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a: Ta có: \(x^2-x+1\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{4}\)
b: Ta có: \(x^2+y^2-4x+y+5\)
\(=\left(x^2-4x+4\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=\left(x-2\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x,y\)
Dấu '=' xảy ra khi x=2 và \(y=-\dfrac{1}{2}\)
9x2 - 6x + 1 = (3x + 1 )2
-x2 + 6x - 9 = ( -x - 3 )2
25x2 + 30x + 9 = ( 5x + 3 )2
-1/4 + 2x - x2 = ( - 1/2 - x )2
4x2 - 12x + 9 = ( 2x +3 )2
a, 9x2-6x+1
=(3x)2-2.3x.1+12
=(3x-1)2
b, -x2+6x-9
=-(x2-6x+9)
=-(x2-2x.3+32)
=-(x-3)2
c, 25x2+30x+9
=(5x)2+2.5x.3+32
=(5x+3)2
d, Mik nghĩ là đề sai chỗ +2x phải là +x (ko tin bn có thể thế đại 1 số rùi thử lại nhoa!!!)
e, 4x2-12x+9
=(2x)2-2.2x.3+32
=(2x-3)2
Chúc bn học giỏi nhoa!!!
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
a: 2x^2y-50xy=2xy(x-25)
b: 5x^2-10x=5x(x-2)
c: 5x^3-5x=5x(x^2-1)=5x(x-1)(x+1)
d: \(x^2-xy+x=x\left(x-y+1\right)\)
e: x(x-y)-2(y-x)
=x(x-y)+2(x-y)
=(x-y)(x+2)
f: 4x^2-4xy-8y^2
=4(x^2-xy-2y^2)
=4(x^2-2xy+xy-2y^2)
=4[x(x-2y)+y(x-2y)]
=4(x-2y)(x+y)
f1: x^2ỹ-y^2+y
=(x-y)(x+y)+(x+y)
=(x+y)(x-y+1)
c) 37 . 43
= (40 - 3) . ( 40+3)
= 40^2 - 3^2
= 1600 - 0
= 1591
BÀI 3: viết biểu thức sau dưới dạng tổng, hiệu 2 bình phương.
4 x^2 - 12x - y^2 + 2y + 1
= (2x)^2 -12x +9-y^2 +2y +1
= (2x-3)^2 - (y+1)^2
a) \(3x^2-3xy-5x+5y\)
\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
b) \(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left[x^2-\left(y+1\right)^2\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
c) \(x^2+1+2x-y^2\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
d) \(x^2+4x-2xy-4y+y^2\)
\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+4\right)\)
e) \(x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(=x\left(x-1\right)^2\)
f) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
a: =3x(x-y)-5(x-y)
=(x-y)(3x-5)
b: \(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
d:
Sửa đề: x^2+4x-2xy-4y+y^2
=x^2-2xy+y^2+4x-4y
=(x-y)^2+4(x-y)
=(x-y)(x-y+4)
e: =x(x^2-2x+1)
=x(x-1)^2
f: =2(x^2+2x+1-y^2)
=2[(x+1)^2-y^2]
=2(x+1+y)(x+1-y)
a/ \(4x^2+2y^2-4xy+4x-2y+5=0\)
\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+2\left(2x-y\right)+1+4=0\)
\(\Leftrightarrow\left(2x-y\right)^2+2\left(2x-y\right)+1+4=0\)
\(\Leftrightarrow\left(2x-y+1\right)^2+4=0\)
Với mọi x, y ta có :
\(\left(2x-y+1\right)^2\ge0\Leftrightarrow\left(2x-y+1\right)^2+4>0\)
\(\Leftrightarrow pt\) vô nghiệm
a) \(4x^2-4xy+2y^2+2y+1=\left(2x-y\right)^2+\left(y+1\right)^2\)
b) \(x^2+2x+y^2-4y+5=\left(x+1\right)^2+\left(y-2\right)^2\)
c) bạn ktra lại đề