Ko cần đâu bn à mk mong bn đấy

a)\(\left(3x-1\right)\left(5-\frac{1}{2}x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)

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

    \(2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)

    \(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{29}{12}\\x=-\frac{13}{12}\end{cases}}\)

a)\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\)3x - 1 = 0      hay      \(\frac{-1}{2}\)x + 5 = 0
\(\Leftrightarrow\)3x     = 1         I\(\Leftrightarrow\)\(\frac{-1}{2}\)x     = -5
\(\Leftrightarrow\)  x     = \(\frac{1}{3}\)  I\(\Leftrightarrow\)            x     = 10

b) 2 I \(\frac{1}{2}x-\frac{1}{3}\)I - \(\frac{3}{2}\)=\(\frac{1}{4}\)
\(\Leftrightarrow\) 2 I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{4}\)
\(\Leftrightarrow\)    I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{7}{8}\)          hay     \(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{-7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x\)           = \(\frac{29}{24}\)        I\(\Leftrightarrow\)\(\frac{1}{2}x\)           = \(\frac{-13}{24}\)
\(\Leftrightarrow\)      x              = \(\frac{29}{12}\)        I\(\Leftrightarrow\)      x              = \(\frac{-13}{12}\)

c) (2x +\(\frac{3}{5}\))² - \(\frac{9}{25}\)= 0
\(\Leftrightarrow\)(2x +\(\frac{3}{5}\))²       = \(\frac{9}{25}\)
\(\Leftrightarrow\) 2x +\(\frac{3}{5}\)         = \(\frac{3}{5}\)    hay      2x +\(\frac{3}{5}\)= \(\frac{-3}{5}\)
\(\Leftrightarrow\) 2x                    = 0           I \(\Leftrightarrow\)2x           = \(\frac{-6}{5}\)
\(\Leftrightarrow\)   x                    = 0           I \(\Leftrightarrow\) x           = \(\frac{-3}{5}\)

d) 3(x -\(\frac{1}{2}\)) - 5(x +\(\frac{3}{5}\)) = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)3x - \(\frac{3}{2}\)- 5x - 3 = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)-2x + x - \(\frac{9}{2}\)- \(\frac{1}{5}\)= 0
\(\Leftrightarrow\)-x = \(\frac{-47}{10}\)
\(\Leftrightarrow\) x = \(\frac{47}{10}\)

a: =>4/3x=7/9-4/9=1/3

=>x=1/4

b: =>5/2-x=9/14:(-4/7)=-9/8

=>x=5/2+9/8=29/8

c: =>3x+3/4=8/3

=>3x=23/12

hay x=23/36

d: =>-5/6-x=7/12-4/12=3/12=1/4

=>x=-5/6-1/4=-10/12-3/12=-13/12

em moi lop 4 mà

\(d,-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Leftrightarrow-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-4}{12}\)

\(\Leftrightarrow-\dfrac{5}{6}-x=\dfrac{1}{4}\)

\(\Leftrightarrow-x=-\dfrac{5}{6}-\dfrac{1}{4}\)

\(\Leftrightarrow-x=-\dfrac{13}{12}\)

\(\Leftrightarrow x=\dfrac{13}{12}\)

\(c,3x+\dfrac{3}{4}=2\dfrac{2}{3}\)

\(\Leftrightarrow3x+\dfrac{3}{4}=\dfrac{8}{3}\)

\(\Leftrightarrow3x=\dfrac{8}{3}-\dfrac{3}{4}\)

\(\Leftrightarrow3x=\dfrac{23}{12}\)

\(\Leftrightarrow x=\dfrac{23}{12}:3\)

\(\Leftrightarrow x=\dfrac{23}{36}\)

f(x)=9x³-1/3x+3x²-3x+1/3x²-1/9x³-3x²-9x+27+3x

    = 9x³-1/9x³+3x²+1/3x²-3x²-1/3-3x-9x+3x+27

   = 80/9x³+1/3x²-28/3x+27

\(A=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\cdot...\cdot\left(1-\dfrac{1}{9801}\right)\)

