(4x+5)⋮(2x+1)
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12 tháng 2 2021
a) Ta có: \(\left(x-2\right)\cdot x=2x\cdot\left(x+5\right)\)
\(\Leftrightarrow x\cdot\left(x-2\right)-2x\left(x+5\right)=0\)
\(\Leftrightarrow x\cdot\left[x-2-2\left(x+5\right)\right]=0\)
\(\Leftrightarrow x\left(x-2-2x-10\right)=0\)
\(\Leftrightarrow x\left(-x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)
Vậy: S={0;-8}
b) Ta có: \(\left(2x-5\right)\left(x+11\right)=\left(5-2x\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)-\left(5-2x\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+11+2x+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(3x+12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\3x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{5}{2};-4\right\}\)
c) Ta có: \(x^2+6x+9=4x^2\)
\(\Leftrightarrow\left(x+3\right)^2-\left(2x\right)^2=0\)
\(\Leftrightarrow\left(x+3-2x\right)\left(x+3+2x\right)=0\)
\(\Leftrightarrow\left(-x+3\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x+3=0\\3x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-3\\3x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy: S={3;-1}
d) Ta có: \(\left(x+2\right)\left(5-4x\right)=x^2+4x+4\)
\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(5-4x-x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(-5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\-5x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{5}\right\}\)
LV
1
13 tháng 3 2022
\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}.\\x=1.\end{matrix}\right.\)
6 tháng 2 2022
a: \(\Leftrightarrow x\left(2x+10\right)-x\left(x-2\right)=0\)
=>x(2x+10-x+2)=0
=>x(x+12)=0
=>x=0 hoặc x=-12
b: \(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)
=>(2x-5)(3x+12)=0
=>x=5/2 hoặc x=-4
c: \(\Leftrightarrow\left(2x\right)^2-\left(x+3\right)^2=0\)
=>(x-3)(3x+3)=0
=>x=3 hoặc x=-1
d: \(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(5-4x-x-2\right)=0\)
=>(x+2)(-5x+3)=0
=>x=-2 hoặc x=3/5
6 tháng 2 2022
\(a,\left(x-2\right)x=2x\left(x+5\right)\)
\(\Leftrightarrow\left(x-2\right)x-2x\left(x+5\right)=0\)
\(\Leftrightarrow x.\left(x-2-2x-10\right)=0\)
\(\Leftrightarrow x\left(-x-12\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+12=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-12\end{matrix}\right.\)
HA
1
26 tháng 10 2021
1) ĐKXĐ: \(x\ge\dfrac{5}{2}\)
\(\sqrt{x^2}=2x-5\\ \Rightarrow\left|x\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x=2x-5\\x=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)
2) ĐKXĐ: \(x\ge3\)
\(\sqrt{25x^2-10x+1}=2x-6\\ \Rightarrow\left|5x-1\right|=2x-6\\ \Rightarrow\left[{}\begin{matrix}5x-1=2x-6\\5x-1=6-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
3) ĐKXĐ: \(x\ge\dfrac{5}{2}\)
