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SO
18 tháng 10 2018
1/x(x-y)-4x+4y
=x(x-y)-4(x-y)
=(x-y)(x-4)
2/a)x^2-x=0
x(x-1)=0
<=>x=0 hoặc x-1=0
x =1
=>S={0;1}
b)(x+2)(x-3)-x-2=0
(x+2)(x-3)-(x+2)=0
(x+2)(x-3-1)=0
(x+2)(x-4)=0
<=>x+2=0 hoặc x-4=0
x =-2 x =4
=>{-2;4}
c)36^2-49=0
(6x-7)(6x+7)=0
<=>6x-7=0 hoặc 6x+7=0
6x =7 6x =-7
x =7/6 x =-7/6
=>{7/6;-7/6}
3/(n+7)^2-(n-5)^2
=(n+7-n+5)(n+7+n-5)
=12(2n+2)
=12*2(n+1)
=24(n+1) chia hết cho 24
=>(n+7)^2-(n-5)^2 chia hết cho 24.
TH
1
N
1
4 tháng 8 2018
\(4x^2+4x-3=0\)
\(\left[\left(2x\right)^2+2.2x.1+1\right]-4=0\)
\(\left(2x+1\right)^2-2^2=0\)
\(\left(2x+1-2\right).\left(2x+1+2\right)=0\)
\(\left(2x-1\right).\left(2x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}\)
\(x^4-3x^3-x+3=0\)
\(x^3.\left(x-3\right)-\left(x-3\right)=0\)
\(\left(x-3\right).\left(x^3-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^3-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(x^2.\left(x-1\right)-4x^2+8x-4=0\)
\(x^2.\left(x-1\right)-\left[\left(2x\right)^2-2.2x.2+2^2\right]=0\)
\(x^2.\left(x-1\right)-\left(2x-2\right)^2=0\)
\(x^2.\left(x-1\right)-4.\left(x-1\right)^2=0\)
\(\left(x-1\right).\left[x^2-4.\left(x-1\right)\right]=0\)
\(\left(x-1\right).\left[x^2-2.x.2+2^2\right]=0\)
\(\left(x-1\right).\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
Vậy \(\begin{cases}x=1\\x=2\end{cases}\)
Tham khảo nhé~
N
0
CV
4
15 tháng 11 2018
a. 3x(x-2)-x+2=0
3x(x-2)-(x-2)=0
(3x-1)(x-2)=0
=>\(\hept{\begin{cases}3x-1=0\\x-2=0\end{cases}}\)
=> \(\hept{\begin{cases}3x=1\\x=2\end{cases}}\)
=>\(\hept{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)
vậy x thuộc (1/3;2)
HT
2
HL
3 tháng 7 2017
a) \(x\left(x+2\right)-3x-6=0\)
\(x\left(x+2\right)-3\left(x+2\right)=0\)
\(\left(x+2\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
b) \(\left(x^3+3x^2+3x+1\right)-3x^2-3x=0\)
\(x^3+1=0\)
\(\left(x+1\right)\left(x^2-x+1\right)=0\)
\(x=-1\)
c) \(4x^2-25=0\)
\(\left(2x-5\right)\left(2x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
TN
3 tháng 7 2017
\(a,x\left(x+2\right)-3x-6=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)\(b,\left(x^3+3x^2+3x+1\right)-3x^2-3x=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-3x^2-3x=0\)
\(\Leftrightarrow x^3+1=0\)
\(\Rightarrow x^3=1\Rightarrow x=1\)
\(c,4x^2-25=0\)
\(\Leftrightarrow\left(2x+5\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}2x+5=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
NT
1
24 tháng 9 2017
mk giúp bn câu a) còn các câu khác tương tự thôi:
a)= x2 - 2.2x +4 -1 = (x-2)2 -12 = (x-2 -1)(x-2+1) =0
x -3 =0 => x =3
x -1=0 => x = 1
( bn phân tich thành nhân tử r đưa nó về pt tích như mk làm đó. mk có cách giải theo lop9 hay hơn nhiu nhưng lop8 chưa học tới)
14 tháng 8 2020
a) Ta có: 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{19}{24}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{19}{24}\right\}\)
b) Ta có: \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\)
\(\Leftrightarrow8x=15\)
hay \(x=\frac{15}{8}\)
Vậy: \(x=\frac{15}{8}\)
c) Ta có: \(3x\left(2-x\right)+2x\left(x-1\right)=5x\left(x+3\right)\)
\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\)
\(\Leftrightarrow-x^2+4x-5x^2-15x=0\)
\(\Leftrightarrow-6x^2-11x=0\)
\(\Leftrightarrow6x^2+11x=0\)
\(\Leftrightarrow x\left(6x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-11}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{-11}{6}\right\}\)
d) Ta có: \(3x\left(x+1\right)-5x\left(3-x\right)+6\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow3x^2+3x-15x+5x^2+6x^2+12x+18=0\)
\(\Leftrightarrow14x^2+18=0\)
\(\Leftrightarrow14x^2=-18\)
mà \(14x^2\ge0\forall x\)
nên \(x\in\varnothing\)
Vậy: \(x\in\varnothing\)
27 tháng 9 2020
a) \(3x^3-12x=0\)
=> \(3x\left(x^2-4\right)=0\)
=> \(\orbr{\begin{cases}3x=0\\x^2-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)
b) \(x^2\left(x-3\right)+12-4x=0\)
=> \(x^2\left(x-3\right)+\left(-4x+12\right)=0\)
=> \(x^2\left(x-3\right)-4x+12=0\)
=> \(x^2\left(x-3\right)-4\left(x-3\right)=0\)
=> \(\left(x-3\right)\left(x^2-4\right)=0\Rightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}\)
c) \(\left(3x-1\right)^2-\left(2x-3\right)^2=0\)
=> \(\left[3x-1-\left(2x-3\right)\right]\left(3x-1+2x-3\right)=0\)
=> \(\left(3x-1-2x+3\right)\left(3x-1+2x-3\right)=0\)
=> \(\left(x+2\right)\left(5x-4\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{4}{5}\end{cases}}\)
d) \(x^2-4x-21=0\)
=> \(x^2+3x-7x-21=0\)
=> \(x\left(x+3\right)-7\left(x+3\right)=0\)
=> (x + 3)(x - 7) = 0 => x = -3 hoặc x = 7
e) 3x2 - 7x - 10 = 0
=> 3x2 + 3x - 10x - 10 = 0
=> 3x(x + 1) - 10(x + 1) = 0
=> (x + 1)(3x - 10) = 0
=> x = -1 hoặc x = 10/3
F
27 tháng 9 2020
a) \(3x^3-12x=0\)
\(\Leftrightarrow3x\left(x^2-4\right)=0\)
\(\Leftrightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow x\in\left\{-2;0;2\right\}\)
b) \(x^2\left(x-3\right)+12-4x=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow x\in\left\{-2;2;3\right\}\)
c) \(\left(3x-1\right)^2-\left(2x-3\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(5x-4\right)=0\)
\(\Leftrightarrow x\in\left\{-2;\frac{4}{5}\right\}\)
a,x^2+3x=0
=> x.(x+3)=0
=> +)x=0
+) x+3=0 => x=-3
b,x^3-4x=0
=> x.(x^2-2^2)=0
=> x.(x-2).(x+2)=0
=> +) x=0
+) x-2=0 => x=2
+) x+2=0 => x= -2
a) \(x^2+3x=0\)
\(x\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
vay \(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
b) \(x^3-4x=0\)
\(x\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=4\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
vay \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)