\(a) x^2+x-xy-2y^3+y \)
\(b) x^2 +4xy+2x +3y^2+6y\)
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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)
\(a,x^2-x-xy-2y^2+2y\)
\(=xy+x^2-x-2y^2-2xy+2y\)
\(=\left(xy-2y^2\right)+\left(x^2-2xy\right)-\left(x-2y\right)\)
\(=y\left(x-2y\right)+x\left(x-2y\right)-\left(x-2y\right)\)
\(=\left(x-2y\right)\left(y+x-1\right)\)
\(b,x^2+4xy+2x+3y^2+6y\)
\(=3y^2+xy+3xy+x^2+6y+2x\)
\(=y\left(3y+x\right)+x\left(3y+x\right)+2\left(3y+x\right)\)
\(=\left(3y+x\right)\left(y+x+2\right)\)
a) \(x^2-x-xy-2y^2+2y\)
\(=x^2-x-xy-y^2-y^2+y+y\)
\(=\left(x^2-y^2\right)-\left(x-y\right)-\left(xy+y^2-y\right)\)
\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)-y\left(x+y-1\right)\)
\(=\left(x-y\right)\left(x+y-1\right)-y\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x-2y\right)\)
Câu a, b, c thì đơn giản òi. Câu d phải chú ý điểm rơi:v
d) Ta có: \(D=\left(x-\frac{1}{2}\right)^4+\frac{1}{2}\left(3x^2-3x+\frac{15}{8}\right)\)
\(=\left(x-\frac{1}{2}\right)^4+\frac{3}{2}\left(x-\frac{1}{2}\right)^2+\frac{9}{16}\ge\frac{9}{16}\)
Đẳng thức xảy ra khi x = 1/2
2x^2+xy+2y^2 = 5/4.(x+y)^2 + 3/4. (x-y)^2 >= 5/4. (x+y)^2
=> cbh(2x^2+xy+2y^2) >= cbh5 / 2. (x+y)
tương tự với 2 căn còn lại.. cộng vế ta có VT >= cbh5 ( x+y+z) = cbh5 : dpcm
dau = cay ra <=> x=y=z=1/3
\(1,=\left(x-2\right)\left(5-y\right)\\ 2,=2\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(2x-2y-z\right)\\ 3,=5xy\left(x-2y\right)\\ 4,=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-4z^2\right]\\ =3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\\ 6,=-\left(6x^2-3x-4x+2\right)=-\left(2x-1\right)\left(3x-2\right)\\ 7,=\left(2x+y\right)\left(2x+y+x\right)=\left(2x+y\right)\left(3x+y\right)\\ 8,=\left(x-y\right)\left(x+5\right)\\ 9,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 10,=\left(x^2-9\right)x=x\left(x-3\right)\left(x+3\right)\\ 11,=\left(x-2\right)\left(y+1\right)\\ 12,=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\\ 13,=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)
a) \(x^2-x-xy-2y^2+2y\)
\(=x^2-x-2xy+xy-2y^2+2y\)
\(=\left(-2y^2-2xy+2y\right)+\left(xy+x^2-x\right)\)
\(=2y\left(-y-x+1\right)+x\left(y+x-1\right)\)
\(=-2y\left(y+x-1\right)+x\left(y+x-1\right)\)
\(=\left(x-2y\right)\left(y+x-1\right)\)
b) \(x^2+4xy+2x+3y^2+6y\)
\(=x^2+3xy+xy+2x+3y^2+6y\)
\(=\left(3y^2+3xy+6y\right)+\left(xy+x^2+2x\right)\)
\(=3y\left(y+x+2\right)+x\left(y+x+2\right)\)
\(=\left(3y+x\right)\left(y+x+2\right)\)
a) \(3x^2+x-2=3x^2+3x-2x-2=3x\left(x+1\right)-2\left(x+1\right)=\left(x+1\right)\left(3x-2\right)\)
b)Đề sai, số quá xấu
Xin một slot tí làm full
Bạn tth làm câu a) rồi, câu b) sai đề nên mình làm tiếp nhé :
c) \(2x^2-3xy-2y^2=2x^2+xy-4xy-2y^2\)
\(=x.\left(2x+y\right)-2y.\left(2x+y\right)\)
\(=\left(x-2y\right).\left(2x+y\right)\)
d) \(x^2-x-xy-2y^2+2y=x^2-x+xy-2xy-2y^2+2y\)
\(=x.\left(-1+x+y\right)-2y.\left(-1+x+y\right)\)
\(=\left(x-2y\right).\left(-1+x+y\right)\)
e) \(x^2+4xy+2x+3y^2+6y=x^2+xy+2x+3xy+3y^2+6y\)
\(=x.\left(x+y+2\right)+3y.\left(x+y+2\right)\)
\(=\left(x+3y\right).\left(x+y+2\right)\)
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