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a) 1 - 2y + y2
= (1-y)2
b) ( x + 1 )2 - 25
=( x + 1 )2 - 52
=(x+1+5)(x+1-5)
a. \(1-2y+y^2=\left(1-y\right)^2\)
b. \(\left(x+1\right)^2-25=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)
c. \(1-4x^2=\left(1+2x\right)\left(1-2x\right)\)
d. \(8-27x^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
e. \(27+27x+9x^2+x^3=\left(x+3\right)^3\)
f, \(8x^3-12x^2y+6xy^2-y^3=\left(2x-y\right)^3\)
g, \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
\(\left(a\right)1-2y+y^2\)
\(\Leftrightarrow y^2-2y+1\)
\(\Leftrightarrow\left(y-1\right)^2\)
\(\left(b\right)\left(x+1\right)^2-25\)
\(\Leftrightarrow\left(x+1\right)^2-5^2\)
\(\Leftrightarrow\left(x-4\right)\left(x+6\right)\)
\(\left(c\right)1-4x^2\)
\(\Leftrightarrow1-\left(2x\right)^2\)
\(\Leftrightarrow\left(1-2x\right)\left(1+2x\right)\)
\(\left(d\right)8-27x^3\)
\(\Leftrightarrow2^3-\left(3x\right)^3\)
\(\Leftrightarrow\left(2-3x\right)\left(4+6x+9x^2\right)\)
\(\left(e\right)27+27x+9x^2+x^3\)
\(\Leftrightarrow\left(x+3\right)^3\)
\(\left(f\right)8x^3-12x^2y+6xy^2-y^3\)
\(\Leftrightarrow\left(2x\right)^3-12x^2y+6xy^2-y^3\)
\(\Leftrightarrow\left(2x-y\right)^3\)
\(\left(g\right)x^3+8y^3\)
\(\Leftrightarrow\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
a) \(A=\left(5x+2y\right)^2\)
\(=\left(5x\right)^2+2\cdot5x\cdot2y+\left(2y\right)^2\)
\(=25x^2+20xy+4y^2\)
b) \(B=\left(4x-y\right)^2\)
\(=\left(4x\right)^2-2\cdot4xy+y^2\)
\(=16x^2-8xy+y^2\)
c) \(C=9x^2-25\)
\(=\left(3x-5\right)\left(3x+5\right)\)
d) \(D=\left(x+2\right)^3\)
\(=x^3+3x^2\cdot2+3x\cdot2^2+2^3\)
\(=x^3+6x^2+3x\cdot4+8\)
\(=x^3+6x^2+12x+8\)
a, \(A=\left(5x+2y\right)^2=25x^2+20xy+4y^2\)
b, \(B=\left(4x-y\right)^2=16x^2-8xy+y^2\)
c, \(C=9x^2-25=\left(3x-5\right)\left(3x+5\right)\)
d, \(D=\left(x+2\right)^3=x^3+6x^2+12x+8\)
e, \(E=\left(3x-1\right)^3=27x^3-9x^2+9x-1\)
g, \(G=x^3+64=\left(x+4\right)\left(x^2-4x+16\right)\)
h, \(H=27x^3-1=\left(3x-1\right)\left(9x^2+3x+1\right)\)
\(1.\)
\(a.\)
\(x^2-2x=x\left(x-2\right)\)
b.
\(3y^3+6xy^2+3x^2y\)
\(=3y\left(y^2+2xy+x^2\right)\)
\(=3y\left(x+y\right)^2\)
\(c.\)
\(x^2-2xy-xy+2y^2\)
\(=x\left(x-2y\right)-y\left(x-2y\right)\)
\(=\left(x-y\right)\left(x-2y\right)\)
\(2.\)
\(a.\)
\(x^2-y^2+5x-5y\)
\(=\left(x-y\right)\left(x+y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+5\right)\)
\(b.\)
\(x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(c.\)
\(x^2-6xy+9y^2-16\)
\(=\left(x^2-6xy+9y^2\right)-4^2\)
\(=\left(x-3\right)^2-4^2\)
\(=\left(x-3-4\right)\left(x-3+4\right)\)
\(=\left(x-7\right)\left(x+1\right)\)
Tương tự câu \(d,e,g\)
\(3.\)
\(a.\)
\(x^3-2x=0\)
\(\Rightarrow x\left(x^2-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{2}\end{matrix}\right.\)
\(b.\)
\(x\left(x-4\right)+\left(x-4\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
\(c.\)
\(x\left(x-3\right)+4x-12=0\)
\(\Rightarrow x\left(x-3\right)+3\left(x-3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Tương tự \(d,e,g\)
a) A = (x + 1)(y - 2) - (2 - y)2
= -[(x + 1)(2 - y) + (2 - y)2]
= -[(x + 1 - 2 + y)(2 - y)]
= -[(x - 1 + y)(2 - y)]
= (x - 1 + y)(y - 2)
Bài 2:
a) \(A=\left(x+1\right)\left(y-2\right)-\left(2-y\right)^2\)
\(A=\left(x+1\right)\left(y-2\right)-\left(y-2\right)^2\)
\(A=\left(y-2\right)\left(x+1-y+2\right)\)
\(A=\left(y-2\right)\left(x-y+3\right)\)
b) \(B=x^2-6xy+9y^2+4x-12y\)
