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a) \(2x-2y-x^2+2xy-y^2\)
\(=2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
\(=2\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(2-x+y\right)\)
b) \(9x^2+6xy+y^2-25\)
\(=\left(3x\right)^2+6xy+y^2-25\)
\(=\left(3x+y\right)^2-5^2\)
\(=\left(3x+y+5\right)\left(3x+y-5\right)\)
\(25-9x^2-6xy-y^2\)
\(=25-\left(9x^2+6xy+y^2\right)\)
\(=5^2-\left(3x+y\right)^2\)
\(=\left(5-3x-y\right)\left(5+3x+y\right)\)
Bài làm:
Ta có: \(9x^2y^2+y^2-6xy+y+2\)
\(=\left(9x^2y^2-6xy+1\right)+\left(y^2+y+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(3xy-1\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)
=> BT lớn hơn hẳn ko
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) 1 - 2y + y2
= (1-y)2
b) ( x + 1 )2 - 25
=( x + 1 )2 - 52
=(x+1+5)(x+1-5)
\(9x^2+6xy+y^2-25\)
\(=\left(3x+y\right)^2-25\)
\(=\left(3x+y-5\right)\left(3x+y+5\right)\)
\(9x^2+6xy+y^2-25=\left(3x+y\right)^2-25=\left(3x+y-5\right)\left(3x+y+5\right)\)