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
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)
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
=> \(\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
b) (3x - 5)2 - 4 = 0
=> (3x - 5)2 = 0 + 4
=> (3x - 5)2 = 4
=> (3x - 5)2 = 22
=> \(\orbr{\begin{cases}3x-5=2\\3x-5=-2\end{cases}}\)
=> \(\orbr{\begin{cases}3x=7\\3x=3\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{7}{3}\\x=1\end{cases}}\)
b) ( 3x - 5 )2 - 4 = 0
=> ( 3x - 5 )2 = 4
=> ( 3x - 5 )2 = 22
=> 3x - 5 = 2
=> 3x = 7
=> x = 7/3
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 , biết :
a. \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
b. \(2x^3-50x=0\)
c.\(5x^2-4\left(x^2-2x+1\right)-5=0\)
d. \(x^3-x=0\)
e. \(27x^3-27x^2+9x-1=1\)
a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x+25=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
b) Ta có: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)
d) Ta có: \(x^3-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) Ta có: \(27x^3-27x^2+9x-1=1\)
\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)
\(\Leftrightarrow\left(3x-1\right)^3=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow3x=2\)
hay \(x=\dfrac{2}{3}\)
a \(\left(x-1\right)^2-\left(y+1\right)^2=0\)
\(x+3y-5=0\)
b \(xy-2x-y+2=0\)
3x+y=8
c \(\left(x+y\right)^2-4\left(x+y\right)=12\)
\(\left(x-y\right)^2-2\left(x-y\right)=3\)
d \(2x-y=1\)
\(2x^2+xy-y^2-3y=-1\)
a.
\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)
b.
\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
TH1:
\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
c.
\(\left\{{}\begin{matrix}\left(x+y\right)^2-4\left(x+y\right)-12=0\\\left(x-y\right)^2-2\left(x-y\right)=3\end{matrix}\right.\)
Xét pt:
\(\left(x+y\right)^2-4\left(x+y\right)-12=0\)
\(\Leftrightarrow\left(x+y+2\right)\left(x+y-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y+2=0\\x+y-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=-x-2\\y=6-x\end{matrix}\right.\)
TH1: \(y=-x-2\) thế vào \(\left(x-y\right)^2-2\left(x-y\right)=3\)
\(\Rightarrow\left(2x+2\right)^2-2\left(2x+2\right)=3\)
\(\Leftrightarrow4x^2+4x-3=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\Rightarrow y=-\dfrac{5}{2}\\x=-\dfrac{3}{2}\Rightarrow y=-\dfrac{1}{2}\end{matrix}\right.\)
TH2: \(y=6-x\) thế vào...
\(\left(2x-6\right)^2-2\left(2x-6\right)=3\)
\(\Leftrightarrow4x^2-28x+45=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\Rightarrow y=\dfrac{7}{2}\\y=\dfrac{9}{2}\Rightarrow y=\dfrac{3}{2}\end{matrix}\right.\)
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) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
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
a) \(=x^4-2x^3-3x^2+4x+4+x^2-4x+4\)
\(=x^4-2x^3-2x^2+8\)
\(=x^3\left(x-2\right)-2x\left(x-2\right)-4\left(x-2\right)\)
\(=\left(x^3-2x-4\right)\left(x-2\right)\)
\(=\left[x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\right]\left(x-2\right)\)
\(=\left(x-2\right)^2\left(x^2+2x+2\right)\)
b) \(=x^4-x+2019\left(x^2+x+1\right)\)
\(=x\left(x^3-1\right)+2019\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2019\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2019\right)\)\
c)\(x^4+2x^3+5x^2+4x-5\\=x^4+x^3+x^3-x^2+x^2+5x^2-x+5x-5\\ =x^2\left(x^2+x-1\right)+x\left(x^2+x-1\right)+5\left(x^2+x-1\right)=\left(x^2+x-1\right)\left(x^2+x+5\right)\)
Thực hiện phép tính
a, \(A=\left(3x^2y-11x^2-5y\right)\left(8xy-5x+6\right)\)
b,\(B=\left(-4x^2y-5x^2+3y^2\right)\left(2x^2-xy+3y^2\right)\)
c,\(C=5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(3x-1\right)\left(3x+1\right)\)
A= 3x2 y-11x2-5y.8xy-5+6
=(3-11-5.8-5+6).(x2.x2.x).(y.y.y)
=-47x5y3
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, \(x^2-2018x-2019=0\)