Đề 1
Bài 1 :phân tích đa thức sau thành nhân tử
a)8x-4y
b)5y(x-1)-3x(x-1)
c)-6x-6y+x(x+y)
d)x2-49
Bài 2:tìm x biết
a)5x2-x=0
b)x2-25=0
c)x4+2x3+6x-9=0
Đề 2
Bài 1:phân tích đa thức sau thành nhân tử
a)8x-4y
b)5x(y-5)-3y(y-5)
c)-5a-5b+c(a+b)
d)a2-64
Bài 2:tìm x biết
a)5y2-2y=0
b)y2-81=0
c)y4+2y3+6y-9=0
giúp mk luông với nha mai mk phải nộp rồi
Đề 1 :
Bài 1 .a) 8x -4y = 4( 2x -y)
b) 5y.(x-1) - 3x.(x-1) = ( x-1).(5y-3x)
c) -6x -6y + x(x+y) = -6.( x+y) + x(x+y) = ( x -6).(x+y)
d) x2 - 49 = x2 - 72 = ( x-7).(x+7)
Bài 2. a) 5x2 -x = 0
x( 5x -1)=0
*) x =0
*) 5x = 1 -> x = \(\dfrac{1}{5}\)
b) x2 - 25 =0
x2 - 52 =0
( x-5).(x+5)=0
*) x =5
*) x = -5
c) x4 + 2x3 + 6x -9 =0
(x2)2 - 32 + 2x( x2 + 3) =0
( x2 +3).(x2 - 3) + 2x( x2 + 3) =0
( x2 + 3).( x2 +2x -3) =0
*) x2 = -3 -> x vô nghiệm
*) x2 + 2x - 3 =0 -> x2 + 2x + 6 -9 =0
-> ( x+3).(x-3) + 2.(x + 3) =0
-> ( x+ 3).( x - 1) =0
-> x = -3 ; x =1
Đề 1
Bài 1:Phân tích đa thức sau thành nhân tử
a) \(\text{8x-4y}\)
\(=4.\left(2x-y\right)\)
b)\(\text{5y(x-1)-3x(x-1)}\)
\(=\left(x-1\right).\left(5y-3x\right)\)
c)\(\text{-6x-6y+x(x+y)}\)
\(=-6.\left(x+y\right)+x.\left(x+y\right)\)
\(=\left(x-6\right)\left(x+y\right)\)
d)\(x^2-49\)
\(=\left(x-7\right)\left(x+7\right)\)
Bài 2:Tìm x biết
a)\(5x^2-x=0\)
\(\Leftrightarrow x.\left(5x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\5x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)
b)\(x^2-25=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
c)\(x^4+2x^3+6x-9=0\)
\(\Leftrightarrow\left(x^2\right)^2+2x\left(x^2+3\right)-3^2=0\)
\(\Leftrightarrow\left(x^2+3\right)\left(x^2-3\right)+2x\left(x^2+3\right)=0\)
\(\Leftrightarrow\left(x^2+3\right)\left(x^2-3+2x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+3=0\\x^2+2x-3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=-3\left(khongthoaman\right)\\x^2+3x-x-3\end{matrix}\right.\)
\(\Rightarrow x\left(x-1\right)+3.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Bài 1:
a, \(8x-4y=4\left(2x-y\right)\)
b, \(5x\left(y-5\right)-3y\left(y-5\right)=\left(y-5\right)\left(5x-3y\right)\)
c, \(-5a-5b+c\left(a+b\right)=-5\left(a+b\right)+c\left(a+b\right)\)
\(=\left(a+b\right)\left(c-5\right)\)
d, \(a^2-64=a^2-8^2=\left(a-8\right)\left(a+8\right)\)
Bài 2:
Giống:
a, \(y^2-81=0\Rightarrow\left(y-9\right)\left(y+9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y-9=0\\y+9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
b, \(y^4+2y^3+6y-9=0\)
\(\Rightarrow\left(y^4-9\right)+\left(2y^3+6y\right)=0\)
\(\Rightarrow\left(y^2-3\right)\left(y^2+3\right)+2y\left(y^2+3\right)=0\)
\(\Rightarrow\left(y^2+3\right)\left(y^2+2y-3\right)=0\)
\(\Rightarrow\left(y^2+3\right)\left(y^2-y+3y-3\right)=0\)
\(\Rightarrow\left(y^2+3\right)\left[y\left(y-1\right)+3\left(y-1\right)\right]=0\)
\(\Rightarrow\left(y^2+3\right)\left(y-1\right)\left(y+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y^2+3=0\\y-1=0\\y+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y\in\varnothing\\y=1\\y=-3\end{matrix}\right.\)
Khác:
a, \(5x^2-x=0\Rightarrow x\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)
b, \(5y^2-2y=0\Rightarrow y\left(5y-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y=0\\5y-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{2}{5}\end{matrix}\right.\)
Chúc bạn học tốt!!!