\(B=\left[x^2-2\cdot x\cdot3y+\left(3y\right)^2\right]+4\left(x-3y\right)\)
\(B=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(B=\left(x-3y\right)\left(x-3y+4\right)\)
Bài 3:
a) \(3\left(x-2\right)\left(x+3\right)-x\left(3x+1\right)=2\)
\(\left(3x^2+3x-18\right)-\left(3x^2+x\right)-2=0\)
\(3x^2+3x-18-3x^2-x-2=0\)
\(2x-20=0\)
\(x=10\)
b) \(6x^2+13x+5=0\)
\(6x^2+10x+3x+5=0\)
\(2x\left(3x+5\right)+\left(3x+5\right)=0\)
\(\left(3x+5\right)\left(2x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x+5=0\\2x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{3}\\x=\frac{-1}{2}\end{cases}}}\)
Bài 1:tìm x ,biết:
a) (2x - 1)(3x + 2) - 6x(x + 1) = 0
\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)
\(\Leftrightarrow-5x=2\)
\(\Leftrightarrow x=\frac{-2}{5}\)
b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)
\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)
\(\Leftrightarrow-10x=-4\)
\(\Leftrightarrow x=\frac{2}{5}\)
c) \(4x^2-1=2\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-2\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)
2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)
\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)
b) \(4x^2-1-2x\left(2x-1\right)=\left(2x-1\right)\left(2x+1\right)-2x\left(2x-1\right)\)
\(=1.\left(2x-1\right)\)
c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)
\(=\left(x-4-2y\right)\left(x-4+2y\right)\)
d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)
\(=\left(3x-2-y\right)\left(3x-2+y\right)\)
e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)
\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)
\(=\left(2x+\frac{5+3\sqrt{5}}{2}\right)\left(2x+\frac{5-3\sqrt{5}}{2}\right)\)
a) \(a^2x+a^2y-9x-9y\)
\(=\left(a^2x+a^2y\right)-\left(9x+9y\right)\)
\(=a^2\left(x+y\right)-9\left(x+y\right)\)
\(=\left(x+y\right)\left(a^2-9\right)\)
\(=\left(x+y\right)\left(a-3\right)\left(a+3\right)\)
b) \(x^2-x-12\)
\(=x^2-4x+3x-12\)
\(=\left(x^2-4x\right)+\left(3x-12\right)\)
\(=x\left(x-4\right)+3\left(x-4\right)\)
\(=\left(x-4\right)\left(x+3\right)\)
c) \(x^2\left(x-3\right)+12-4x\)
\(=x^2\left(x-3\right)-\left(4x-12\right)\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
d) \(4x\left(x-y\right)+6y\left(x-y\right)\)
\(=\left(x-y\right)\left(4x+6y\right)\)
\(=2\left(x-y\right)\left(2x+3y\right)\)
e) \(5\left(x+y\right)-xy-y^2\)
\(=5\left(x+y\right)-\left(xy+y^2\right)\)
\(=5\left(x+y\right)-y\left(x+y\right)\)
\(=\left(x+y\right)\left(5-y\right)\)
\(1-2y+y^2=\left(1-y\right)^2\)
\(\left(x+1\right)^2-25=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
\(27+27x+9x^2=9\left(3+3x+x^2\right)\)
\(8x^3-12x^2y+6xy^2-y^3=\left(2x-y\right)^3\)
\(3x^2-6xy+9y^2=3\left(x^2-2xy+3y^2\right)\)
\(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
\(x^2-4x-5=x^2+x-5x-5=x\left(x+1\right)-5\left(x+1\right)=\left(x-5\right)\left(x+1\right)\)
a ) \(1-2y+y^2=y^2-2y+1=\left(y-1\right)^2\)
b ) \(\left(x+1\right)^2-25=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right).\)
c ) \(1-4x^2=\left(1-2x\right)\left(1+2x\right).\)
d ) \(27+27x+9x^2=9\left(3+3x+x\right)=9\left(3+4x\right).\)
e ) \(8x^3-12x^2y+6xy^2-y^3=\left(2x-y\right)^3\)
f ) \(3x^2-6xy+9y^2=3\left(x^2-2xy+3y^2\right).\)
g ) \(x^2+4x+3==x^2+3x+x+3=\left(x+1\right)\left(x+3\right)\)
h ) \(x^2-4x-5=x^2+x-5x-5=\left(x-5\right)\left(x+1\right).\)