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B3) a) x(x-5)-4(x-5)=0
<=> (x-4)(x-5)=0
TH1 :x-4=0
<=.x=4
TH2 : x-5=0
<=>x=5
b) x(x-6)-7x-42=0
<=>x(x+6)-7(x+6)=0
<=>(x-7)(x+6)=0
th1;x-7=0
<=>x=7
th2; x+6=0
<=>x=-6
c)x^3-5x^2+x-5=0
<=> x(x^2+1)-5(x^2+1)=0
<=> (x-5)(x^2+1)=0
th1:x-5=0
<=>x=5
TH2 : x^2+1=0
<=> x^2=-1 ( vo li )
=> th2 ko tồn tại
nho thick nha
Bài 3
a, x(x-5)-4(x-5)=0
(x-4)(x-5)=0
=>\(\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
b,x(x+6)-7(x+6)=0
(x-7)(x+6)=0\(\Rightarrow\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}\)
c,x^2(x-5)+(x-5)=0
(x^2+1)(x-5)=0
\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\in\Phi\\x=5\end{cases}}\)
\(1,=\left(x-2\right)\left(5-y\right)\\ 2,=2\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(2x-2y-z\right)\\ 3,=5xy\left(x-2y\right)\\ 4,=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-4z^2\right]\\ =3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\\ 6,=-\left(6x^2-3x-4x+2\right)=-\left(2x-1\right)\left(3x-2\right)\\ 7,=\left(2x+y\right)\left(2x+y+x\right)=\left(2x+y\right)\left(3x+y\right)\\ 8,=\left(x-y\right)\left(x+5\right)\\ 9,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 10,=\left(x^2-9\right)x=x\left(x-3\right)\left(x+3\right)\\ 11,=\left(x-2\right)\left(y+1\right)\\ 12,=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\\ 13,=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)
\(\text{Tìm x:}\)
\(a.x\left(x-1\right)-3x+3x=0\)
\(x\left(x-1\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}}\)
\(b.3x\left(x-2\right)+10-5x=0\)
\(3x^2-6x+10-5x=0\)
\(3x^2-11x+10=0\)
\(3x^2-11x=-10\)(bn xem lại đề nhé)
\(c.x^3-5x^2+x-5=0\)
\(x^3-5x^2+x=5\)
\(d.x^4-2x^3+10x^2-20x=0\)
bài 1:phân tích thành phân tử
a> x^2-6x-y^2+9
= (x-3)^2 -y^2
= (x-3 -y) (x-3+y)
b>x^2-xy-8x+8y
= x(x-y) - 8(x-y)
= (x-8) (x-y)
c>25-4x^2-4xy-y^2
= 5^2 - (2x + y)^2
= (5 - 2x -y) (5 +2x+y)
d>xy-xz-y+z
= x(y-z) - (y-z)
= (x-1) (y-z)
e>x^2-xz-yz+2xy+y^2
= (x+y)^2 - z(x+y)
= (x+y-z) (x+y)
g>x^2-4xy+4y^2-z^2-4zt-4t^2
= (x-2y)^2 - (z + 2t)^2
= (x-2y -x-2t) (x-2y + z +2t)
bài 2:tìm X bt
a>x.(x-1)-3x+3x=0
x (x-1) =0
\(\Rightarrow\hept{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy x=0 và x=1
b>3x.(x-2)+10-5x=0
3x(x-2) - 5 (x-2)=0
(3x-5) (x-2) =0
\(\Rightarrow\hept{\begin{cases}3x-5=0\\x-2=0\end{cases}\Rightarrow\hept{\begin{cases}3x=5\\x=2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}}}\)
c>x^3-5x^2+x-5=0
x^2 (x-5) + (x-5) =0
(x^2 +1)(x-5) =0
\(\Rightarrow\hept{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Rightarrow\hept{\begin{cases}x^2=-1\\x=5\end{cases}\Rightarrow}\hept{\begin{cases}x\in\varphi\\x=5\end{cases}}}\)
Vậy x=5
d>x^4-2x^3+10x^2-20x=0
x^3 (x-2) + 10x(x-2) =0
(x^3 + 10x) (x-2) =0
x(x^2 + 10) (x-2) =0
\(\Rightarrow\hept{\begin{cases}x=0\\x^2+10=0\\x-2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=-10\\x=2\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x\in\varphi\\x=2\end{cases}}}}\)
Vậy x=0 và x=2
Thời gian có hạn copy cái này hộ mình vào google xem nha: :
Link : https://lazi.vn/quiz/d/16491/nhac-edm-la-loai-nhac-the-loai-gi
Vào xem xong các bạn nhận được 1 thẻ cào mệnh giá 100k nhận thưởng bằng cách nhắn tin vs mình và 1 phần thưởng bí mật là chiếc áo đá bóng,....