\(\sqrt{25-10x+x^2}=2x-5\\ \Rightarrow\left|x-5\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x-5=2x-5\\x-5=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{10}{3}\left(tm\right)\end{matrix}\right.\)
4) ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(\sqrt{1-2x+x^2}=2x-1\\ \Rightarrow\left|x-1\right|=2x-1\\ \Rightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)
KH
22 tháng 9 2018
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
KB
14 tháng 9 2018
a ) \(\left(3x^2-4x+5\right)\left(2x^2-4\right)-2x\left(3x^3-4x^2+8\right)\)
\(=\left(3x^2-4x+5\right).2x^2-4\left(3x^2-4x+5\right)-6x^4+8x^3-16x\)
\(=6x^4-8x^3+10x^2-12x^2+16x-20-6x^4+8x^3-16x\)
\(=\left(6x^4-6x^4\right)+\left(8x^3-8x^3\right)-\left(12x^2-10x^2\right)+\left(16x-16x\right)-20\)
\(=-2x^2-20\)
b ) \(\left(1-3x+x^2\right)\left(2-4x\right)+2x\left(2x^2+5\right)\)
\(=2\left(1-3x+x^2\right)-4x\left(1-3x+x^2\right)+4x^3+10x\)
\(=2-6x+2x^2-4x+12x^2-4x^3+4x^3+10x\)
\(=\left(4x^3-4x^3\right)+\left(12x^2+2x^2\right)+\left(10x-6x-4x\right)+2\)
\(=14x^2+2\)
H9
HT.Phong (9A5)
CTVHS
31 tháng 7 2023
v) \(\left(-\dfrac{1}{2}x+3\right)\left(2x+6-4c^3\right)\)
\(=-\dfrac{1}{2}\left(2x+6-4c^3\right)+3\left(2x+6-4c^3\right)\)
\(=-x^2-3x+2c^3x+6x+18-12c^3\)
\(=-x^2+3x+2c^3x+18-12c^3\)
f) \(\left(2x-5\right)\left(x^2-x+3\right)\)
\(=2x\left(x^2-x+3\right)-5\left(x^2-x+3\right)\)
\(=2x^3-2x^2+6x-5x^2+5x-15\)
\(=2x^3-7x^2+11x-15\)
w) \(\left(3x+1\right)\left(x^2-2x-5\right)\)
\(=3x\left(x^2-2x-5\right)+\left(x^2-2x-5\right)\)
\(=3x^3-6x^2-15x+x^2-2x-5\)
\(=3x^3-5x^2-17x-5\)
x) \(\left(6x-3\right)\left(x^2+x-1\right)\)
\(=6x\left(x^2+x-1\right)-3\left(x^2+x-1\right)\)
\(=6x^3+6x^2-6x-3x^2-3x+3\)
\(=6x^3+3x^2-9x+3\)
y) \(\left(5x-2\right)\left(3x+1-x^2\right)\)
\(=5x\left(3x+1-x^2\right)-2\left(3x+1-x^2\right)\)
\(=15x^2+5x-5x^3-6x-2+2x^2\)
\(=-5x^3+17x^2-x-2\)
z) \(\left(\dfrac{3}{4}x+1\right)\left(4x^2+4x+4\right)\)
\(=\dfrac{3}{4}x\left(4x^2+4x+4\right)+\left(4x^2+4x+4\right)\)
\(=3x^3+3x^2+3x+4x^2+4x+4\)
\(=3x^3+7x^2+7x+4\)
31 tháng 7 2023
f: =2x^3-2x^2+6x-5x^2+5x-15
=2x^3-7x^2+11x-15
w: =3x^3-6x^2-15x+x^2-2x-5
=3x^3-5x^2-17x-5
x: =6x^3+6x^2-6x-3x^2-3x+3
=6x^3+3x^2-9x+3
y: =(5x-2)(-x^2+3x+1)
=-5x^3+15x^2+5x+2x^2-6x-2
=-5x^3+17x^2-x-2
z: =3x^3+3x^2+3x+4x^2+4x+4
=3x^3+7x^2+7x+4
TN
1
20 tháng 7 2019
a) \(2x\left(x-3\right)+4\left(x^2-5\right)=6x^2+1\)
\(\Rightarrow2x^2-6x+4x^2-20-6x^2-1=0\)
\(\Rightarrow-6x=21\)
\(\Rightarrow x=\frac{-7}{2}\)
Vậy..........
b) \(\left(2x-3\right)2x+\left(5-4x\right)x=3x\left(1+4x\right)\)
\(\Rightarrow4x^2-6x+5x-4x^2=3x+12x^2\)
\(\Rightarrow-12x^2-4x=0\)
\(\Rightarrow-4x\left(3x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{3}\end{cases}}\)
Vậy.........
Ta có ( 4x + 5 ) ⋮ ( 2x + 1 )
⇒ ( 4x + 2 + 3 ) ⋮ ( 2x + 1 )
Vì ( 4x + 2 ) ⋮ ( 2x + 1 ) ⇒ 3 ⋮ ( 2x + 1 ) hay ( 2x + 1 ) ϵ Ư( 3 ) = { 1; 3 }
Nếu 2x + 1 = 1 ⇒ x = 0
Nếu 2x + 1 = 3 ⇒ x = 1
Vậy x ϵ { 0; 1 } để ( 4x + 5 ) ⋮ ( 2x + 1 )