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a, Đặt \(A=16x^2-24x+9\)
⇒ \(A=(4x-3)^2\)
Vs x = 0
=> A = \((-3)^2=9\)
Vs \(x=\frac{1}{4}\)
⇒ \(A=\left(1-3\right)^2=4\)
Vs \(x=12\)
=> \(A=\left(48-3\right)^2=45^2=2025\)
Vs \(x=\frac{3}{4}\)
⇒ A = 0
2.
a, \(=4x^2-12x+9\)
b, \(=\frac{25}{16}-\frac{5}{2}x+x^2\)
c, \(=4x^2+12xy+9y^2\)
d, \(=9x^2+4xyz+\frac{4}{9}y^2z^2\)
e, \(=\left(\frac{x^2y^2}{4}-\frac{x^2y^2}{9}\right)\) (bỏ ngoặc hộ mình nhé <3)
f, \(=4x^2+y^2+z^2-4xy+4xz-2yz\)
a, A=2x2+y2-2xy-2x+3
= (x2-2xy+y2)+(2x2-2x+2)+1
=(x-y)2+2(x-1)2+1
vì (x-y)2 ≥0 ∀x,y
(x-1)2 ≥ 0 ∀x
=> (x-y)2+2(x-1)2+1 ≥1 ∀x,y
=> A ≥1
= > GTNN A = 1 khi
x-1=0
=> x=1
x-y=0
=> 1-y=0
=> y=1
vậy GTNN A =1 khi x=y=1
a) \(A=x^2y+y+xy^2-x\) (hẳn đề là vậy)
\(A=xy\left(x+y\right)+\left(y-x\right)\)
\(A=\left(-5\right).2\left(-5+2\right)+2+5\)
\(A=30+7=37\)
b) \(B=3x^3-2y^3-6x^2y^2+xy\)
\(B=3.\left(\frac{2}{3}\right)^3-2.\left(\frac{1}{2}\right)^3-6.\left(\frac{2}{3}\right)^2.\left(\frac{1}{2}\right)^2+\frac{2}{3}.\frac{1}{2}\)
\(B=\frac{8}{9}-\frac{1}{4}-\frac{2}{3}+\frac{1}{3}\)
\(B=\frac{11}{36}\)
c) \(C=2x+xy^2-x^2y-2y\)
\(C=2.\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right).\left(-\frac{1}{3}\right)^2-\left(-\frac{1}{2}\right)^2.\left(-\frac{1}{3}\right)-2.\left(-\frac{1}{3}\right)\)
\(C=-1-\frac{1}{18}+\frac{1}{12}+\frac{2}{3}\)
\(C=-\frac{11}{36}\)
a: \(=n^3+2n^2+3n^2+6n-n-2-n^3+5\)
\(=5n^2+5n+3⋮̸5\)
b:\(=6n^2+30n+n+5-6n^2+3n-10n+5\)
\(=24n+10=2\left(12n+5\right)⋮2\)
d: \(=4x^2y^2-2x^2y+2xy^2-xy-4x^2y^2+xy\)
\(=-2\left(x^2y-xy^2\right)⋮2\)
\(a,=x^2-10x+25\\ b,=4x^2+12x+9\\ c,=4x^2-12x+9\\ d,=9x^2+12x+4\\ e,=x^3y^3-8\\ d,=x^2y^2-25\)