Có 500 giải nhanh nha đã có 200 người nhận rồi. Mình là phụ trách
OK N
x2 - x - y2 - y
= (x - y)(x + y) - (x + y)
= (x + y)(x - y - 1)
***
9x2 + y2 - 16z2 + 6xy
= (3x + y)2 - (4z)2
= (3x + y - 4z)(3x + y + 4z)
***
a3 - a2x - ay + xy
= a2(a - x) - y(a - x)
= (a - x)(a2 - y)
***
2x2 - 8y2 + 3x + 6y
= 2(x2 - 4y2) + 3(x + 2y)
= 2(x - 2y)(x + 2y) + 3(x + 2y)
= (x + 2y)(2x - 4y + 3)
***
xy(x + y) + yz(y + z) + xz(x + z) + 2xyz
= xy(x + y + z) + yz(x + y + z) + xz(x + z)
= y(x + y + z)(x + z) + xz(x + z)
= (x + z)(xy + y2 + yz + xz)
= (x + z)[y(x + y) + z(x + y)]
= (x + z)(x + y)(y + z)
A = xy + y - 2x - 2
= y( x + 1 ) - 2( x + 1 )
= ( x + 1 )( y - 2 )
B = x2 - 3x + xy - 3y
= x( x - 3 ) + y( x - 3 )
= ( x - 3 )( x + y )
C = 3x2 - 3xy - 5x + 5y
= 3x( x - y ) - 5( x - y )
= ( x - y )( 3x - 5 )
D = xy + 1 + x + y
= y( x + 1 ) + ( x + 1 )
= ( x + 1 )( y + 1 )
E = ax - bx + ab - x2
= ( ax - x2 ) + ( ab - bx )
= x( a - x ) + b( a - x )
= ( a - x )( x + b )
F = x2 + ab + ax + bx
= ( ax + x2 ) + ( ab + bx )
= x( a + x ) + b( a + x )
= ( a + x )( x + b )
G = a3 - a2x - ay + xy
= a2( a - x ) - y( a - x )
= ( a - x )( a2 - y )
Bonus : = ( a - x )[ a2 - ( √y )2 ]
= ( a - x )( a - √y )( a + √y )
H = 2xy + 3z + 6y + xz
= ( 6y + 2xy ) + ( 3z + xz )
= 2y( 3 + x ) + z( 3 + x )
= ( 3 + x )( 2y + z )
A = xy + y - 2x - 2 = y(x + 1) - 2(x + 1) = (y - 2)(x + !1
B = x2 - 3x + xy - 3y = x(x - 3) + y(x - 3) = (x + y)(x - 3)
C = 3x2 - 3xy - 5x + 5y = 3x(x - y) - 5(x - y) = (3x - 5)(x - y)
D = xy + 1 + x + y = xy + x + y + 1 = x(y + 1) + (y + 1) = (x + 1)(y + 1)
E = ax - bx + ab - x2 = ax - x2 + ab - bx = a(a - x) - b(a - x) = (a - b)(a - x)
F = x2 + ab + ax + bx = ab + ax + bx + x2 = a(b + x) + x(b + x) = (a + x)(b + x)
G = a3 - a2x - ay + xy = a2(a - x) - y(a - x) = (a2 - y)(a - x)
H = 2xy + 3z + 6y + xz = 2xy + 6y + 3z + xz = 2y(x + 3) + z(x + 3) = (2y + z)(x + 3)
1) \(4x^2-7x-2=4x^2-8x+x-2=\left(4x^2-8x\right)+\left(x-2\right)\)
\(=4x\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(4x+1\right)\)
2) \(4x^2+5x-6=4x^2+8x-3x-6=\left(4x^2+8x\right)-\left(3x+6\right)\)
\(=4x\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(4x-3\right)\)
3) \(5x^2-18x-8=5x^2-20x+2x-8=\left(5x^2-20x\right)+\left(2x-8\right)\)
\(=5x\left(x-4\right)+2\left(x-4\right)=\left(x-4\right)\left(5x+2\right)\)
4) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)
\(=xy\left(x+y\right)-y^2z-yz^2+x^2z-xz^2\)
