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\(a,VT=\left(a+b+c\right)\left(a-b+c\right)\)
\(=\left(a+c+b\right)\left(a+c-b\right)\)
\(=\left(a+c\right)^2-b^2\)
\(=a^2+2ac+c^2-b^2=VP\)
\(b,VT=\left(3x+2y\right)\left(3x-2y\right)-\left(4x-2y\right)\left(4x+2y\right)\)
\(=9x^2-4y^2-16x^2+4y^2=-7x^2=VP\)
\(c,VT=x^3-1-x^3-1=-2=VP\)
\(d,VT=8x^3+1-8x^3+1=2=VP\)
\(e,VT=\left(x^2+2xy+4y^2\right)\left(x-2y-2x+1\right)\)
\(=\left(x^2+2xy+4y^2\right)\left(-x-2y+1\right)\)
\(=-x^3-2x^2y+x^2-2x^2y-4xy^2+2xy-4xy^2-8y^3+4y^2\)
( bn kiểm tra lại đề nhé)
a ) \(x^2-x+1\)
\(\Leftrightarrow\left(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right)+\dfrac{3}{4}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có : \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN là \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}.\)
Bài 2:
a: \(=-\left(x^2+2x-100\right)\)
\(=-\left(x^2+2x+1-101\right)\)
\(=-\left(x+1\right)^2+101< =101\)
Dấu = xảy ra khi x=-1
b: \(=-3\left(x^2-\dfrac{1}{3}x\right)\)
\(=-3\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}-\dfrac{1}{36}\right)\)
\(=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{1}{12}< =\dfrac{1}{12}\)
Dấu = xảy ra khi x=1/6
c: \(=-\left(3x^2+4y^2-18x+8y-12\right)\)
\(=-\left(3x^2-18x+27+4y^2+8y+4-43\right)\)
\(=-3\left(x-3\right)^2-4\left(y+1\right)^2+43< =43\)
Dấu = xảy ra khi x=3 và y=-1
Tìm GTNN chủa biểu thức:
a, A=x2+6y2-2xy-12x+2y+45
b, B=x2-2xy+3y2-2xy-10y+20
c, C=x2+4y2-2xy-10x+4y+32
Bài làm:
a) x2 - y2 - 2x + 2y = (x - y)(x + y) - (2x - 2y)
= (x - y)(x + y) - 2(x - y)
= [(x + y) - 2].(x - y)
= (x + y - 2)(x - y)
c)3a2 - 6ab + 3b2 - 12c2 = (3a2 - 6ab + 3b2) - 12c2
= 3(a2 - 2ab + b2) - 12c2
= 3[(a - b)2] - 12c2
= 3[(a - b)2 - 4c2]
= 3[(a - b)2 - (2c)2]
= 3[(a - b - c) - (a - b + c)]
= 3(a - b - c - a + b - c)
= 3(-2c)
= -6c
d)x2 - 5 + y2 + 2xy = (x2 + 2xy + y2) - 5
= (x + y)2 - 5
= (x + y)2 -(\(\sqrt{5}\))2
= (x + y - \(\sqrt{5}\)) - (x + y + \(\sqrt{5}\))
= x + y - \(\sqrt{5}\) - x - y -\(\sqrt{5}\)
= -2\(\sqrt{5}\)
e) a2 + 2ab + b2 - ac - bc = (a2 + 2ab + b2) - (ac + bc)
= (a + b)2 - c(a + b)
= [(a + b) - c].(a + b)
= (a + b - c)(a + b)
Còn câu b) và câu f) Vàng sẽ nghĩ sau :v
Tiếp câu f luôn !
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
a) \(-\left(x+2\right)\cdot\left(x^2-1x+4\right)\)
\(=-\left(x+2\right)\cdot\left(x^2-x+4\right)\)
\(=-\left(x^3-x^2+4x+2x^2-2x+8\right)\)
\(=-\left(x^3+x^2+2x+8\right)\)
\(=-x^3-x^2-2x-8\)
b) \(-\left(x+2y\right)\cdot\left(x^2-2xy+y^2\right)\)
\(=-\left(x^3-2x^2y+xy^2+2x^2y-4xy^2+2y^3\right)\)
\(=-\left(x^3-3xy^2+2y^3\right)\)
\(=-x^3+3xy^2-2y^3\)
c) \(-\left(5-a\right)\cdot\left(25+5a+a^2\right)\)
\(=-\left(125-a^3\right)\)
\(=-125+a^3\)
d) \(-\left(x-2y\right)\cdot\left(x^2+2xy+4y^2\right)\)
\(=-\left(x^3-8y^3\right)\)
\(=-x^3+8y^3\)
Nguyễn Huy Túhình như câu b cũng sai đề nà