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3 tháng 9 2019
-3x^3+5x^2-9x+15 -3x-5 x^2 -3x^3-5x^2 - 10x^2-9x+15 -(10/3)x 10x^2+(50/3)x - -(23/3)x+15 +23/9 -(23/3)x-115/9 - 250/9
Chả biết có sai ko @@
3 tháng 9 2019
x^4-2x^3 +2x-1 x^2-1 x^2-2x x^4 -x^2 - -2x^3+x^2+2x-1 -2x^3 +2x - x^2-1 +1 x^2-1 - 0
AQ
Thực hiện phép tính:
a. ( 2x - 6 )( 12x2 + 9x + 36 )
b. ( 2x4 + x3 - 3x2 + 5x - 2 ) : ( 5x2 - 5x + 5 )
0
30 tháng 4 2017
a) \(45+x^3-5x^2-9x\)
\(\Leftrightarrow\left(45-9x\right)+\left(x^3-5x^2\right)\)
\(\Leftrightarrow-9\left(x-5\right)+x^2\left(x-5\right)\)
\(\Leftrightarrow\left(x-5\right)\left(x-3\right)\left(x+3\right)\)
TK NKA !!!
3 tháng 10 2017
a. \(5x^2\left(2x-3\right)+\left(2x^2+3x+3\right)\left(3-2x\right)=6x^3-9x^2\Leftrightarrow5x^2\left(2x-3\right)-\left(2x^2+3x+3\right)\left(2x-3\right)=3x^2\left(2x-3\right)\Leftrightarrow5x^2\left(2x-3\right)-\left(2x^2+3x+3\right)\left(2x-3\right)-3x^2\left(2x-3\right)=0\Leftrightarrow\left[5x^2-\left(2x^2+3x+3\right)-3x^2\right]\left(2x-3\right)=0\Leftrightarrow\left(5x^2-2x^2-3x-3-3x^2\right)\left(2x-3\right)=0\Leftrightarrow\left(2x-3\right)\left(-3x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\-3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\-3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\end{matrix}\right.\)
b. \(\left(4x^2+2x\right)\left(x^2-x\right)+\left(4x^2+6\right)\left(x-x^2\right)=0\Leftrightarrow\left(4x^2+2x\right)\left(x^2-x\right)-\left(4x^2+6\right)\left(x^2-x\right)=0\Leftrightarrow\left(x^2-x\right)\left[\left(4x^2+2x\right)-\left(4x^2+6\right)\right]=0\Leftrightarrow\left(x^2-x\right)\left(4x^2+2x-4x^2-6\right)=0\Leftrightarrow x\left(2x-6\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-6=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=6\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=0\\x=3\\x=1\end{matrix}\right.\)
DT
29 tháng 9 2018
a ) \(5x^2\left(2x-3\right)+\left(2x^2+3x+3\right)\left(3-2x\right)=6x^3-9x^2\)
\(\Rightarrow5x^2\left(2x-3\right)-\left(2x^2+3x+3\right)\left(2x-3\right)=3x^2\left(2x-3\right)\)
\(\Rightarrow5x^2\left(2x-3\right)-\left(2x^2+3x+3\right)\left(2x-3\right)-3x^2\cdot\left(2x-3\right)=0\)
\(\Rightarrow\left(5x^2-2x^2-3x-3-2x+3\right)\left(2x-3\right)=0\)
\(\Rightarrow\left(3x^2-5x\right)\left(2x-3\right)=0\)
\(\Rightarrow x\left(3x-5\right)\left(2x-3\right)=0\)
\(\Rightarrow\) +) x = 0
+) 3x - 5 = 0\(\Rightarrow x=\dfrac{5}{3}\)
+ )\(2x-3=0\Rightarrow x=\dfrac{3}{2}\)
vậy x \(=0;x=\dfrac{3}{2};x=\dfrac{5}{3}\)
b) \(8x^3+12x^2+6x+7-3\left(2x+1\right)^2=6\)
\(\Rightarrow\left(2x\right)^3+3.2x.1+3.2x.1^2+1^2+6-3\left(2x+1\right)^2-6=0\)
\(\Rightarrow\left(2x+1\right)^3-3\left(2x+1\right)^2=0\)
\(\Rightarrow\left(2x+1\right)^2\left(2x+1-3\right)=0\)
\(\Rightarrow\left(2x+1\right)^2\left(2x-2\right)=0\Rightarrow\left(2x+1\right)^2\left(x-1\right)2=0\)
\(\Rightarrow\) +)\(\left(2x+1\right)^2=0\Rightarrow2x+1=0\Rightarrow x=\dfrac{-1}{2}\)
+) x - 1 = 0 \(\Rightarrow x=1\)
Vậy x = \(\dfrac{-1}{2}\) hoặc x = 1
12 tháng 10 2020
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
1 tháng 11 2019
A.5X2+3(X+Y)2-5Y2
= 5.(x2 - y2) + 3(x+y)2
=5.( x-y).(x+y) +3(x+y)2
= ( x+y).[(5.( x-y) + 3(x+y)]
=( x+y).( 5x-5y +3x + 3y)
=( x+y).( 8x - 2y)
=( x+y).2(4x - y)
C.81x2 - 6yz - 9y2 - z2
=( 9x)2 - (z2 + 6yz+ 9y2)
=( 9x)2 -( z +3y)2
=(9x - z -3y).(9x + z+3y)
D.x2y-x3-9y+9x
=9.(x - y) + x2(y - x)
= 9.(x-y) - x2(x-y)
=(x+y).[32- x2]
=(x+y).(3-x).(x+2)
E.x3+9x2-4x-36
=(x3-4x) + (-36 +9x2)
=x.(x2-4) + 9.(x2-4)
=(x2-4).(x+9)
H.2x3+x2-8x-4
= x2.(2x+1)-4.(2x +1)
=(2x +1).(x2 -4)
= (2x +1).(x-2).(x+2)
Cậu xem trước nhé
\(5x^3-9x^2=2x\)
\(\Leftrightarrow x\left(5x^2-9x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-\dfrac{1}{5}\end{matrix}\right.\)