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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
\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\in\Phi\\x=5\end{cases}}\)
a, -x - y2 + x2 - y = (x2 - y2) - (x + y)
= (x - y)(x + y) - (x + y)
= (x + y)(x - y - 1)
b, x( x + y ) - 5x - 5y = x(x + y) - 5(x + y)
= (x - 5)(x + y)
c, x2 - 5x + 5y - y2 = (x - y)(x + y) - 5(x - y)
= (x - y)(x + y - 5)
d, 5x3 - 5x2y - 10x2 + 10xy = 5x2(x - y) - 10x(x - y)
= 5x(x - y)(x - 2)
e, 27x3 - 8y3 = (3x - 2y)(9x2 + 6xy + 4y2)
f, x2 - y2 - x - y = (x - y)(x + y) - (x + y)
= (x + y)(x - y - 1)
g, x2 - y2 - 2xy + y2 = (x2 - 2xy + y2) - y2
= (x - y)2 - y2
= (x - y - y)(x - y + y) = x(x - 2y)
h, x2 - y2 + 4 - 4x = (x2 - 4x + 4) - y2
= (x - 2)2 - y2
= (x - y - 2)(x + y - 2)
i, x3 + 3x2 + 3x + 1 - 27z3 = (x + 1)3 - 27z3
= (x+1-3z)(x2+2x+1+3xz+3z+9z2)
k, 4x2 + 4x - 9y2 + 1 = (2x + 1)2 - 9y2
= (2x - 3y + 1)(2x + 3y + 1)
m, x2 - 3x + xy - 3y = x(x - 3) + y(x - 3)
= (x - 3)(x + y)
1)2xy+3z+6y+xz
= x(2y + z) + 3(z + 2y)
= (x + 3)(2y + z)
2)x^4-9x^3+x^2-9x
= x^2(x^2 + 1) - 9x(x^2 + 1)
= (x^2 + 1)(x^2 - 9x)
= x(x - 9)(x^2 + 1)
3)x^2-xy+x-y
= x(x - y) + (x - y)
= (x + 1)(x - y)
4)xz+yz-5(x+y)
= z(x + y) - 5(x + y)
= (z - 5)(x + y)
5)3x^2-3xy-5x+5y
= 3x(x - y) - 5(x - y)
= (3x - 5)(x - y)
6)x^2+4x-y^2+4y
= (x - y)(x + y) + 4(x + y)
= (x - y + 4)(x + y)
a) \(\left(5x-4\right)^2-49x^2\)
\(=\left(5x-4\right)^2-\left(7x\right)^2\)
\(=\left(12x-4\right)\left(-2x-4\right)\)
\(=-6\left(3x-1\right)\left(x+2\right)\)
c) \(x^2-y^2-x+y\)
\(=\left(x+y\right)\left(x-y\right)-\left(x-y\right)\)
\(=\left(x+y-1\right)\left(x-y\right)\)
d)\(4x^2-9y^2+4x-6y\)
\(=\left(2x-3y\right)\left(2x+3y\right)+2\left(2y-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+2\right)\)
e) \(-x^2+5x+2xy-5y-y^2\)
\(=-\left(x^2-2xy+y^2\right)+\left(5x-5y\right)\)
\(=-\left(x-y\right)^2+5\left(x-y\right)\)
\(=\left(x-y\right)\left(y-x+5\right)\)
f) \(y^2\left(x^2+y\right)-zx^2-zy\)
\(=y^2\left(x^2+y\right)-z\left(x^2+y\right)\)
\(=\left(y^2-z\right)\left(x^2+y\right)\)
a) x2-x-y2-y=(x2-y2)-(x+y)=(x+y)(x-y)-(x+y)=(x+y)(x-y-1)
b)x2-2xy+y2-z2=(x-y)2-z2=(x-y-z)(x-y+z)
c)5x-5y+ax-ay=5(x-y)+a(x-y)=(x-y)(5+a)
d)a3-a2x-ay+xy=a2(a-x)-y(a-x)=(a-x)(a2-y)
e)4x2-y2+4x+1=[(2x)2+2.2x.1+12]-y2=(2x+1)2-y2=(2x+1-y)(2x+1+y)
Chúc bạn học tốt!
a) 6x2 - 12x
= 6x(x - 2)
b) x2 + 2x + 1 - y2
= (x2 + 2x + 1) - y2
= (x + 1)2 - y2
= (x + 1 - y)(x + 1 + y)
c) x + y + z + x2 + xy + xz
= (x + x2) + (y + xy) + (z + xz)
= x(1 + x) + y(1 + x) + z(1 + x)
= (x + y + z)(x + 1)
d) xy + xz + y2 + yz
= (xy + xz) + (y2 + yz)
= x(y + z) + y(y + z)
= (x + y)(x + z)
e) x3 + x2 + x + 1
= (x3 + x2) + (x + 1)
= x2(x + 1) + (x + 1)
= (x2 + 1)(x + 1)
f) xy + y - 2x - 2
= (xy + y) - (2x + 2)
= y(x + 1) - 2(x + 1)
= (y - 2)(x + 1)
g) x3 + 3x - 3x2 - 9
= (x3 - 3x2) + (3x - 9)
= x2(x - 3) + 3(x - 3)
= (x2 + 3)(x - 3)
h) x2 - y2 - 2x - 2y
= (x2 - y2) - (2x + 2y)
= (x + y)(x - y) - 2(x + y)
= (x + y)(x - y - 2)
i) 7x2 - 7xy - 5x = 5y
mk thấy con này sai sai ý
a) \(-x-y^2+x^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right).1\)
\(=\left(x+y\right)\left(x-y-1\right)\)
b) \(x\left(x+y\right)-5x-5y\)
\(=x\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-5\right)\)
c) \(x^2-5x+5y-y^2\)
\(=\left(x^2-y^2\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
d) \(5x^3-5x^2y-10x^2+10xy\)
\(=5x\left(x^2-xy-2x+2y\right)\)
\(=5x\left[x\left(x-y\right)-2\left(x-y\right)\right]\)
\(=5x\left(x-y\right)\left(x-2\right)\)
e) \(27x^3-8y^3\)
\(=\left(3x\right)^3-\left(2y\right)^3\)
\(=\left(3x-2y\right)\left[\left(3x\right)^2+3x2y+\left(2y\right)^2\right]\)
\(=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)
f) \(x^2-y^2-x-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
g) \(x^2-y^2-2xy+y^2\)
\(=\left(x^2-2xy+y^2\right)-y^2\)
\(=\left(x-y\right)^2-y^2\)
\(=\left(x-y-y\right)\left(x-y+y\right)\)
\(=\left(x-y^2\right)x\)
h) \(x^2-y^2+4-4x\)
\(=\left(x^2-4x+4\right)-y^2\)
\(=\left(x^2-2.2x+2^2\right)-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
i) \(x^6-y^6\)
\(=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left[\left(x-y\right)\left(x^2+xy+y^2\right)\right]\left[\left(x+y\right)\left(x^2-xy+y^2\right)\right]\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
a: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}\)
\(=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)
c: \(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)
\(=\dfrac{3\left(1+2x\right)}{2\left(x+4\right)}\)
d: \(=\dfrac{12x}{8x^3}\cdot\dfrac{15y^4}{5y^3}=\dfrac{3}{2x^2}\cdot3y=\dfrac{9y}{2x^2}\)
f: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}\cdot\dfrac{x+4}{2\left(x-2\right)}=\dfrac{x+2}{6}\)
Đề bài là j vậy ?O?