* ĐỀ như dưới phải ko ??? Ai lại ghi x2 bao giờ ??

a) (x + 8)(x + 6) - x² =104

\(\Leftrightarrow\) x² + 6x + 8x + 48 - x² = 104

\(\Leftrightarrow\) 14x + 48 = 104

\(\Leftrightarrow\) 14x = 56

\(\Leftrightarrow\) x = 4

a: \(13329:x=3\)

=>\(x=\dfrac{13329}{3}=4443\)

b: x:8=5678

=>\(x=5678\cdot8=45424\)

\(a.\)

\(\left(x-8\right)\left(x^3+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

\(S=\left\{8,-2\right\}\)

\(b.\)

\(\left(4x-3\right)-\left(x+5\right)=3\cdot\left(10-x\right)\)

\(\Leftrightarrow4x-3-x-5-30+3x=0\)

\(\Leftrightarrow6x-38=0\)

\(\Leftrightarrow x=\dfrac{38}{6}\)

\(S=\left\{\dfrac{38}{6}\right\}\)

a) \(\left(x-8\right)\left(x^3+8\right)=0\)

=>\(x-8=0 => x=8\)

hoặc \(x^3+8=0\)=>\(x=-2\)

b) \(\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\)

\(< =>3x-8=3\left(10-x\right)\)

\(< =>3x-8-30+3x=0\)

\(< =>6x=38=>x=\dfrac{38}{6}=\dfrac{19}{3}\)

a) (x - 8 )( x³ + 8) = 0

\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x^3=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

b)(4x - 3) – ( x + 5) = 3(10 - x)

\(\Leftrightarrow4x-3-x-5=30-3x\)

\(\Leftrightarrow3x-8=30-3x\)

\(\Leftrightarrow3x-8-30+3x=0\)

\(\Leftrightarrow6x-38=0\)

\(\Leftrightarrow x=\dfrac{19}{3}\)

Sửa lại câu `b) :` 

`a)`

`( x-8 )( x^3 + 8 )`

`=> x-8=0` hoặc `x^3+8=0`

`=> x=8` hoặc `x^3 = -8=(-2)^3`

`=> x=8` hoặc `x=-2`

Vậy `x in { -2;8}`

`b)`

`( 4x-3 ) - ( x+5) = 3( 10-x)`

`=> 4x-3-x-5=30-3x`

`=> ( 4x-x)+(-3-5)=30-3x`

`=> 3x-8=30-3x`

`=> 6x=38`

`=> x=19/3`

Vậy `x=19/3` 

`a)`

`( x-8 )( x^3 + 8 )`

`=> x-8=0` hoặc `x^3+8=0`

`=> x=8` hoặc `x^3 = -8=(-2)^3`

`=> x=8` hoặc `x=-2`

Vậy `x in { -2;8}`

`b)`

`( 4x-5 ) - ( x+5) = 3( 10-x)`

`=> 4x-5-x-5=30-3x`

`=> ( 4x-x)+(-5-5)=30-3x`

`=> 3x-10=30-3x`

`=> 6x=40`

`=> x=20/3`

Vậy `x=20/3` 

\(a,\left(8+x\right)\left(6-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}8+x=0\\6-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\\ b,x^2-5x=0\\ \Rightarrow x\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

a) (8+x).(6-x)=0
<=> 8+x = 0 hoặc 6-x = 0
=> x = -8 hoặc x = 6

b) c) x^2 - 5x=0
<=> x^2 = 0 hoặc -5x = 0
=> x = 0 hoặc x = 5

a) [8+x]x[6-x]=0

vì tích của chúng phải bằng 0 nên có 2 trường hợp

TH1:(8+(-8))x(6-(xEN))

TH2:(8+(XEN))x(6-6)

`x :3*5 = 3/4 :(-5/6)`

`x :15 =3/4*(-6/5)=-9/10`

`x = -9/10 *15 =-27/2`

`x-1*2/2 = 8/x -1.2`

`x- 1*1 = 8/x -2`

`x-8/x = -2+1`

`x-8/x =-1`

`x^2 -8x =-x`

`x^2 -8x +x=0`

`x^2 -7x =0`

`x(x-7) =0`

`=>[(x=0),(x=7):}`

`a, x \div 15=-9/10`

`x=-9/10*14`

`x=-27/2`

`b, (x-1*2)/2=8/(x-1*2)`

\(\left(x-1\cdot2\right)\cdot\left(x-1\cdot2\right)=8\cdot2\)

`(x-1*2)^2=16`

`(x-1*2)^2=(+-4)^2`

\(\Rightarrow\left[{}\begin{matrix}x-1\cdot2=4\\x-1\cdot2=-4\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4+2\\x=\left(-4\right)+2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

\(a,\Rightarrow x=19-17=2\\ b,\Rightarrow x+8=28:2=14\\ \Rightarrow x=14-8=6\\ c,\Rightarrow42-x=5^2=25\\ \Rightarrow x=42-25=17\)

a)x=2

b)x+8=14

x=6

c)\(42-x=5^2\)

\(42-x=25\)

\(-x=-17\)

\(x=17\)

a)x+17=19

⇔x=2

b)2(x+8)=28

⇔x+8=14

⇔x=6

c)42-x=5⁸:5⁶

⇔42-x=5²

⇔42-x=25

⇔x=17

b: y=9

\(a,\Leftrightarrow\left(x+8\right)\left(8x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=\dfrac{1}{8}\end{matrix}\right.\\ b,\Leftrightarrow\left(y-9\right)^2=0\Leftrightarrow y=9\)