\(5,4x^2-36=0\\ \Leftrightarrow\left(2x\right)^2-6^2=0\\ \Leftrightarrow\left(2x-6\right)\left(2x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Vậy \(S=\left\{3;-3\right\}\)

\(7,\left(3x+1\right)^2-16=0\\ \Leftrightarrow\left(3x+1\right)^2-4^2=0\\ \Leftrightarrow\left(3x+1-4\right)\left(3x+1+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-3=0\\3x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)

Vậy \(S=\left\{1;-\dfrac{5}{3}\right\}\)

\(8,\left(2x-3\right)^2-49=0\\ \Leftrightarrow\left(2x-3\right)^2-7^2=0\\ \Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-10=0\\2x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Vậy \(S=\left\{-2;5\right\}\)

\(-1,3.\left(0,2-2,1\right)+2,1.\)\(\left(-1,3+0,2\right)\) 

\(=-1,3.\left(-1,9\right)+2,1.\left(-1,1\right)\)

\(=2,47-2,31\)

\(=0,16\)

-1,3.( 0,2 - 2,1 ) +2,1.( -1,3 + 0,2)

= -1,3.( -1,9 ) + 2,1.( -1.1 )

= 2,47 + ( -2,31)

=0,16

C= \(3|x-2|+|3x+1|\)

Vì \(|x-2|\ge0\Rightarrow3|x-2|\ge0\)với mọi x  mà \(|3x+1|\)\(\ge\)0 nên C= \(3|x-2|+|3x+1|\ge0\)với mọi x

  suy ra C \(\ge0\) với mọi x

    Dấu "=" khi : \(|x-2|+|3x+1|\) = 0 => \(\hept{\begin{cases}x=-2\\x=1\end{cases}}\)

  Gtnn của C là 0 khi x= -2 và x=1

Chết mình nhầm

\(x^4+x^3+3x^2+2x+2=0\)

\(\Leftrightarrow x^4-x^3+x^2+2x^2-2x+2\)

\(\Leftrightarrow x^2\left(x^2-x+1\right)+2\left(x^2-x+1\right)\)

\(\Leftrightarrow\left(x^2+2\right)\left(x^2-x+1\right)\)

\(\Leftrightarrow\left(x^2+2\right)\left(x^2-x+\frac{1}{4}+\frac{3}{4}\right)\)

\(\Leftrightarrow\left(x^2+2\right)\left[x-\frac{1}{2}^2\right]+\frac{3}{4}\)

Ta co: \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>\frac{3}{4}\)

\(\left(x^2+2\right)\left[\left(x-\frac{1}{2}^2\right)+\frac{3}{4}\right]\ge\frac{3}{2}\le0\)

\(\left(\frac{-3}{2x}+\frac{1}{4}\right):\frac{1}{6}=\frac{-3}{2}\)

\(\Rightarrow\frac{-3}{2x}+\frac{1}{4}=\frac{-3}{2}.\frac{1}{6}=-\frac{1}{4}\)

\(\Rightarrow\frac{-3}{2x}=-\frac{1}{4}-\frac{1}{4}=-\frac{1}{2}\)

\(\Rightarrow\frac{-3}{2x}=\frac{-3}{6}\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=3\)

(-3/2x + 0,25) : 1/6 = -3/2

=> -3/2x + 0,25 = -3/2 . 1/6

=> -3/2x + 0,25 = -1/4

=> -3/2x = -1/4 - 0,25

=> -3,2x = -1/2

=> x = -1/2 : (-3/2)

=> x = 1/3

(-3/2x +0,25) : 1/6 = -3/2

    -3/2x  + 0,25 = -3/2 x  1/6

    -3/2x +0,25= -1/4

     -3/2x = -1/4-o,25

     -3/2x=-1/2

         x= -1/2 :(-3/2) 

        x=1/3

Mk k hơn 3 tk lun đó ! Mk còn là bn của bn ấy !

Mình k bạn ầy ở nick phụ của mình rồi, nick này mình mới lập! K mik nha

minhf kick rùi đó

các bn giúp mk với mk đang cần gấp lắm ! Thanks các bn nhìu

(2,5x - 4/3)² - 5/6 = -7/18

=> (2,5x - 4/3)² = -7/18 + 5/6

=> (2,5x - 4/3)² = 4/9

=> (2,5x - 4/3)² = (2/3)²

=> \(\orbr{\begin{cases}2,5x-\frac{4}{3}=\frac{2}{3}\\2,5x-\frac{4}{3}=-\frac{2}{3}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{4}{5}\\x=\frac{4}{15}\end{cases}}\)

\(\left(2,5x-\frac{4}{3}\right)^2-\frac{5}{6}=-\frac{7}{18}\)

\(\left(2,5x-\frac{4}{3}\right)^2=-\frac{7}{18}+\frac{5}{6}\)

\(\left(2,5x-\frac{4}{3}\right)^2=\frac{4}{9}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{4}{5}\\x=\frac{4}{15}\end{cases}}\)

em ko giai duoc ,em lop 2

-3/2 - | 3x + 4 |.2/5= 2

<=> | 3x + 4 | = (2+3/2)/(2/5)

<=>  | 3x + 4 | = 35/4

<=>3x + 4 =35/4

Hoặc 3x + 4 = - 35/4

Tự giải nốt nha

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

\(|3x+4|.\frac{2}{5}=\frac{-7}{2}\)

\(|3x+4|=\frac{-7}{2}:\frac{2}{5}\)

\(|3x+4|=\frac{-35}{4}\)

\(\Leftrightarrow3x+4=\frac{-35}{4}\)        hoặc          \(3x+4=\frac{35}{4}\)

         x=\(\frac{-17}{4}\)                                                x=\(\frac{19}{12}\)

Vậy x=\(\frac{-17}{4}\) hoặc    x= \(\frac{19}{12}\)