D
2
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
23 tháng 8 2017
Câu 1:
Ta có:\(x\left(x^2-y\right)+x\left(y^2-y\right)-x\left(x^2+y^2\right)\)
\(=x\left(x^2-y+y^2-y-x^2-y^2\right)\)
\(=-2xy\)
Tại \(x=\frac{1}{2};y=-100\) PT có dạng:
\(=-2.\frac{1}{2}.\left(-100\right)=100\)
TA
26 tháng 7 2017
a) \(\left(x-3\right).\left(x^2+3x+9\right)-x.\left(x+4\right)\left(x-4\right)=21\)
\(\Leftrightarrow x^3-27-x.\left(x^2-16\right)=21\) \(\Leftrightarrow x^3-27-x^3+16x=21\)
\(\Leftrightarrow16x=21+27\) \(\Leftrightarrow16x=48\) \(\Leftrightarrow x=3\)
b) \(\left(x+2\right)\left(x^2-2x+4\right)-x.\left(x^2+2\right)=4\)
\(\Leftrightarrow x^3+8-x^3-2x=4\) \(\Leftrightarrow-2x=4-8\) \(\Leftrightarrow-2x=-4\) \(\Leftrightarrow x=2\)
17 tháng 3 2020
\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
\(x^3-2x^2+4x+2x^2-4x+8-x^3+2x=15\)
\(2x+8=15\)
\(2x=7\)
\(x=\frac{7}{2}\)
17 tháng 3 2020
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)
\(\Leftrightarrow9x+7=17\)
\(\Leftrightarrow9x=10\)
\(\Leftrightarrow x=\frac{10}{9}\)
TM
3 tháng 7 2017
3) \(\left(x-3\right)\left(x+3\right)\left(x^2+9\right)-\left(x^2-2\right)\left(x^2+2\right)\)
\(=\left(x^2-9\right)\left(x^2+9\right)-\left(x^4-4\right)\)
\(=\left(x^4-81\right)-\left(x^4-4\right)\)
\(=x^4-81-x^4+4\)
=-77 =>đpcm
4)\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2\)
\(=\left(3x+1-3x-5\right)^2\)
=(-4)2
=16 => đpcm
TM
3 tháng 7 2017
1)\(\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)=\left(x^2-4x+4\right)-\left(x^2-4x+3\right)=1\)
=>đpcm
2)\(\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)
\(=\left(x-1-x-1\right)\left[\left(x-1\right)^2+\left(x-1\right)\left(x+1\right)+\left(x+1\right)^2\right]+6\left(x^2-1\right)\)
\(=\left(-2\right)\left(x^2-2x+1+x^2-1+x^2+2x+1\right)+6x^2-6\)
\(=\left(-2\right)\left(3x^2+1\right)+6x^2-6=-6x^2-2+6x^2-6=-8\) => đpcm
SM
19 tháng 9 2018
Bài 1 :
1) 4x2 - y2 = ( 2x + y ) ( 2x - y )
2) 9x2 - 4y2 = ( 3x - 2y ) ( 3x + 2y )
3) 4x2 + y2 + 4xy = ( 2x + y )2
Bài 2:
1) 2x2 + 8x = 0
=> 2x ( x + 4 ) = 0
=> \(\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
2) 3 ( x - 4 ) + x2 - 4x = 0
=> 3 ( x - 4 ) + x ( x - 4 ) = 0
=> ( x - 4 ) ( 3 + x ) = 0
=> \(\orbr{\begin{cases}x-4=0\\3+x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)
3) 3 ( x - 2 ) = x2 - 2x
=> 3 ( x - 2 ) - x2 + 2x = 0
=> 3 ( x - 2 ) - x ( x - 2 ) = 0
=> ( x - 2 ) ( 3 - x ) = 0
=> \(\orbr{\begin{cases}x-2=0\\3-x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
4) x ( x - 2 ) - 6 ( 2 - x ) = 0
=> x ( x - 2 ) + 6 ( x - 2 ) = 0
=> ( x - 2 ) ( x + 6 ) = 0
=> \(\orbr{\begin{cases}x-2=0\\x+6=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)
