Phan tích thanh nhan tu ( lm nhanh dùm mình nha )
a) \(x^5-x^4-1\)
b) \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
c) \(x^2+2xy+y^2-x-y-12\)
K
Khách
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5 tháng 9 2017
Câu a :
\(\left(x-5\right)^2+\left(x-5\right)\left(x+5\right)-\left(5-x\right)\left(2x+1\right)\)
\(=x^2-10x+25+x^2-25-10x-5+2x^2+x\)
\(=4x^2-19x-5\)
Câu b :
\(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=12x^2-9x-8x+6-2x+2+3x^2-3x-6x^2-6x+4x+4\)
\(=9x^2-24x+2\)
30 tháng 6 2017
a VT=.\(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)
=\(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}:\frac{x-1+x\left(x-1\right)+2}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^2+2x+1-x^2+2x-1}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{x^2+2x+1}\)
\(=\frac{4x}{\left(x+1\right)^2}\)=VP
b.VT\(=\frac{2+x}{2-x}.\frac{\left(2-x\right)^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{\left(x+2\right)\left(x^2-2x+4\right)}.\frac{4-2x+x^2}{2-x}\right)\)
=\(\frac{4-x^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{4-x^2}\right)=\frac{4-x^2}{4x^2}.\frac{2\left(2+x\right)-4}{4-x^2}\)
=\(\frac{2x}{4x^2}=\frac{1}{2x}\)=VP
c VT=.\(\left[\left(\frac{3}{x-y}+\frac{3x}{x^2-y^2}\right).\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)
\(=\left[\frac{3\left(x+y\right)+3x}{\left(x+y\right)\left(x-y\right)}.\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)
\(=\frac{3\left(2x+y\right)\left(x+y\right)^2}{\left(x+y\right)\left(x-y\right)\left(2x+y\right)}.\frac{x-y}{3}\)
\(=x+y=\)VP
Vậy các đẳng thức được chứng minh
=
I
2 tháng 10 2017
Ta có (x^2 + y^2 )^3 + (z^2 – x^2 )^3 – (y^2 + z^2 )^3
= (x^2 + y^2 )^3 + (z^2 – x^2 )^3 + (-y^2 - z^2 )^3
Ta thấy x^2 + y^2 + z^2 – x^2 – y^2 – z^2 = 0
=> áp dụng nhận xét ta có: (x^2+y^2 )^3+ (z^2 -x^2 )^3 -y^2 -z^2 )^3
= 3(x^2 + y^2 ) (z^2 –x^2 ) (-y^2 – z^2 )
= 3(x^2+y^2 ) (x+z)(x-z)(y^2+z^2 )
NC
0
DH
8 tháng 6 2017
a, \(x^2+2xy+y^2-x-y-12\)
\(=\left(x^2+2xy+y^2\right)-\left(x+y\right)-12\)
\(=\left(x+y\right)^2-\left(x+y\right)-12\)
\(=\left(x+y\right)^2+3\left(x+y\right)-4\left(x+y\right)-12\)
\(=\left[\left(x+y\right)^2+3\left(x+y\right)\right]-\left[4\left(x+y\right)+12\right]\)
\(=\left(x+y\right).\left[\left(x+y\right)+3\right]-4.\left[\left(x+y\right)+3\right]\)
\(=\left[\left(x+y\right)+3\right].\left[\left(x+y\right)-4\right]\)
b,B = \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(t=x^2+x+1\Rightarrow t+1=x^2+x+2\)
\(\Rightarrow B=t.\left(t+1\right)-12\)
\(B=t^2+t-12\)
\(B=t^2-3t+4t-12\)
\(B=\left(t^2-3t\right)+\left(4t-12\right)\)
\(B=t.\left(t-3\right)+4.\left(t-3\right)=\left(t-3\right).\left(t-4\right)\)
Mà \(t=x^2+x+1\) nên
\(B=\left(x^2+x+1-3\right).\left(x^2+x+1-4\right)\)
\(B=\left(x^2+x-2\right).\left(x^2+x-3\right)\)
\(B=\left(x^2-x+2x-2\right).\left(x^2+x-3\right)\)
\(B=\left[\left(x^2-x\right)+\left(2x-2\right)\right].\left(x^2+x-3\right)\)
\(B=\left[x.\left(x-1\right)+2.\left(x-1\right)\right].\left(x^2+x-3\right)\)
\(B=\left(x-1\right).\left(x+2\right).\left(x^2+x-3\right)\)
Chúc bạn học tốt!!!
a)=x5-x4+x3-x3-1
=x3(x2-x+1)-(x3+1)
=x3(x2-x+1)-(x+1)(x2-x+1)
=(x2-x+1)(x3-x-1)
b)=(x2+x)2-2(x2+x).1+1-16
=(x2+x-1)2-16
=(x2+x-1+4)(x2+x-1-4)
=(x2+x+3)(x2+x-5)
kho the thi ai lam duoc day