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a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
3)
\(A=\dfrac{5}{x^2-2x+5}\)
ta có x2-2x+5
=x2-2x+1+4
=(x2-2x+1)+4
=(x-1)2+4
=> A=\(\dfrac{5}{\left(x-1\right)^2+4}\)
do \(\left(x-1\right)^2\ge0\forall x\)
=> \(\left(x-1\right)^2+4\ge4\)
=> \(\dfrac{5}{\left(x-1\right)^2+4}\le\dfrac{5}{4}\)
=> A\(\le\dfrac{5}{4}\)
GTLN của A =\(\dfrac{5}{4}\)
khi x-1=0
=> x=1
vậy GTLN của A=\(\dfrac{5}{4}\) khi x=1
a. \(x^3+x^2-4x-4=x^2\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x^2-4\right)=\left(x+1\right)\left(x+2\right)\left(x-2\right)\)
b. \(x^2-y^2-4x+4=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2=\left(x+y-2\right)\left(x-y-2\right)\)
c. \(\left(x^2+9\right)^2-36x^2=\left(x^2+6x+9\right)\left(x^2-6x+9\right)=\left(x+3\right)^2\left(x-3\right)^2\)
d. \(25-x^2+2xy-y^2=25-\left(x-y\right)^2=\left(5+x-y\right)\left(5-x+y\right)\)
còn lại làm tương tự
a) \(x^3+x^2-4x-4=x^2\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
b) \(x^2-y^2-4x+4=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)
c) \(\left(x^2+9\right)^2-36x^2=\left(x^2+9\right)^2-\left(6x\right)^2=\left(x^2-6x+9\right)\left(x^2+6x+9\right)\)
\(=\left(x-3\right)^2\left(x+3\right)^2\)
d) \(25-x^2+2xy-y^2=5^2-\left(x-y\right)^2=\left(5-x+y\right)\left(5+x-y\right)\)
e) \(x^3-4x^2+4x-1=\left(x-1\right)\left(x^2+x+1\right)-4x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1-4x\right)=\left(x-1\right)\left(x^2-3x+1\right)\)
f) \(3x-3y-x^2+2xy-y^2=3\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3-x+y\right)\)
g) \(2x^2-9x+10=2x^2-4x-5x+10=2x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(2x-5\right)\)
h) \(x^2-5x-14=x^2-7x+2x-14=x\left(x-7\right)+2\left(x-7\right)=\left(x-7\right)\left(x+2\right)\)
i) \(x^3-3x^2+2=x^3-2x^2-x^2+2=x^2\left(x-1\right)-2\left(x^2-1\right)\)
\(=x\left(x-1\right)-2\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x-2x-2\right)\)
k) \(x^4+4=\left(x^2\right)^2+2\cdot x^2\cdot2+2^2-2\cdot x^2\cdot2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
a) Ta có: \(-3x^2\left(2x^2-\frac{1}{3}x+2\right)\)
\(=-6x^4+x^3-6x^2\)
b) Ta có: \(2xy^2\left(x-3y+xy\right)\)
\(=2x^2y^2-6xy^3+2x^2y^3\)
c) Ta có: \(\left(5x^2-4x\right)\left(x-2\right)\)
\(=5x^3-10x^2-4x^2+8x\)
\(=5x^3-14x^2+8x\)
d) Ta có: \(-\left(2-x\right)\left(2x+3\right)\)
\(=\left(x-2\right)\left(2x+3\right)\)
\(=2x^2+3x-4x-6\)
\(=2x^2-x-6\)
e) Ta có: \(\left(3x^3-2x^2+x\right):\left(-2x\right)\)
\(=\frac{-3}{2}x^2+x-\frac{1}{2}\)
f) Ta có: \(\left(15x^2y^2-21x^3y+2x^2y\right):\left(3x^2y\right)\)
\(=5y-7x+\frac{2}{3}\)
g) 
câu 20
\(\)\(C_{20}=\left(a^2+1\right)^2-4a^2=\left(a^2+1\right)^2-\left(2a\right)^2=\left[\left(a^2+1\right)-2a\right]\left[\left(a^2+1\right)+2a\right]\)\(C_{20}=\left[a^2-2a+1\right]\left[a^2+2a+1\right]=\left(a-1\right)\left(a-1\right)\left(a+1\right)\left(a+1\right)\)
\(C_{20}=\left(a-1\right)\left(a-1\right)\left(a+1\right)\left(a+1\right)\)
1: =(4x-1)^2-3(4x-1)
=(4x-1)(4x-1-3)
=4(x-1)(4x-1)
2: =-8x^4y^5(2y+3x)
3: =(a-5)^2-4b^2
=(a-5-2b)(a-5+2b)
5: =x^2-mx-nx+mn
=x(x-m)-n(x-m)
=(x-m)(x-n)
6: =(4a^2-3a-18-4a^2-3a)(4a^2-3a-18+4a^2+3a)
=(-6a-18)(8a^2-18)
=-6(2a-3)(2x+3)(a+3)
\(\cdot x^2-4x-5=x^2-5x+x-5=\left(x^2-5x\right)+\left(x-5\right)=x\left(x-5\right)+\left(x-5\right)=\left(x-5\right)\left(x+1\right)\)
\(x^2-y^2-2xy+y^2\\=x^2-2xy+y^2-y^2\\ =\left(x-y\right)^2-y^2\\ =\left(x-y-y\right)\left(x-y+y\right) \)