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\(a.\)
\(\left(x-8\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
\(S=\left\{8,-2\right\}\)
\(b.\)
\(\left(4x-3\right)-\left(x+5\right)=3\cdot\left(10-x\right)\)
\(\Leftrightarrow4x-3-x-5-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{38}{6}\)
\(S=\left\{\dfrac{38}{6}\right\}\)
Bài 2
P(x) + Q(x) = x3 – 6x + 2 + 2x2 - 4x3 + x - 5 = - 3x3 + 2x2 – 5x - 3
P(x) - Q(x) = x3 – 6x + 2 - 2x2 + 4x3 - x + 5 = 5x3 − 2x2 − 7x+7
a) `(x-8)(x^3+8)=0`
`<=>(x-8)(x+2)(x^2-2x+4)=0`
`<=>` \(\left[ \begin{array}{l}x=8\\x=-2\end{array} \right.\) (Vì `x^2-2x+4 \ne 0 forall x)`
Vậy `A={8;-2}`.
b) `(4x-3)-(x+5)=3(10-x)`
`,=>4x-3-x-5=30-3x`
`<=>3x-8=30-3x`
`<=>6x=38`
`<=>x=19/3`
Vậy `S={19/3}`.
\(a,\left(x-8\right)\left(x^3+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
\(b,\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\\ \Leftrightarrow4x-3-x-5=30-3x\\ \Leftrightarrow3x-8-30+3x=0\\ \Leftrightarrow6x-38=0\\ \Leftrightarrow x=\dfrac{19}{3}\)
TK
`a.(x-8)(x+8)=0`
`⇔³{x−8=0x³+8=2 `
`⇔³³{x=8x³=−2³ `
`⇔{x=8x=−2`
Vậy ` x = 8;-2`
`b. ( 4 x − 3 ) − ( x + 5 ) = 3 . ( 10 − x )`
`⇔ 4 x − 3 − x − 5 = 30 − 3 x`
`⇔ 3 x − 8 = 30 − 3 x`
`⇔ 3 x + 3 x = 30 + 8`
`⇔ 6 x = 38`
`⇔ x = 19/ 3`
Vậy ` x = 19/ 3`
Bài 3:
\(\left(x-8\right)\left(x3+8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x3+8=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=\dfrac{-8}{3}\end{matrix}\right.\)
\(\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\)
\(4x-3-x-5=30-3x\)
\(4x-x+3x=30+3+5\)
\(6x=38\)
\(x=38:6\)
\(x=\dfrac{19}{3}\)
Bài 1.
a.\(\left(x-8\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
b.\(\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\)
\(\Leftrightarrow4x-3-x-5=30-3x\)
\(\Leftrightarrow4x-x+3x=30+5+3\)
\(\Leftrightarrow6x=38\)
\(\Leftrightarrow x=\dfrac{19}{3}\)
Bài 1:
a. $(x-8)(x^3+8)=0$
$\Rightarrow x-8=0$ hoặc $x^3+8=0$
$\Rightarrow x=8$ hoặc $x^3=-8=(-2)^3$
$\Rightarrow x=8$ hoặc $x=-2$
b.
$(4x-3)-(x+5)=3(10-x)$
$4x-3-x-5=30-3x$
$3x-8=30-3x$
$6x=38$
$x=\frac{19}{3}$
\(a,\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow4x-3-x-5=30-3x\\ \Leftrightarrow4x-x+3x=30+5+3\\ \Leftrightarrow6x=38\\ \Leftrightarrow x=\dfrac{19}{3}\)
c: =>x+y-xy=-16
=>x+y-xy-1=-17
=>x(1-y)-(1-y)=-17
=>(1-y)(x-1)=-17
=>(x-1;y-1)=17
=>(x-1;y-1) thuộc {(1;17); (17;1); (-1;-17); (-17;-1)}
=>(x,y) thuộc {(2;18); (18;2); (0;-16); (-16;0)}
b: Tham khảo:
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`