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\(1.A=\left(x-2y\right)^3-\left(x+2y\right)^3+12x^2y-xy\\ =x^3-3x^2.2y+3x\cdot\left(2y\right)^2-\left(2y\right)^3-\left[x^3+3x^2\cdot2y+3x\cdot\left(2y\right)^2+\left(2y\right)^3\right]+12x^2y-xy\)
\(=x^3-6x^2y+12xy^2-8y^3-x^3-6x^2y-12xy^2-8y^3+12x^2y-xy\\ =xy-16y^3\)
2 . Tìm x :
\(a)\left(2x-3\right)^2-\left(x+2\right)^2=0\\ \left[\left(2x-3\right)-\left(x+2\right)\right]\cdot\left[\left(2x-3\right)+\left(x+2\right)\right]=0\\ \left(2x-3-x-2\right)\cdot\left(2x-3+x+2\right)=0\\ \left(x-5\right)\left(3x-1\right)=0\\ 4x-x-15x+5=0\\ -18x=-5\Rightarrow x=\dfrac{5}{18}\)
\(b)\left(x-3\right)^3=x\cdot\left(x^2-9x+2\right)\\ x^3-9x^2+27x-27=x^3-9x^2+2x\\ \Leftrightarrow27x-27=2x\\ \Leftrightarrow27x-2x=27\\ \Leftrightarrow25x=27\\ \Leftrightarrow x=\dfrac{27}{25}\)
\(c)36x^2-49=0\\ \Leftrightarrow36x^2=49\\ \Leftrightarrow x^2=\dfrac{49}{36}\\ \Leftrightarrow x^2=\left(\pm\dfrac{7}{6}\right)^2\\ \Rightarrow x=\pm\dfrac{7}{6}\)
Dài dữ trời :V Về sau gửi từng bài một thôi, nhìn hoa mắt quá @@
B1: Phân tích thành nhân tử:
a) \(6x^2+9x=3x\left(2x+3\right)\)
b) \(4x^2+8x=4x\left(x+2\right)\)
c) \(5x^2+10x=5x\left(x+2\right)\)
d) \(2x^2-8x=2x\left(x-4\right)\)
e) \(5x-15y=5\left(x-3y\right)\)
f) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
g) \(x^2-2x+1-4y^2=\left(x-1\right)^2-4y^2\)
\(=\left(x-1-2y\right)\left(x-1+2y\right)\)
h) \(x^2-100=\left(x-10\right)\left(x+10\right)\)
i) \(9x^2-18x+9=\left(3x-3\right)^2\)
k) \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
l) \(x^2+6xy^2+9y^4=\left(x+3y\right)^2\)
m) \(4xy-4x^2-y^2=-\left(4x^2-4xy+y^2\right)\)
\(=-\left(2x-y\right)^2\)
n) \(\left(x-15\right)^2-16=\left(x-15-16\right)\left(x-15+16\right)\)
\(=\left(x-31\right)\left(x+1\right)\)
o) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3+x\right)\)
\(=\left(2+x\right)\left(8+x\right)\)
p) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)
\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)
\(=\left(5x-5\right)\left(9x-3\right)\)
Bài 1 :
a ) \(6x^2+9x=3x\left(x+3\right)\)
b ) \(4x^2+8x=4x\left(x+2\right)\)
c ) \(5x^2+10x=5x\left(x+2\right)\)
d ) \(2x^2-8x=2x\left(x-4\right)\)
e ) \(5x-15y=5\left(x-3y\right)\)
f ) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
g ) \(x^2-2x+1-4y^2=\left(x-1\right)^2-\left(2y\right)^2=\left(x-1-2y\right)\left(x-1+2y\right)\)
h ) \(x^2-100=x^2-10^2=\left(x-10\right)\left(x+10\right)\)
i ) \(9x^2-18x+9=\left(3x-3\right)^2\)
k ) \(x^3-8=\left(x-2\right)\left(x^2+2x+2^2\right)\)
l ) \(x^2+6xy^2+9y^4=\left(x+3y^2\right)^2\)
m ) \(4xy-4x^2-y^2=-\left(2x-y\right)^2\)
n ) \(\left(x-15\right)^2=x^2-30x+15^2\)
o ) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)
p ) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)\)
Bài 2 :
a ) \(3x^3-6x^2+3x^2y-6xy=3x\left(x^2-2x+xy-2y\right)\)
b ) \(x^2-2x+xy-2y=x\left(x-2\right)+y\left(x-2\right)=\left(x-2\right)\left(x+y\right)\)
c ) \(2x+x^2-2y-2xy=......................\)
d ) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
e ) \(x^2+y^2-2xy-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
f )\(2xy-x^2-y^2+9=-\left(x-y\right)^2+9=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)
Câu 1:
a: \(C=a^2+b^2=\left(a+b\right)^2-2ab=23^2-2\cdot132=265\)
b: \(D=x^3+y^3+3xy\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\)
\(=1-3xy+3xy=1\)
Bài 3 :
a ) \(x\left(x-1\right)+x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy...........
b ) \(3\left(x-3\right)-4x+12=0\)
\(\Leftrightarrow3\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\) \(\left(x-3\right)=0\Rightarrow x=3\)
Vậy............
Các câu sau tương tự
a) ta có: \(|4x^2-1|\ge0\forall x\)
\(|2x-1|\ge0\forall x\Leftrightarrow3x|2x-1|\ge0\forall x\)
Mà \(|4x^2-1|+3x|2x-1|=0\)
=> I4x^2-1I và 3xI2x-1I=0
=> 4x^2-1=0 và 3x=0 hoặc 2x-1=0
=> 4x^2=1 và x=0 hoặc 2x=1
=> x^2=1/4 và x=0 hoặc x=1/2
=> x=\(\pm\frac{1}{2}\)và x=0 hoặc x=1/2
Vậy x=\(\pm\frac{1}{2}\); x=0
a) 2x2 + 4x + xy + 2y
= (2x2 + xy) + (4x + 2y)
= x(2x + y) + 2(2x + y)
= (x + 2)(2x + y)
b) x2 + xy - 7x - 7y
= x(x + y) - 7(x + y)
= (x - y)(x + y)
a) \(2x+2y\)
\(=2\left(x+y\right)\)
b) \(5x+20y\)
\(=5\left(x+4y\right)\)
c) \(6xy-30y\)
\(=6y\left(x-5\right)\)
d) \(5x\left[x-110-10y\left(x-11\right)\right]\)
\(=5x\left(x-110-10xy+110\right)\)
\(=5x\left(x-10xy\right)\)
\(=5x^2\left(1-10y\right)\)
e) \(x^3-4x^2+x\)
\(=x\left(x^2-4x+1\right)\)
f) \(x\left(x+y\right)-\left(2x+2y\right)\)
\(=x\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-2\right)\)
h) \(5x\left(x-2y\right)+2\left(2y-x\right)\)
\(=5x\left(x-2y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\left(5x-2\right)\)
i) \(x^2y^3-\dfrac{1}{2}x^4y^8\)
\(=x^2y^3\left(1-\dfrac{1}{2}xy^5\right)\)
j) \(a^2b^4+a^3b-abc\)
\(=ab\left(ab^3+a^2-c\right)\)
c, 36 x^2 - 49 = 0
-> x^2=49/36
x=7/6 hoặc là -7/6