Đề 1
Câu 1:
\(\text{a) }8x-4y=4\left(2x-y\right)\)
\(\text{b) }5y\left(x-1\right)-3x\left(x-1\right)=\left(5y-3x\right)\left(x-1\right)\)
\(\text{c) }-6x-6y+x\left(x+y\right)\\ =-\left(6x+6y\right)-x\left(x+y\right)\\ =-6\left(x+y\right)+x\left(x+y\right)\\ =\left(x-6\right)\left(x+y\right)\)
\(\text{d) }x^2-49=\left(x-7\right)\left(x+7\right)\)
Câu 2:
\(\text{a) }5x^2-x=0\\ \Leftrightarrow5x\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ \text{Vậy }x=0\text{ hoặc }x=1\)
\(\text{b) }x^2-25=0\\ \Leftrightarrow x^2=25\\ \Leftrightarrow x=\pm5\\ \text{Vậy }x=\pm5\)
\(\text{c) }x^4+2x^3+6x-9=0\\ \Leftrightarrow x^4+3x^3-x^3+9x-3x-9+3x^2-3x^2=0\\ \Leftrightarrow\left(x^4+3x^3+3x^2+9x\right)-\left(x^3+3x^2+3x+9\right)=0\\ \Leftrightarrow x\left(x^3+3x^2+3x+9\right)-\left(x^3+3x^2+3x+9\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^3+3x^2+3x+9\right)=0\\ \Leftrightarrow\left(x-1\right)\left[\left(x^3+3x^2\right)+\left(3x+9\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left[x^2\left(x+3\right)+3\left(x+3\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(TM\right)\\x^2=-3\left(Vô\text{ }lí\right)\\x=-3\left(TM\right)\end{matrix}\right.\\ \text{Vậy }x=1\text{ }\text{hoặc }x=-3\)
Đề 2 .
Bài 1 . a) 8x - 4y = 4( 2x - y)
b)5x( y -5) - 3y( y -5) = ( 5x - 3y).(y-5)
c) -5a - 5b +c.( a+b) = -5.( a+b) + c.( a+b) = ( c -5).(a+b)
d) a2 - 64 = a2 - 82 = ( a-8).( a+8)
Bài 2. a) 5y2 - 2y =0
y( 5y -2) =0
*) y =0
* y= \(\dfrac{2}{5}\)
b) y2 - 81 =0
y2 - 92 =0
( y+9).(y-9) =0
*) y = -9
*) y =9
c) y4 + 2y3 + 6y -9 =0
( y2)2 - 32 + 2y( y2 + 3) =0
( y2 - 3).( y2 + 3)+ 2y( y2 + 3) =0
( y2 + 3).( y2 - 3 + 2y) = 0
*) y2 = -3 -> y vô nghiệm
*) y2 - 3 + 2y =0 -> y2 - 9 + 6 + 2y=0
-> ( y-3).(y+3) + 2( y +3) =0
-> ( y+3).(y - 1) =0
-> y = -3 ; y=1
Đề 2. Tương tự (đề 1) bạn luyện tập làm thử nhé!
