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a) = (x + 3)2 - y2 = (x + 3 - y)(x + 3 + y)
b) = x2(x - 3) -4(x - 3) = (x - 3)(x2 - 4) = (x - 3)(x - 2)(x + 2)
c) = 3x(x - y) - 5(x - y) = (x - y)(3x - y)
d) Nhầm đề. tui sửa lại x3 + y3 + 2x2 - 2xy + 2y2
= x3 + y3 + 2(x2 - xy + y2) = (x + y)(x2 - xy + y2) + 2(x2 - xy + y2) = (x2 - xy + y2)(x + y + 2)
e) = x4 - x3 - x3 + x2 - x2 + x + x - 1 = x3(x - 1) - x2(x - 1) - x(x - 1) + x - 1 = (x - 1)(x3 - x2 - x + 1) = (x - 1)(x - 1)(x2 - 1) = (x - 1)3(x + 1)
f) = x3 - 3x2 - x2 + 3x + 9x - 27 = x2(x - 3) - x(x - 3) + 9(x - 3) = (x-3)(x2 - x + 9)
g) chắc là 3xyz
= x2y + xy2 + y2z + yz2 + x2z + xz2 + 3xyz = x2y + xy2 + xyz + y2z + yz2 + xyz + x2z + xz2 + xyz = (x + y + z)(xy + yz + xz)
h) = 23 -(3x)3 = (2 - 3x)(4 + 6x + 9x2)
i) = (x + y - x + y)(x + y + x - y) = 2y*2x = 4xy
k) = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy +y2)(x + y)(x2 - xy +y2).
a) \(P=-x^2+13x+2012\)
\(\Leftrightarrow P=-x^2+2.x.\dfrac{13}{2}-\left(\dfrac{13}{2}\right)^2+2054,25\)
\(\Leftrightarrow P=-\left[x^2-2.x.\dfrac{13}{2}+\left(\dfrac{13}{2}\right)^2\right]+2054,25\)
\(\Leftrightarrow P=-\left(x-\dfrac{13}{2}\right)^2+2054,25\)
Vậy GTLN của \(P=2054,25\) khi \(x=\dfrac{13}{2}\)
b) \(A=x^2-2x+2\)
\(\Leftrightarrow A=x^2-2x+1+1\)
\(\Leftrightarrow A=\left(x-1\right)^2+1\)
Vậy GTNN của \(A=1\) khi \(x=1\)
1
a,\(x^2+5x+5xy+25y\)
\(=\left(x^2+5x\right)+\left(5xy+25y\right)\)
\(=x\left(x+5\right)+5y\left(x+5\right)\)
\(=\left(x+5y\right)\left(x+5\right)\)
b,Mình chưa làm được.
c,\(x^2-24x-25\)
\(=x^2+25x-x-25\)
\(=\left(x^2-x\right)+\left(25x-25\right)\)
\(=x\left(x-1\right)+25\left(x-1\right)\)
\(=\left(x+25\right)\left(x-1\right)\)
d,\(4x-8y\)
\(=4\left(x-2y\right)\)
e,\(x^2+2xy+y^2-16\)
\(=\left(x+y\right)^2-4^2\)
\(=\left(x+y-4\right)\left(x+y+4\right)\)
f,\(3x^2+5x-3xy-5y\)
\(=\left(3x^2-3xy\right)+\left(5x-5y\right)\)
\(=3x\left(x-y\right)+5\left(x-y\right)\)
\(=\left(3x+5\right)\left(x-y\right)\)
a:Ta có: \(A=-4x^2+x-1\)
\(=-4\left(x^2-\dfrac{1}{4}x+\dfrac{1}{4}\right)\)
\(=-4\left(x^2-2\cdot x\cdot\dfrac{1}{8}+\dfrac{1}{64}+\dfrac{63}{64}\right)\)
\(=-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{63}{16}\le-\dfrac{63}{16}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{8}\)
b: Ta có: \(B=-3x^2+5x+6\)
\(=-3\left(x^2-\dfrac{5}{3}x-2\right)\)
\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{97}{36}\right)\)
\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{12}\le\dfrac{97}{12}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{5}{6}\)
c: Ta có: \(C=-x^2+3x+4\)
\(=-\left(x^2-3x-4\right)\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{25}{4}\right)\)
\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)