{2}\) = \(\dfrac{7}{2}\)\(x\) - \(\dfrac{1}{2}\)\(x\) 

    \(\dfrac{3}{4}\) + \(\dfrac{10}{4}\) = 3\(x\)

    3\(x\)          = \(\dfrac{13}{4}\)

     \(x\)          = \(\dfrac{13}{4}\) : 3

     \(x=\dfrac{13}{12}\) 

Vậy \(x=\dfrac{13}{12}\)

2)

(x:3-4)*5=15

=>x=15:5+4*3=3+7=10

a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)

\(\left(\frac{4}{13}.\frac{8}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)

\(\frac{32}{65}.\left(2x+1\right)^2=\frac{10}{13}\)

\(\left(2x+1\right)^2=\frac{10}{13}\div\frac{32}{65}\)

\(\left(2x+1\right)^2=\frac{25}{16}\)

\(\Rightarrow2x+1\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)

\(\hept{\begin{cases}2x+1=\frac{5}{4}\\2x+1=-\frac{5}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{9}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{8}\\x=-\frac{9}{8}\end{cases}}}\)

Vậy \(x\in\left\{\frac{1}{8};-\frac{9}{8}\right\}\)

\(x^3-\frac{9}{16}.x=0\)

\(x\left(x^2-\frac{9}{16}\right)=0\)

\(\hept{\begin{cases}x=0\\x^2-\frac{9}{16}=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=\frac{9}{16}\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\pm\frac{3}{4}\end{cases}}}\)

Vậy \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)

a)

\(\frac{x}{y+z+1}=\frac{y}{x+z+1}=\frac{z}{x+y+1}=\frac{x+y+z}{2\left(x+y+z\right)+3}=x+y+z\)

=> 2(x+y+z) +3 =1=> x+y+z=-1

Luôn đùng Vói mọi x;y;z khác o  sao cho x+y+z = -1

b)\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}=\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{3}+\frac{5}{4}}=\frac{49}{\frac{49}{12}}=12\)

x= 3/2 .12=18

y= 4/3 .12=16

z=5/4 .12=15

\(\left(x-3\right)\left(x-12\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=12\end{cases}}\)

\(\Rightarrow x\in\left\{3;12\right\}\)

\(\left(x^2-81\right)\left(x^2+9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2-81=0\\x^2+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x\in\varnothing\end{cases}}\Leftrightarrow x=9\)

\(\Rightarrow x=9\)

\(\left(x-4\right)\left(x+2\right)< 0\)

\(\Rightarrow\hept{\begin{cases}x-4\\x+2\end{cases}}\)trái dấu

\(TH1:\hept{\begin{cases}x-4>0\\x+2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x< -2\end{cases}}\Leftrightarrow x\in\varnothing\)

\(TH2:\hept{\begin{cases}x-4< 0\\x+2>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 4\\x>-2\end{cases}}\Leftrightarrow x\in\left\{-1;0;1;2;3\right\}\)

Vậy \(x\in\left\{-1;0;1;2;3\right\}\)