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

\(b,\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\\ \Leftrightarrow4x-3-x-5=30-3x\\ \Leftrightarrow3x-8-30+3x=0\\ \Leftrightarrow6x-38=0\\ \Leftrightarrow x=\dfrac{19}{3}\)

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`

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

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\\left(x+2\right)\left(x^2-2x+4\right)=0\end{matrix}\right.\)

Ta có: \(x^2-2x+4=x^2-2x+1+3=\left(x-1\right)^2+3\ge3>0\)
\(\Rightarrow x=-2\)

Vậy \(S=\left\{-2;8\right\}\)

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

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

\(\Leftrightarrow6x=38\)

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

Vậy \(S=\left\{\dfrac{19}{3}\right\}\)

\(a,\Rightarrow 2x^2-10x-3x-2x^2=26\\ \Rightarrow -13x=26\\ \Rightarrow x=-2\\ b, \Rightarrow -2x^2+3x+3-3x-3+2x^2-x=18\\ \Rightarrow -x=18\Rightarrow x=-18\)

Không rõ đề lắm bạn ơi

câu hỏi của bạn có sai ko ạ

a) \(x=\dfrac{1}{2}+\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{2}{4}+\dfrac{3}{4}\\ \Rightarrow x=\dfrac{5}{4}\)

b) \(\dfrac{x}{5}=\dfrac{1}{2}-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{x}{5}=\dfrac{3}{6}-\dfrac{4}{6}\)

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

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

\(\Rightarrow x=-\dfrac{5}{6}\)

c) \(x=\dfrac{1}{2}-\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{3}{6}-\dfrac{4}{6}\)

\(\Rightarrow x=-\dfrac{1}{6}\)

d) \(\dfrac{x}{3}=\dfrac{2}{3}-\dfrac{1}{7}\)

\(\Rightarrow\dfrac{x}{3}=\dfrac{14}{21}-\dfrac{3}{21}\)

\(\Rightarrow\dfrac{x}{3}=\dfrac{11}{21}\)

\(\Rightarrow x=\dfrac{11}{21}.3\)

\(\Rightarrow x=\dfrac{33}{21}\)

a:Ta có: \(16-\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(x+3\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

a: Ta có: \(4\left(x+1\right)^2+\left(2x+1\right)^2-8\left(x-1\right)\left(x+1\right)-11=0\)

\(\Leftrightarrow4x^2+8x+4+4x^2+4x+1-8x^2+8-11=0\)

\(\Leftrightarrow12x=-2\)

hay \(x=-\dfrac{1}{6}\)

b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)-1=0\)

\(\Leftrightarrow x^2+6x+9-x^2-4x+32-1=0\)

\(\Leftrightarrow2x=-40\)

hay x=-20

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.

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

$4x-3-x-5=30-3x$

$3x-8=30-3x$

$6x=38$
$x=\frac{19}{3}$

Bài 2:

$f(x)=(x-1)(x+2)=0$

$\Leftrightarrow x-1=0$ hoặc $x+2=0$

$\Leftrightarrow x=1$ hoặc $x=-2$

Vậy $g(x)$ cũng có nghiệm $x=1$ và $x=-2$

Tức là:

$g(1)=g(-2)=0$

$\Rightarrow 1+a+b+2=-8+4a-2b+2=0$

$\Rightarrow a=0; b=-3$

\(a,\Leftrightarrow\left(x-5\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)