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Bài 2:
Ta có: \(n^4-5n^3-3n^2+17n-17⋮n-5\)
\(\Leftrightarrow n^4-5n^3-3n^2+15n+2n-10-7⋮n-5\)
\(\Leftrightarrow n-5\in\left\{1;-1;7;-7\right\}\)
hay \(n\in\left\{6;4;12;-2\right\}\)
3: Ta có: \(x^3-2x^2-x+m+2⋮x+3\)
\(\Leftrightarrow x^3+3x^2-5x^2-15x+14x+42+m-40⋮x+3\)
=>m-40=0
hay m=40
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: =>(x-1-2)(x-1+2)=0
=>(x+1)(x-3)=0
=>x=3 hoặc x=-1
b: =>(x-3)(2x-x-3)=0
=>(x-3)(x-3)=0
=>x=3
c: =>x^3-1=5x+x^3-5-x^2
=>-x^2+5x-5=1
=>-x^2+5x-6=0
=>x^2-5x+6=0
=>x=2 hoặc x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(\Leftrightarrow4x^2-4x+1-4x^2-16x-16=9\)
=>-20x-15=9
=>-20x=24
=>x=-6/5
b: \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)
=>9x=18
=>x=2
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bài 1:
a) 2m(x-y) + x-y = 2m(x-y) + (x-y) = (2m+1)(x-y)
b) x(y-2) + y\(^2\) - 2y = x(y-2) + y(y-2) = (x+y)(y-2)
c) x\(^2\) +xy - 2x - 2y = x(x+y) - 2(x+y) = (x-2)(x+y)
d) x + x\(^2\) - x\(^3\) - x\(^4\) = x(1 + x - x\(^2\) - x\(^3\))
e) 2+2x-xy-y = 2(1+x) - y(x+1) = (2-y)(x+1)
f) x\(^2\) + 2y - 1 -2x + 1 - y\(^2\) = (x\(^2\) -2x+1) - (y\(^2\)-2y+1) = (x-1)\(^2\) - (y-1)\(^2\)
g) (x+1)\(^2\) -x-1 = (x+1)\(^2\) -(x+1) =(x+1)(x+1-1) = (x+1)x
Bài 3:
\(P=x^2-4x+4+5=\left(x-2\right)^2+5>=5\)
Dấu = xảy ra khi x=2
Bài 4:
a: \(=-\left(x^2-4x-5\right)\)
\(=-\left(x^2-4x+4-9\right)\)
\(=-\left(x-2\right)^2+9< =9\)
Dấu = xảy ra khi x=2
b: \(=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}< =\dfrac{1}{4}\)
Dấu = xảy ra khi x=1/2
c: \(=x^2-6x+9+3=\left(x-3\right)^2+3>=3\)
Dấu '=' xảy ra khi x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
2) Bạn làm phép chia đa thức cho đa thức, kẻ hẳn dấu chia ra như tiểu học ấy. Được kết quả là \(\left(4y^2+1\right)\) dư (-2y+6) nhé.
3) a) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow x=\pm3\)
b) \(\left(x^2+1\right)\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow x^2+1=0\) hoặc x-3=0 hoặc x+2=0
Trường hợp 1 loại vì \(x^2\) không âm, hai trường hợp còn lại tìm được x=3 và x = -2.
4) a)\(x^2-y^2+2y-1=x^2-\left(y^2-2y+1\right)=x^2-\left(y-1\right)^2=\left(x-y+1\right)\left(x+y-1\right)\)
b) \(5x^2-10xy-20z^2+5y^2\)
= \(5\left(x^2-2xy-4z^2+y^2\right)\)
= \(5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
= 5 ( x-y-2z ) ( x-y+2z )
5) \(x^3=x\Leftrightarrow x=\pm1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1.
a) ( x - 2)2 - ( x + 3)( x - 3)= 17
=> x2 - 4x + 4 - x2 + 9 - 17 = 0
=> -4x - 4 = 0
=> -4( x + 1 ) = 0
=> x = -1
Vậy,...
b)4( x - 3)2 - ( 2x - 1)( 2x + 1) = 10
=> 4( x2 - 6x + 9) - 4x2 + 1 - 10 = 0
=> - 24x + 36 - 9 = 0
=> -24x + 27 = 0
=> -3( 8x - 9) = 0
=> x = \(\dfrac{9}{8}\)
Vậy,...
c) ( x - 4)2 - ( x - 2)( x + 2)= 36
=> x2 - 8x + 16 - x2 + 4 - 36 = 0
=> -8x - 16 = 0
=> -8( x + 2) = 0
=> x = -2
d) ( 2x + 3)2 - ( 2x + 1)( 2x - 1) = 10
=> 4x2 + 12x + 9 - 4x2 + 1 - 10 = 0
=> 12x = 0
=> x = 0
Vậy,...
Bài 2.
\(\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}\)
a) ĐKXĐ : ( x + 1)( 2x - 6) # 0
=> 2( x + 1)( x - 3) # 0
=> x # -1 ; x # 3
Vậy,...
b) Để P = 1
=> \(\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=1\)
=> \(\dfrac{3x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{3x}{2\left(x-3\right)}=1\)
=> 3x = 2x - 6
=> x = -6 ( thỏa mãn ĐKXĐ)
Vậy,...
Bài 3.
P = \(\dfrac{x}{x-1}+\dfrac{x^2+1}{1-x^2}\)
a) Để P có nghĩa tức P xác định .
ĐKXĐ : x - 1 # 0 => x # 1
* 1 - x2 # 0 => x # 1 ; x # -1
Vậy,...
b) P = \(\dfrac{x}{x-1}+\dfrac{x^2+1}{1-x^2}\)
P = \(\dfrac{x^2+x-x^2-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x+1}\)( x# 1; x# -1)
c) Để P = -1 thì :
\(\dfrac{1}{x+1}=-1\)
=> -x - 1 = 1
=> x = -2 ( thỏa mãn ĐKXĐ )
Vậy,...
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 6:
a) \(x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b) \(5x\left(x-3\right)-x+3=0\)
\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
c) \(3x\left(x-5\right)-\left(x-1\right)\left(2+3x\right)=30\)
\(\Leftrightarrow3x^2-15x-2x-3x^2+2+3x=30\)
\(\Leftrightarrow-14x+2=30\)
\(\Leftrightarrow-14x=28\)
\(\Leftrightarrow x=-2\)
d) \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)
\(\Leftrightarrow2x+16=0\)
\(\Leftrightarrow2x=-16\)
\(\Leftrightarrow x=-8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
\(VT=\left(a-2\right)\left(a^2+2a+4\right)\left(a-1\right)\left(a^2+a+1\right)\)
\(=\left(a^3-8\right)\left(a^3-1\right)\)
\(=a^6-9a^3+8\)
Bài 3:
\(\Leftrightarrow x^3+8-x\left(x^2-9\right)=26\)
\(\Leftrightarrow x^3+8-x^3+9x=26\)
=>9x=18
hay x=2