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14 tháng 2 2022
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
14 tháng 2 2022
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
28 tháng 2 2021
a. (3x - 1)2 - (x + 3)2 = 0
\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow4x+2=0\) hoặc \(2x-4=0\)
1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)
2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
S=\(\left\{-\dfrac{1}{2};2\right\}\)
28 tháng 2 2021
b. \(x^3=\dfrac{x}{49}\)
\(\Leftrightarrow49x^3=x\)
\(\Leftrightarrow49x^3-x=0\)
\(\Leftrightarrow x\left(49x^2-1\right)=0\)
\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(7x+1=0\) hoặc \(7x-1=0\)
1. x=0
2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)
3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
LP
2
5 tháng 3 2022
1.\(\left(x+2\right)\left(2x-3\right)=x^2-4\)
\(\Leftrightarrow\left(x+2\right)\left(2x-3\right)-\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-3-x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
2.\(x^2+3x+2=0\)
\(\Leftrightarrow x^2+x+2x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
3.\(2x^2+5x+3=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{2}\end{matrix}\right.\)
4.\(x^3+x^2-12x=0\)
\(\Leftrightarrow x\left(x^2+x-12\right)=0\)
\(\Leftrightarrow x\left(x+4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=3\end{matrix}\right.\)
5 tháng 3 2022
a: \(\Leftrightarrow\left(x+2\right)\left(2x-3\right)-\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-3-x+2\right)=0\)
=>(x+2)(x-1)=0
=>x=-2 hoặc x=1
b: =>(x+1)(x+2)=0
=>x=-1 hoặc x=-2
c: =>(2x+3)(x+1)=0
=>x=-1 hoặc x=-3/2
d: =>x(x+4)(x-3)=0
hay \(x\in\left\{0;-4;3\right\}\)
PT
1
CM
30 tháng 1 2019
(x – 1)( x 2 + 5x – 2) – ( x 3 – 1) = 0
⇔ (x – 1)( x 2 + 5x – 2) – (x – 1)( x 2 + x + 1) = 0
⇔ (x – 1)[( x 2 + 5x – 2) – ( x 2 + x + 1)] = 0
⇔ (x – 1)( x 2 + 5x – 2 – x 2 – x – 1) = 0
⇔ (x – 1)(4x – 3) = 0 ⇔ x – 1 = 0 hoặc 4x – 3 = 0
x – 1 = 0 ⇔ x = 1
4x – 3 = 0 ⇔ x = 0,75
Vậy phương trình có nghiệm x = 1 hoặc x = 0,75
x^3 - 4x^2 +5x=0
<=> x^3 +x^2 - 5x^2 +5x = 0
<=>x^2.(x+1) - 5x.(x+1) = 0
<=>(x+1) .(x^2- 5x) = 0
<=>(x+1) .x(x- 5) = 0
TH1: x=0
TH2: x+1=0 => x= -1
TH3: x-5=0 => x=5
Vậy ....
x^ 3 - 4x^2 +5x=0
=> x (x^2 - 4x+5) = 0
Đến đây rút ra x =0