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\(\left(x-3\right)\left(x+3\right)-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)\left(x+3-x+3\right)=0\Leftrightarrow6\left(x-3\right)=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Ta có: \(\left(x-3\right)\left(x+3\right)-\left(x-3\right)^2=0\)
\(\Leftrightarrow x^2-9-x^2+6x-9=0\)
\(\Leftrightarrow6x=18\)
hay x=3
\(a,\left(x+3\right)\left(x-3\right)+x\left(3-x\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-3\right)-x\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+3-x\right)=0\)
\(\Rightarrow3\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
\(b,x\left(x-3\right)+x-3=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\hept{\begin{cases}x-3=0\\x+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=-1\end{cases}}}\)
Bài giải
\(\left(x-3\right)2-\left(x-3\right)\left(x+3\right)=0\)
\(\left(x-3\right)\left(x+3-2\right)=0\)
\(\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{3\text{ ; }-1\right\}\)
\(\left(x-3\right).2-\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left(x-3\right)\left[2-\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(2-x-3\right)=0\Leftrightarrow\left(x-3\right)\left[\left(-1\right)-x\right]\). Xét 2 trường hợp
Xét 2 trường hợp. \(TH1:x-3=0\Leftrightarrow x=0+3=3\)
\(TH2:\left(-1\right)-x=0\Leftrightarrow x=\left(-1\right)-0=-1\). Vậy \(x\in\left\{-1;3\right\}\)
2x(x - 3) + (x - 3) = 0
<=> (2x + 1)(x - 3) = 0
<=> \(\left[{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-0,5\\x=3\end{matrix}\right.\)
5x(x – 3) – x + 3 = 0
⇔ 5x(x – 3) – (x – 3) = 0
(Xuất hiện nhân tử chung x – 3)
⇔ (x – 3)(5x – 1) = 0
⇔ x – 3 = 0 hoặc 5x – 1= 0
+ x – 3 = 0 ⇔ x = 3
+ 5x – 1 = 0 ⇔ 5x = 1 ⇔ x = 1/5
Vậy x = 3 hoặc x = 1/5.
\(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2-12=0\)
\(\Leftrightarrow3x=40\)
hay \(x=\dfrac{40}{3}\)
a: Ta có: \(x\left(x-3\right)-x^2+5=0\)
\(\Leftrightarrow-3x+5=0\)
hay \(x=\dfrac{5}{3}\)
b: Ta có: \(x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
2x(x – 3) + 5(x – 3) = 0
ó (2x + 5)(x – 3) = 0
Vậy x = - 5 2 hoặc x = 3
Đáp án cần chọn là: B
\(x\left(x-3\right)^3+3-x=0\)
\(\Leftrightarrow x\left(x-3\right)^2-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[x\left(x-3\right)-1\ne0\right]=0\)
\(\Leftrightarrow x=3\)
ta có
\(\left(x-3\right)^3-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(x-3\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^2-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=3\\\left(x-2\right)\left(x-4\right)=0\end{cases}}\)pt dưới \(\Leftrightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}\)
vậy x=2 hoặc x=3 hoặc x=4