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DD
3
GH
28 tháng 2 2018
a) \(\left(x-3\right)\left(2x+1\right)\left(4-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\\4-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
Vậy ..................
b) \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left(2x^2-2x-3x+3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x-1\right)-3\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy .................
c) \(\left(x-3\right)^2=\left(2x+1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(2x+1-x+3\right)\left(2x+1+x-3\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy .......................
d) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left[x\left(x-3\right)-4\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)
Vậy ...................
TT
28 tháng 2 2018
a,\(\left(x-3\right)\left(2x+1\right)\left(4-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\\4-5x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
Vậy...
b,\(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left(2x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-3=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\\x=1\end{matrix}\right.\)
Vậy...
c,Sửa đề:
\(\left(x-3\right)^2=\left(2x+1\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-3+2x+1\right)\left(x-3-2x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(-x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\-x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-4\end{matrix}\right.\)
Vậy...
d,\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x+4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x+4=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-4\\x=3\end{matrix}\right.\)
Vậy...
14 tháng 8 2020
a) Ta có: \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
b) Ta có: \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=\left(6x^2+x-2\right)\left(3-x\right)\)
\(=18x^2-6x^3+3x-x^2-6+2x\)
\(=-6x^3+17x^2+5x-6\)
c) Ta có: \(\left(x+3\right)\left(x^2+3x-5\right)\)
\(=x^3+3x^2-5x+3x^2+9x-15\)
\(=x^3+6x^2+4x-15\)
d) Ta có: \(\left(x+1\right)\left(x^2-x+1\right)\)
\(=x^3+1\)
e) Ta có: \(\left(2x^3-3x-1\right)\left(5x+2\right)\)
\(=10x^4+4x^3-15x^2-6x-5x-2\)
\(=10x^4+4x^3-15x^2-11x-2\)
f) Ta có: \(\left(x^2-2x+3\right)\left(x-4\right)\)
\(=x^3-4x^2-2x^2+8x+3x-12\)
\(=x^3-6x^2+11x-12\)
g) Ta có: \(\left(4x-1\right)\left(3x+1\right)-5x\left(x-3\right)-\left(x-4\right)\left(x-3\right)\)
\(=12x^2+4x-3x-1-5x^2+15x-\left(x^2-7x+12\right)\)
\(=7x^2+16x-1-x^2+7x-12\)
\(=6x^2+23x-23\)
h) Ta có: \(\left(5x-2\right)\left(x+1\right)-3x\left(x^2-x-3\right)-2x\left(x-5\right)\left(x-4\right)\)
\(=5x^2+5x-2x-2-3x^3+3x^2+9x-2x\left(x^2-9x+20\right)\)
\(=-3x^3+8x^2+12x-2-2x^3+18x^2-40x\)
\(=-5x^3+26x^2-28x-2\)
N
0
12 tháng 7 2019
\(a,\left(6x+1\right)\left(x+2\right)-2x\left(3x-5\right)\)
\(=6x^2+12x+x+2-6x^2+10x\)
\(=23x+2\)
12 tháng 7 2019
a) (6x + 1)(x + 2) - 2x(3x - 5)
= 6x2 + 12x + x + 2 - 6x2 + 10x
= (6x2 - 6x2) + (12x + x + 10x) + 2
= 23x + 2
b) (2x - 1)2 - (2x - 3)(2x + 3)
= 4x2 - 4x + 1 - 4x2 + 9
= (4x2 - 4x2) - 4x + (1 + 9)
= -4x + 10
c) (2x - 3)3 - (3x + 1)(5 - 4x) - 16x2
= 8x3 - 36x2 + 54x - 15x + 12x2 - 5 + 4x - 16x2
= 8x3 - (36x2 - 12x2 + 16x2) + (54x - 15x + 4x) - 5
= 8x3 - 40x2 + 43x - 5
d) (3x + 2) - (x - 5) - x(3x - 13)
= 3x + 2 - x + 5 - 3x2 + 13x
= (3x - x + 13x) + (2 + 5) - 3x2
= 15x + 7 - 3x2
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}
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.
VK
2
27 tháng 6 2017
1) \(\left(3x-5\right)^2+\left(2x+1\right)^2=\left(5x-6\right)\left(2x+3\right)+3x^2\)
\(\Leftrightarrow\) \(9x^2-30x+25+4x^2+4x+1=10x^2+15x-12x-18+3x^2\)
\(\Leftrightarrow\) \(13x^2-26x+26=13x^2+3x-18\)
\(\Leftrightarrow\) \(-29x+44\) \(\Leftrightarrow\) \(-29x=-44\Leftrightarrow\) \(x=\dfrac{44}{29}\) vậy \(x=\dfrac{44}{29}\)
2) \(\left(7x-4\right)\left(5x+3\right)=\left(6x-5\right)^2-x^2+1\)
\(\Leftrightarrow\) \(35x^2+21x-20x-12=36x^2-60x+25-x^2+1\)
\(\Leftrightarrow\) \(35x^2+x-12=35x^2-60x+26\)
\(\Leftrightarrow\) \(61x-38=0\) \(\Leftrightarrow\) \(61x=38\) \(\Leftrightarrow x=\dfrac{38}{61}\) vậy \(x=\dfrac{38}{61}\)
27 tháng 6 2017
1. \(\left(3x-5\right)^2+\left(2x+1\right)^2=\left(5x-6\right)\left(2x+3\right)+3x^2\)
\(\Leftrightarrow9x^2-30x+25+4x^2+4x+1=10x^2+15x-12x-18+3x^2\)
\(\Leftrightarrow-29x=-44\)
\(\Rightarrow x=\dfrac{44}{29}\)
2. \(\left(7x-4\right)\left(5x+3\right)=\left(6x-5\right)^2-x^2+1\)
\(\Leftrightarrow35x^2+21x-20x-12=36x^2-60x+25-x^2+1\)
\(\Leftrightarrow61x=38\)
\(\Rightarrow x=\dfrac{38}{61}\)
(2x3 - 3x - 1).(5x +2)
= 10x4 + 4x3 - 15x2 - 6x - 5x - 2
= 10x4 + 4x3 - 15x2 - 11x - 2
(2x3 - 3x - 1).(5x +2)
= 10x4 + 4x3 - 15x2 - 6x - 5x - 2
= 10x4 + 4x3 - 15x2 - 11x - 2