\(=xy\left(x+y\right)+\left(x^2z-y^2z\right)-\left(yz^2+xz^2\right)\)
\(=xy\left(x+y\right)+z\left(x^2-y^2\right)-z^2.\left(x+y\right)\)
\(=xy\left(x+y\right)+z\left(x-y\right)\left(x+y\right)-z^2\left(x+y\right)\)
\(=xy\left(x+y\right)+\left(zx-zy\right)\left(x+y\right)-z^2\left(x+y\right)\)
\(=\left(x+y\right)\left(xy+xz-yz-z^2\right)=\left(x+y\right).\left[x\left(y+z\right)-z\left(y+z\right)\right]\)
\(=\left(x+y\right)\left(y+z\right)\left(x-z\right)\)
1) 4x2 - 7x - 2 = 4x2 - 8x + x - 2 = 4x( x - 2 ) + ( x - 2 ) = ( x - 2 )( 4x + 1 )
2) 4x2 + 5x - 6 = 4x2 - 8x + 3x - 6 = 4x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( 4x + 3 )
3) 5x2 - 18x - 8 = 5x2 - 20x + 2x - 8 = 5x( x - 4 ) + 2( x - 4 ) = ( x - 4 )( 5x + 2 )
4) xy( x + y ) - yz( y + z ) + xz( x - z )
= x2y + xy2 - y2z - yz2 + xz( x - z )
= ( x2y - yz2 ) + ( xy2 - y2z ) + xz( x - z )
= y( x2 - z2 ) + y2( x - z ) + xz( x - z )
= y( x - z )( x + z ) + y2( x - z ) + xz( x - z )
= ( x - z )[ y( x + z ) + y2 + xz ]
= ( x - z )( xy + yz + y2 + xz )
= ( x - z )[ ( xy + y2 ) + ( xz + yz ) ]
= ( x - z )[ y( x + y ) + z( x + y ) ]
= ( x - z )( x + y )( y + z )
5) xy( x + y ) + yz + xz( x + z ) + 2xyz ( đề có thiếu không vậy .-. )
a, \(\left(2x+1\right)^2-2\left(2x+1\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left(2x+1-x+3\right)^2=\left(x+4\right)^2\)
b, \(xy+xz+3y+3z=x\left(y+z\right)+3\left(y+z\right)=\left(x+3\right)\left(y+z\right)\)
c, \(xy-xz+y-z=x\left(y-z\right)+\left(y-z\right)=\left(x+1\right)\left(y-z\right)\)
d, \(x^2-xy-8x+8y=\left(x^2-xy\right)-\left(8x-8y\right)\)
\(=x\left(x-y\right)-8\left(x-y\right)=\left(x-8\right)\left(x-y\right)\)
e, \(x^2+2xy+y^2-xz-yz=\left(x^2+2xy+y^2\right)-\left(xz+yz\right)\)
\(=\left(x+y\right)^2-z\left(x+y\right)=\left(x+y+z\right)\left(x+y\right)\)
f, \(25-4x^2-4xy-y^2=25-\left(4x^2+4xy+y^2\right)\)
\(=5^2-\left(2x+y\right)^2=\left(5-2x-y\right)\left(5+2x+y\right)\)
1,
a, (2x + 1- x + 3)2 = (x+4)2
b,\(x\left(y+z\right)+3\left(y+z\right)=\left(y+z\right)\left(x+3\right)\)
c, \(x\left(y-z\right)+\left(y-z\right)=\left(y-z\right)\left(x+1\right)\)
d,\(x\left(x-y\right)+8\left(y-x\right)\)=\(\left(x-y\right)\left(x-8\right)\)
e,\(\left(x+y\right)^2-z\left(x+y\right)\)=\(\left(x+y\right)\left(x+y-z\right)\)
f,\(25-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2\)
\(=\left(5+2x+y\right)\left(5-2x-y\right)\)
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