\(x\left(2x-1\right)\left(3x-126\right)=0\Rightarrow\hept{\begin{cases}x=0\\2x-1=0\\3x-126=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=0\\2x=1\\3x=126\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\frac{1}{2}\\x=42\end{cases}}\)

(X - 1005) x (1 + 3 + 5+  7 + 9 + 11) = 1989 x 1990 x (70 - 35 x 2)

(X - 1005) x (11 + 1)[(11 - 1) : 2 + 1): 2] = 1989 x 1990 x 0

(X - 1005) x 36 = 0

X - 1005 = 0 : 36

X - 1005 = 0

X = 1005

( x - 1005 ) . ( 1 + 3 + 5 + 7 + 9 + 11 ) = 1989 . 1990 . ( 70 - 35 . 2 )

( x - 1005 ) . 36 = 1989 . 1990 . ( 70 - 70 )

( x - 1005 ) . 36 = 1989 . 1990 . 0

( x - 1005 ) . 36 = 0

x - 1005 = 0

x = 1005

(x - 1005) x (1+ 3 + 5 + 7 + 9 + 11) =  1989 x 1990 (70 - 35 x 2)

(x - 1005) x 36 = 1989 x 1990 x 0

(x - 1005) x 36 = 0

X -1005 = 0 : 36

X - 1005 = 0

X = 0

Hok tốt!

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147 -( x x 36) = 29

X x 36 = 147 - 29

X x 36 = 118

X = 118 : 36

X = ...

Chúc bạn hok tốt

oki 

147- ( X x 36 ) =29

X x 36=147 - 29

X x 36 =118

X        =118 : 36

X       = .....

bn ơi, 118:36 bằng mấy vj

\(\left(x+2\right)\left(\dfrac{360}{x}-6\right)=360\)

\(ĐK:x\ne0\)

\(\Leftrightarrow\left(x+2\right)\left(\dfrac{360-6x}{x}\right)=360\)

\(\Leftrightarrow360-6x+\dfrac{720-12x}{x}=360\)

\(\Leftrightarrow360x-6x^2+720-12x=360x\)

\(\Leftrightarrow6x^2+12x-720=0\)

\(\Delta=12^2-4.6.\left(-720\right)\)

    \(=17424>0\)

`->` pt có 2 nghiệm

\(\left\{{}\begin{matrix}x_1=\dfrac{-12-\sqrt{17424}}{12}=-12\\x_2=\dfrac{-12+\sqrt{17424}}{12}=10\end{matrix}\right.\) ( tm )

Vậy \(S=\left\{-12;10\right\}\)

a)\(MinA=1\Leftrightarrow x=\dfrac{1}{4}\)

b)\(MaxP=1\Leftrightarrow x=\dfrac{1}{4}\)

a) \(A=x-\sqrt{x}+\dfrac{5}{4}=\left(x-\sqrt{x}+\dfrac{1}{4}\right)+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+1\ge1\)

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

b) \(P=\dfrac{1}{x-\sqrt{x}+\dfrac{5}{4}}=\dfrac{1}{\left(x-\sqrt{x}+\dfrac{1}{4}\right)+1}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+1}\le\dfrac{1}{1}=1\)

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

a, \(A=x-\sqrt{x}+\dfrac{5}{4}\)

\(=x-\sqrt{x}+\dfrac{1}{4}+1\)

\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+1\ge1\)

\(\Rightarrow minA=1\Leftrightarrow x=\dfrac{1}{4}\)

`48/[x+4]+48/[x-4]=5`           `ĐK: x \ne +-4`

`<=>[48(x-4)+48(x+4)]/[(x-4)(x+4)]=[5(x+4)(x-4)]/[(x-4)(x+4)]`

   `=>48x-192+48x+192=5x^2-80`

`<=>5x^2-96x-80=0`

`<=>5x^2-100+4x-80=0`

`<=>5x(x-20)+4(x-20)=0`

`<=>(x-20)(5x+4)=0`

`<=>` $\left[\begin{matrix} x=20\\ x=\dfrac{-4}{5}\end{matrix}\right.$   (t/m)

Vậy `S={-4/5;20}`

ĐK : \(x\ne\pm4\)

\(\Leftrightarrow\cdot\dfrac{48\left(x+4\right)+48\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}=\dfrac{5\left(x+4\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\)

\(\Leftrightarrow48x+192+48x-192==5x^2-80\)

\(\Leftrightarrow96x=5x^2-80\)

\(\Leftrightarrow5x^2-96x-80=0\)

\(\Leftrightarrow5x^2+4x-100-80=0\)

\(\Leftrightarrow4\left(x-20\right)+5x\left(x-20\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-20=0\\5x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-\dfrac{4}{5}\end{matrix}\right.\)