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a)Ta có : \(\dfrac{x+1}{1-x}\)( giữ nguyên )
\(\dfrac{x^2-2}{1-x}\)( giữ nguyên )
\(\dfrac{2x^2-x}{x-1}=\dfrac{x-2x^2}{1-x}\)
b)Ta có : \(\dfrac{1}{x-1}=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x+1}{x^3-1}\)
\(\dfrac{2x}{x^2+x+1}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x^2-2x}{x^3-1}\)
\(\dfrac{2x-3x^2}{x^3-1}\)(giữ nguyên )
c) MTC = ( x+ 2)2(x - 2)2
Do đó , ta có : \(\dfrac{1}{x^2+4x+4}=\dfrac{1}{\left(x+2\right)^2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)^2\left(x-2\right)^2}\)
\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}\)
\(\dfrac{x}{x^2-4}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x^2-2^2\right)}{\left(x+2\right)^2\left(x-2\right)^2}=\dfrac{x^3-4x}{\left(x+2\right)^2\left(x-2\right)^2}\)
d) MTC = xyz( x - y)( y - z)( x - z)
Do đó , ta có : \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}=\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(\dfrac{1}{y\left(y-x\right)\left(y-z\right)}=\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(\dfrac{1}{z\left(z-x\right)\left(z-y\right)}=\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
Cộng các phân thức lại ta có :
\(\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
= \(\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
c)(x2+x)2-2(x2+x)-15
đặt x2+x=a ta có
a2-2a-15
=a2+3a-5a-15
=(a2+3a)-(5a+15)
=a(a+3)-5(a+3)
=(a+3)(a-5)
thay a=x2+x
(x2+x+3)(x2+x-5)
\(a,\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)
\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)
\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)
a) \(x^4-3x^3-x+3=x^4-x^3-2x^3+2x^2-2x^2+2x-3x+3\)
\(=x^3\left(x-1\right)-2x^2\left(x-1\right)-2x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(x^3-2x^2-2x-3\right)\left(x-1\right)=\left(x^3+x^2+x-3x^2-3x-3\right)\left(x-1\right)\)
\(=\left(x\left(x^2+x+1\right)-3\left(x^2+x+1\right)\right)\left(x+1\right)=\left(x^2+x+1\right)\left(x-3\right)\left(x-1\right)\)
b) \(x^2y^2\left(y-x\right)+y^2z^2\left(x-y\right)-z^2x^2\left(z-x\right)\)
\(=-x^2y^2\left(x-y\right)+y^2z^2\left(x-y\right)-z^2x^2\left(z-x\right)\)
\(=\left(y^2z^2-x^2y^2\right)\left(x-y\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(z^2-x^2\right)\left(x-y\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(z+x\right)\left(z-x\right)\left(x-y\right)-z^2x^2\left(z-x\right)\)
\(=\left(y^2\left(z+x\right)\left(x-y\right)-z^2x^2\right)\left(z-x\right)\)
c) câu này đề có sai o bn
hình như đề là : \(4x^2+4x^2y-8y^2\) mới đúng chứ ?? ?
1
a . X.(X-2)-Y.(X-2)=(X-Y).(X-2)
b .(X2 +1+2X).(X2 +1-2X)
2
3X2 +2X+X2 +2X+1-4X2 -10X+10X+5=(-12)
4X+6= -12
X=9/2
1. a, x2-2x+2y-xy = x(x-2)+y(2-y) = x(x-2)-y(x-2) = (x-y)(x-2)
b, (x2+1)2-4x2 = (x2+1-2x)(x2+1+2x) = (x-1)2(x+1)2
2. x(3x+2)+(x+1)2-(2x-5)(2x+5) = -12
=> (3x2+2x+x2+2x+1)-(2x)2-52 = -12
=> 3x2+2x+x2+2x+1-4x2-25 = -12
=> 4x-24 = -12 => 4x = 12 => x = 3
Có \(\left(x+y+z\right)^3\)
\(=\left[\left(x+y\right)+z\right]^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)^2z+3\left(x+y\right)z^2+z^3\)
\(=x^3+3x^2y+3xy^2+y^3+3\left(x+y\right)\left[\left(x+y\right)z+z^2\right]+z^3\)
\(=x^3+y^3+z^3+3xy\left(x+y\right)+3\left(x+y\right)\left(xz+yz+z^2\right)\)
=\(x^3+y^3+z^3+3\left(x+y\right)\left(xz+xy+yz+z^2\right)\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
Ta có: (x + y + z)3 = x3 + y3 + z3 + 3(x + y)(y + z)(z + x)
\(\Leftrightarrow\) (x + y + z)3 - x3 - y3 - z3 = 3(x + y)(y + z)(z + x)
Phân tích VT ta được:
(x + y + z)3 - x3 - y3 - z3 = \(\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
= (x + y)3 + z3 + 3z(x + y)(x + y + z) - x3 - y3 - z3
= x3 + y3 + 3xy(x + y) + z3 + 3z(x + y)(x + y + z) - x3 - y3 - z3
= 3xy(x + y) + 3z(x + y)(x + y + z)
= 3(x +y)(xy + xz + yz + z2)
= 3(x +y)\(\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
= 3(x + y)(y + z)(z + x) (đpcm)
Bài này cần áp dụng công thức (x + y)3 = x3 + y3 + 3xy(x + y) nhiều lần để phân tích nhé bạn.
Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)