I : PTĐTTNT
a) \(\left(x^2-x-2\right)^2+\left(x-2\right)^2\)
b) \(x^4+2019x^2+2018x+2019\)
c) \(x^4+2x^3+5x^2+4x-5\)
help me
Những câu hỏi liên quan
Phân tích đa thức thành nhân tử
a) \(4x^2-12xy+5x^2\)
b) \(\left(x+y+2z\right)^2+\left(x+y-z\right)^2-9z^2\)
c) \(x^4+2019x^2+2018x+2019\)
toán lớp một mà mình lớp 5 ko giải đc :v
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Xem thêm câu trả lời
tìm x biết : a, \(x^2-5x=0\)
b, \(\left(3x-5\right)^2-4=0\)
c, 2018x - 1 + 2019x( 1-2018x )=0
d, \(\left(x+2\right)^3-x^2\left(x-6\right)=8\)
e, ( 1 - 2x ) ( 1 + 2x ) - x( x+2 ) (x - 2 ) = 0
x2 - 5x = 0
=> x(x - 5) = 0
=> \(\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
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b) (3x - 5)2 - 4 = 0
=> (3x - 5)2 = 0 + 4
=> (3x - 5)2 = 4
=> (3x - 5)2 = 22
=> \(\orbr{\begin{cases}3x-5=2\\3x-5=-2\end{cases}}\)
=> \(\orbr{\begin{cases}3x=7\\3x=3\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{7}{3}\\x=1\end{cases}}\)
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b) ( 3x - 5 )2 - 4 = 0
=> ( 3x - 5 )2 = 4
=> ( 3x - 5 )2 = 22
=> 3x - 5 = 2
=> 3x = 7
=> x = 7/3
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Phân tích các đa thức sau thành nhân tử:
a) \(x^5+x+1\)
b) \(x^4+2019x^2+2018x+2019\)
c) \(\left(x^2-2x+4\right)\left(x^2+3x+4\right)-14x^2\)
d) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
tìm x
a, \(2018x-1+2019x\left(1-2018x\right)=0\)
b, \(\left(x+2\right)^3-x^2\left(x-6\right)=4\)
c, \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
d, \(x^2-5x+4=0\)
Câu a):
ta có (x2-x-2)2+(x-2)2
=((x-2)2(x+1))2+(x-2)2
=(x-2)2(x2+2x+2)
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https://i.imgur.com/DxE4HlK.jpg
https://i.imgur.com/t7lbF4T.jpg
1.a.x\(^2-3x+xy-3y\)
b.\(16\left(2x+3\right)^2-9\left(5x-2\right)^2\)
2.tìm x biết
a.2018x-1+2019x(1-2018x)=0
b.(x+2)\(^3-x^2\left(x-6\right)-4\)
Tìm x biết :
a, 2x ( x - 3 ) = \(\left(3-x\right)^2\)
b, \(x^3-49x=0\)
c, \(\left(x+2\right)^2+x^2-4=0\)
d, \(5x^2-5=4\left(x^2-2x+1\right)\)
e, \(x^2-2018x-2019=0\)
I : PTĐTTNT
A= \(\left(x^2-6x\right)^2-2\left(x-3\right)^2-81\)
B=\(x^4+y^4+z^4-2x^2y^2-2y^2z^2-2x^2z^2\)
help me !!!
\(\left(x^2-6x\right)^2-2\left(x-3\right)^2-81=\left[\left(x^2-6x\right)^2-81\right]-2\left(x-3\right)^2=\left[\left(x^2-6x\right)^2-9^2\right]-2\left(x-3\right)^2=\left(x^2-6x+9\right)\left(x^2-6x-9\right)-2\left(x-3\right)^2=\left(x-3\right)^2\left(x^2-6x-9\right)-2\left(x-3\right)^2=\left(x-3\right)^2\left(x^2-6x+11\right)\)
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=\(\left(x-3\right)^2\left(x^2-6x-11\right)\)
nha
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Tìm x, biết:a, 2x^3-x^2+2x-10b, 2018x-1+2019xleft(1-2018xright)0c,left(x+2right)^3-x^2left(x-6right)-40d,6x^2-left(2x-3right)left(3x+2right)1e,left(x+1right)^3-left(x-1right)left(x^2+x+1right)-20g,7x^2+2x0h,xleft(x+4right)-x^2-6x10i,xleft(x-1right)+2x-20k,left(3x-1right)^2-left(x+5right)^20l,xleft(2x-3right)-2left(3-2xright)0
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Tìm x, biết:
a, \(2x^3-x^2+2x-1=\)0
b, \(2018x-1+2019x\left(1-2018x\right)=0\)
c,\(\left(x+2\right)^3-x^2\left(x-6\right)-4=0\)
d,\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\)
e,\(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-2=0\)
g,\(7x^2+2x=0\)
h,\(x\left(x+4\right)-x^2-6x=10\)
i,\(x\left(x-1\right)+2x-2=0\)
k,\(\left(3x-1\right)^2-\left(x+5\right)^2=0\)
l,\(x\left(2x-3\right)-2\left(3-2x\right)=0\)
Tìm x:
a.\(5x-3\left\{4x-2\left[4x-3(5x-2)\right]\right\}=182\)
b.\(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
c.\(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
Giúp mk vs!!
b) \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)+\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3+x+5\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x+8\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\2x+8=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)
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chán òi ko lm nữa đâu có ng đg bùn mk cx bùn theo lun xl nha
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