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NN
1
GH
5 tháng 3 2018
a) \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left(2x^2-2x-3x+3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x-1\right)-3\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy .................
b) \(\left(x-3\right)^2=\left(2x+1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(2x+1-x+3\right)\left(2x+1+x-3\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ...............
c) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
P/s: tới đây bn tự giải tiếp nha
AM
4
JS
19 tháng 2 2020
a, (3x - 1)(5x + 3) = (2x + 3)(3x - 1)
⇔ 5x + 3 = 2x + 3
⇔ 3x = 0
⇔ x = 0
Vậy phương trình có nghiệm là x = 0
Mình làm lại rồi nhé!
JS
19 tháng 2 2020
a, (3x - 1)(5x + 3) = (2x + 3)(3x - 1)
⇔ 5x + 3 = 2x + 3
⇔ 3x = 0
⇔ x = 0
Vậy phương trình có nghiệm là x = 3.
NN
15 tháng 11 2017
2)
a) \(3x^3-3x=0\)
\(\Leftrightarrow3x\left(x^2-1\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy x=0 ; x=-1 ; x=1
b) \(x^2-x+\dfrac{1}{4}=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
NN
15 tháng 11 2017
1)
a) \(\left(x-2\right)\left(x^2+3x+4\right)\)
\(\Leftrightarrow x^3+3x^2+4x-2x^2-6x-8\)
\(\Leftrightarrow x^3+x^2-2x-8\)
b) \(\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
c) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
d) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=18x^2+12x-9x-6-6x^3-4x^2+3x^2+2x\)
\(=17x^2+5x-6-6x^3\)
26 tháng 5 2016
a) (2x + 1)(3x - 2) = (5x - 8)(2x + 1)
<=> 6x2 - x - 2 = 10x2 - 11x - 8
<=> 6x2 - 10x2 - x + 11x -2 + 8 = 0
<=> -4x2 + 10x + 6 = 0
<=> -2 (2x2 - 5x - 3) = 0
<=> 2x2 - 5x - 3 = 0
<=> 2x2 - 6x + x - 3 = 0
<=> x (2x + 1) - 3 (2x + 1) = 0
<=> (x - 3) (2x + 1) = 0
* x - 3 = 0 => x = 3
* 2x + 1 = 0 => x = -1/2
S = {-1/2; 3}
b) 4x2 – 1 = (2x +1)(3x -5)
<=> 4x2 – 1 - (2x +1)(3x -5) = 0
<=> (2x - 1) (2x + 1) - (2x + 1)(3x - 5) = 0
<=> (2x + 1) (2x - 1 - 3x + 5) = 0
<=> (2x + 1) (-x + 4) = 0
* 2x + 1 = 0 <=> x = -1/2
* -x + 4 = 0 <=> x = 4
S = {-1/2; 4}
c) (x + 1)2 = 4(x2 – 2x + 1)
<=> (x + 1)2 - 4(x2 – 2x + 1) = 0
<=> (x + 1)2 - 4(x2 – 1)2 = 0
* (x + 1)2 = 0 <=> x = -1
* 4(x2 - 1)2 = 0 <=> x = 1 và x = -1
S = {-1; 1}
d) 2x3 + 5x2 – 3x = 0
<=> x (2x2 + 5x - 3) = 0
<=> x (2x2 + 6x - x - 3) = 0
<=> x [x(2x - 1) + 3 (2x - 1)] = 0
<=> x (2x - 1) (x + 3) = 0
* x = 0
* 2x - 1 = 0 <=> x = 1/2
* x + 3 = 0 <=> x = -3
S = { -3; 0; 1/2}
31 tháng 7 2018
(2x)2+2.2x.3+32+x2+2.x.1+12=3x2-6x
<=>4x2+12x+9+x2+2x+1-3x2+6x=0
<=>2x2+20x+10=0
<=>2(x2+10x+5)=0
<=>x2+10x+5=0
<=>x2+2.x.5+52-20=0
<=>(x+5)2-20=0
<=>x+5=+_√20
<=>x=+_√20 -5
31 tháng 7 2018
Ta có: \(\left(2x+3\right)^2+\left(x+1\right)^2=3x\left(x-2\right)\)
⇌ \(4x^2+12x+9+x^2+2x+1=3x^2-6x\)
⇌ \(2x^2+20x+10=0\)
⇌\(x^2+10x+5=0\)
⇌\(\left(x+5\right)^2-20=0\)
⇌\(\left(x+5+\sqrt{20}\right)\left(x+5-\sqrt{20}\right)=0\)
⇌\(\left[{}\begin{matrix}x+5+\sqrt{20}=0\\x+5-\sqrt{20}=0\end{matrix}\right.\)⇌\(\left[{}\begin{matrix}x=-5-\sqrt{20}\\x=\sqrt{20}-5\end{matrix}\right.\)
Vậy x ∈ \(\left\{-5-\sqrt{20},\sqrt{20}-5\right\}\)
2 tháng 8 2018
sorry mk nhầm,từ dòng 4 fải lm như sau:
\(2x^2-4x=-10\)
\(\Leftrightarrow2x^2-4x+10=0\)
\(\Leftrightarrow2\left(x^2-2x+5\right)=0\)
\(\Leftrightarrow x^2-2x+5=0\)
\(\Leftrightarrow x^2-2x+1+4=0\)
\(\Leftrightarrow\left(x-1\right)^2+4=0\)
Vì: \(\left(x-1\right)^2+4\ge4>0\forall x\)
=>PT vô nghiệm
Y
2 tháng 8 2018
\(\left(2x-3\right)^2+\left(x+1\right)^2=3x\left(x-2\right)\)
\(\Rightarrow\left(2x-3\right)^2+\left(x+1\right)^2-3x\left(x-2\right)=0\)
\(\Rightarrow\left(2x\right)^2-2.2x.3+3^2+x^2+2x+1-3x^2+6x=0\)
\(\Rightarrow4x^2-12x+9+x^2+2x+1-3x^2+6x=0\)
\(\Rightarrow2x^2-4x+10=0\)
\(\Rightarrow2\left(x^2-2x+5\right)=0\)
\(\Rightarrow x^2-2x+5=0\)
\(\Rightarrow x^2-2x+1+4=0\)
\(\Rightarrow\left(x-1\right)^2+4=0\)
Vì \(\left(x-1\right)^2\ge0\) với mọi x
\(4>0\)
\(\Rightarrow\left(x-1\right)^2+4>0\)
\(\Rightarrow\) Phương trình vô nghiệm
NV
8 tháng 1 2018
Bài 2: a) \(3x^3-3x=0\Leftrightarrow3x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
\(\Leftrightarrow\left(x^2-3x+3\right)\left(x^2-3x+3+x\right)-2x^2=0\)
\(\Leftrightarrow\left(x^2-3x+3\right)^2+x\left(x^2-3x+3\right)-2x^2=0\)
\(\Leftrightarrow\left(x^2-3x+3\right)^2-x\left(x^2-3x+3\right)+2x\left(x^2-3x+3\right)-2x^2=0\)
\(\Leftrightarrow\left(x^2-3x+3\right)\left(x^2-3x+3-x\right)+2x\left(x^2-3x+3-x\right)=0\)
\(\Leftrightarrow\left(x^2-4x+3\right)\left(x^2-x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\\x^2-x+3=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)