NI
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
DD
8 tháng 9 2016
\(\text{Tìm x:}\)
\(a.x\left(x-1\right)-3x+3x=0\)
\(x\left(x-1\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}}\)
\(b.3x\left(x-2\right)+10-5x=0\)
\(3x^2-6x+10-5x=0\)
\(3x^2-11x+10=0\)
\(3x^2-11x=-10\)(bn xem lại đề nhé)
\(c.x^3-5x^2+x-5=0\)
\(x^3-5x^2+x=5\)
\(d.x^4-2x^3+10x^2-20x=0\)
CW
8 tháng 9 2016
bài 1:phân tích thành phân tử
a> x^2-6x-y^2+9
= (x-3)^2 -y^2
= (x-3 -y) (x-3+y)
b>x^2-xy-8x+8y
= x(x-y) - 8(x-y)
= (x-8) (x-y)
c>25-4x^2-4xy-y^2
= 5^2 - (2x + y)^2
= (5 - 2x -y) (5 +2x+y)
d>xy-xz-y+z
= x(y-z) - (y-z)
= (x-1) (y-z)
e>x^2-xz-yz+2xy+y^2
= (x+y)^2 - z(x+y)
= (x+y-z) (x+y)
g>x^2-4xy+4y^2-z^2-4zt-4t^2
= (x-2y)^2 - (z + 2t)^2
= (x-2y -x-2t) (x-2y + z +2t)
bài 2:tìm X bt
a>x.(x-1)-3x+3x=0
x (x-1) =0
\(\Rightarrow\hept{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy x=0 và x=1
b>3x.(x-2)+10-5x=0
3x(x-2) - 5 (x-2)=0
(3x-5) (x-2) =0
\(\Rightarrow\hept{\begin{cases}3x-5=0\\x-2=0\end{cases}\Rightarrow\hept{\begin{cases}3x=5\\x=2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}}}\)
c>x^3-5x^2+x-5=0
x^2 (x-5) + (x-5) =0
(x^2 +1)(x-5) =0
\(\Rightarrow\hept{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Rightarrow\hept{\begin{cases}x^2=-1\\x=5\end{cases}\Rightarrow}\hept{\begin{cases}x\in\varphi\\x=5\end{cases}}}\)
Vậy x=5
d>x^4-2x^3+10x^2-20x=0
x^3 (x-2) + 10x(x-2) =0
(x^3 + 10x) (x-2) =0
x(x^2 + 10) (x-2) =0
\(\Rightarrow\hept{\begin{cases}x=0\\x^2+10=0\\x-2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=-10\\x=2\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x\in\varphi\\x=2\end{cases}}}}\)
Vậy x=0 và x=2
Q
9 tháng 10 2017
a) \(3x^2\left(x-1\right)+\left(1-x\right)\)
\(=3x^2\left(x-1\right)+\left(-x+1\right)\)
\(=3x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(3x^2-1\right)\left(x-1\right)\)
b) \(25-x^2+2xy-y^2\)
\(=-\left(x^2-2xy+y^2\right)+5^2\)
\(=-\left(x-y\right)^2+5^2\)
\(=-\left(x-y+5\right)\left(x-y-5\right)\)
c) \(x^2+4xy^2\)
\(=x\left(x+4y^2\right)\)
TM
1
15 tháng 12 2017
a) \(2xy\left(3x^2-4xy+y^2\right)\)
\(=6x^3y-8x^2y^2+2xy^3\)
b) \(\left(x-3\right)\left(2x^2-x+1\right)\)
\(=2x^3-x^2+x-6x^2+3x-3\)
\(=2x^3-7x^2+4x-3\)
c) \(-3x\left(x-2\right)\left(5-x\right)\)
\(=\left(-3x^2+6x\right)\left(5-x\right)\)
\(=-15x^2+3x^3+30-6x^2\)
\(=3x^3-21x^2+30\)
9 tháng 8 2017
B3) a) x(x-5)-4(x-5)=0
<=> (x-4)(x-5)=0
TH1 :x-4=0
<=.x=4
TH2 : x-5=0
<=>x=5
b) x(x-6)-7x-42=0
<=>x(x+6)-7(x+6)=0
<=>(x-7)(x+6)=0
th1;x-7=0
<=>x=7
th2; x+6=0
<=>x=-6
c)x^3-5x^2+x-5=0
<=> x(x^2+1)-5(x^2+1)=0
<=> (x-5)(x^2+1)=0
th1:x-5=0
<=>x=5
TH2 : x^2+1=0
<=> x^2=-1 ( vo li )
=> th2 ko tồn tại
nho thick nha
9 tháng 8 2017
Bài 3
a, x(x-5)-4(x-5)=0
(x-4)(x-5)=0
=>\(\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
b,x(x+6)-7(x+6)=0
(x-7)(x+6)=0\(\Rightarrow\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}\)
c,x^2(x-5)+(x-5)=0
(x^2+1)(x-5)=0
\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\in\Phi\\x=5\end{cases}}\)
24 tháng 7 2018
a) \(3x^2-3y^2-x-y\)
\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)
\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(\Leftrightarrow3\left(x-y\right)\)
d) \(3x^2-7x+4\)
\(\Leftrightarrow3x^2-7x+7-3\)
\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)
\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)
e) \(-2x^2+3x-1\)
\(\Leftrightarrow\left(-2x^2-1^2\right)+3x\)
\(\Leftrightarrow\left(-2x-1\right)\left(-2x+1\right)+3x\)
f) \(x^2+2xy+y^2-2x-2y\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
k) \(2x^2+5x+3\)
\(\Leftrightarrow2x^2+2x+3x+3\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)
l) \(x^2-2x-y^2+1\)
\(\Leftrightarrow\left(x^2-2x+1\right)-y^2\)
\(\Leftrightarrow\left(x-1\right)^2-y^2\)
\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)
24 tháng 7 2018
a) \(3x^2-3y^2-x-y\)
\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)
\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(\Leftrightarrow3\left(x-y\right)\)
d) \(3x^2-7x+4\)
\(\Leftrightarrow3x^2-7x+7-3\)
\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)
\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)
e) \(-2x^2+3x-1\)
\(\Leftrightarrow\left(-2x^2-1^2\right)+3x\)
\(\Leftrightarrow\left(-2x-1\right)\left(-2x+1\right)+3x\)
f) \(x^2+2xy+y^2-2x-2y\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
k) \(2x^2+5x+3\)
\(\Leftrightarrow2x^2+2x+3x+3\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)
l) \(x^2-2x-y^2+1\)
\(\Leftrightarrow\left(x^2-2x+1\right)-y^2\)
\(\Leftrightarrow\left(x-1\right)^2-y^2\)
\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)