5) 2x ( x + 5 ) = x2 + 5x
=> 2x ( x + 5 ) - x2 - 5x = 0
=> 2x ( x + 5 ) - x ( x + 5 ) = 0
=> ( x + 5 ) ( 2x - x ) = 0
=> \(\orbr{\begin{cases}x+5=0\\2x-x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-5\\x=0\end{cases}}\)
6 ) ( x - 2 )2 - x ( x + 3 ) = 9
=> x2 - 4x + 4 - x2 - 3x = 9
=> - 7x + 4 = 9
=> - 7x = 5
=> x = \(-\frac{5}{7}\)
D
19 tháng 9 2018
\(1,4x^2-y^2=\left(2x\right)^2-y^2=\left(2x-y\right)\left(2x+y\right)\)
\(2,9x^2-4y^2=\left(3x\right)^2-\left(2y\right)^2=\left(3x-2y\right)\left(3x+2y\right)\)
\(3,4x^2+y^2+4xy=\left(2x\right)^2+2.2x.y+y^2=\left(2x+y\right)^2\)
\(1,2x^2+8x=0\Rightarrow2x\left(x+4\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
\(2,3\left(x-4\right)+x^2-4x=0\)
\(\Rightarrow3\left(x-4\right)+x\left(x-4\right)=0\)
\(\Rightarrow\left(3+x\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3+x=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
\(3,3\left(x-2\right)=x^2-2x\)
\(\Rightarrow3\left(x-2\right)-x^2+2x=0\)
\(\Rightarrow3\left(x-2\right)-x\left(x-2\right)=0\)
\(\Rightarrow\left(3-x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3-x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
\(4,x\left(x-2\right)-6\left(2-x\right)=0\)
\(\Rightarrow x\left(x-2\right)+6\left(x-2\right)=0\)
\(\Rightarrow\left(x+6\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+6=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-6\\x=2\end{cases}}\)
DV
2
29 tháng 7 2017
x + 1 = ( x + 1 )2
x + 1 = x2 + 2x + 1
x - 2x - x2 = - 1 + 1
- x - x2 = 0
- x ( x + 1) = 0
TH1: - x = 0 suy ra x = 0
TH2: x + 1 = 0 suy ra x = - 1
Vậy x = 0 hoặc x = - 1.
13 tháng 8 2020
Dài quá ! Nên vẫn phải làm ^_^.
Bài 1:
+) \(A=x^2-2x+6=x^2-2x+1+5=\left(x-1\right)^2+5\ge5\)
Min A = 5 \(\Leftrightarrow x=1\)
+) \(B=x^2+6x+12=x^2+6x+9+3=\left(x+3\right)^2+3\ge3\)
Min B = 3 \(\Leftrightarrow x=-3\)
+) \(C=4-x^2+2x=-\left(x^2-2x+4\right)=-\left[\left(x-1\right)^2+3\right]=-\left(x-1\right)^2-3\le-3\)
Max C = -3 \(\Leftrightarrow x=1\)
+) \(D=-x^2+6x=-\left(x^2-6x+9-9\right)=-\left(x-3\right)^2+9\le9\)
Max D = 9 \(\Leftrightarrow x=3\)
13 tháng 8 2020
Bài 2 :
a) \(x^2-x-3x+3=0\)
\(\Leftrightarrow x^2-4x+4-1=0\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
b) \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
c) Xem lại đề hộ mình nha
d) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\in\left\{0;-3;2\right\}\)
( x + 3 ).( x2 - 3x + 9 ) - x.( x - 2 ).( x + 2 ) = 15
x3 - 3x2 + 9x + 3x2 - 9x + 27 - x.( x2 - 4 ) = 15
x3 - x3 + 4x = 15 - 27
4x = -12
x = -3
Vậy x = - 3
Lời giải
\(x^3-3x^2+9x+3x^2-9x+27-x.\left(x^2-4\right)\)
\(x^3-x^3+4x=15-27\)
\(4x=-12\)
\(x=-12:4\)
\(x=-3\)