\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\left(1-\dfrac{1}{99}\right)\left(1+\dfrac{1}{99}\right)\)

\(=\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{98}{99}\right)\cdot\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{100}{99}\right)\)

\(=\dfrac{1}{99}\cdot\dfrac{100}{2}=\dfrac{50}{99}\)

`a)25/(x+1)-1 1/6=-1/3-0,5`

`=>25/(x+1)=-1/3-1/2+1+1/6`

`=>25/(x+1)=1/3`

`=>75=x+1`

`=>x=74`

Vậy `x=74`

`b)(2x+25 3/5)^2-9/25=0`

`=>(2x+128/5)=9/25`

`**2x+128/5=3/5`

`=>2x=-125/5=-25`

`=>x=-25/2`

`**2x+128/5=-3/5`

`=>2x=-131/5`

`=>x=-131/10`

Giải:

a) \(\dfrac{25}{x+1}-1\dfrac{1}{6}=\dfrac{-1}{3}-0,5\) 

              \(\dfrac{25}{x+1}=\dfrac{-5}{6}+\dfrac{7}{6}\) 

              \(\dfrac{25}{x+1}=\dfrac{1}{3}\) 

\(\Rightarrow1.\left(x+1\right)=25.3\)  

\(\Rightarrow x+1=75\) 

\(\Rightarrow x=75-1\) 

\(\Rightarrow x=74\) 

b) \(\left(2x+25\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\) 

              \(\left(2x+\dfrac{128}{5}\right)^2=0+\dfrac{9}{25}\) 

             \(\left(2x+\dfrac{128}{5}\right)^2=\dfrac{9}{25}\) 

\(\Rightarrow\left[{}\begin{matrix}\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\\\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{128}{5}=\dfrac{3}{5}\\2x+\dfrac{128}{5}=\dfrac{-3}{5}\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-25}{2}\\x=\dfrac{-131}{10}\end{matrix}\right.\) 

Chúc bạn học tốt!

a) => 4/3x = 7/9 - 4/9 = 1/3

=> x = 1/3 : 4/3 = 1/4

b) => 5/2 - x = 9/14 : (-4/7) = -9/8

=> x = 5/2 - (-9/8) = 5/2 + 9/8 = 29/8

c) => 3x = 2 và 2/3 - 3/4 = 8/3 - 3/4 = 23/12

=> x = 23/12 : 3 = 23/36

D) => -5/6 - x = 1/4

=> x = -5/6 - 1/4 = -13/12

a) \(\dfrac{4}{9}+\dfrac{4}{3}x=\dfrac{7}{9}\)

\(\dfrac{4}{3}x=\dfrac{7}{9}-\dfrac{4}{9}=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}:\dfrac{4}{3}\)

\(x=\dfrac{1}{4}\)

b) \(\left(\dfrac{5}{2}-x\right)\left(-\dfrac{4}{7}\right)=\dfrac{9}{14}\)

\(\dfrac{5}{2}-x=\dfrac{9}{14}:\left(-\dfrac{4}{7}\right)=-\dfrac{9}{8}\)

\(x=\dfrac{5}{2}-\left(-\dfrac{9}{8}\right)\)

\(x=\dfrac{29}{8}\)

c) \(3x+\dfrac{3}{4}=2\dfrac{2}{3}\)

\(3x+\dfrac{3}{4}=\dfrac{8}{3}\)

\(3x=\dfrac{8}{3}-\dfrac{3}{4}=\dfrac{23}{12}\)

\(x=\dfrac{23}{12}:3\)

\(x=\dfrac{23}{36}\)

d) \(-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(-\dfrac{5}{6}-x=\dfrac{1}{4}\)

\(x=-\dfrac{5}{6}-\dfrac{1}{4}\)

\(x=-\dfrac{13}{12}\)

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

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

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

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

giúp mik với đi ạ mik thực sự đang cần gấp

1+2 . (25 + 9) - 4 MŨ 3 

= 1 + 2 . 34 - 64

= 1 + 68 - 64

= 69 - 64

= 5

Cảm ơn bạn nha ^^

ko có j