viết sai đề hết rồi

rưa mi chỉnh t cấy

Biến đổi mỗi đa thức theo hướng làm xuất hiện thừa số x+y-2 \(M=x^3+x^2y-2x^2-xy-y^2+3y+x-1\)

\(M=x^3+x^2y-2x^2-xy-y^2+\left(2y+y\right)+x-\left(-2+1\right)\)

\(M=\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(x+y-2\right)+1\)

\(M=\left(x^2.x+x^2.y-2x^2\right)-\left(x.y+y.y-2y\right)+\left(x+y-2\right)+1\)

\(M=x^2.\left(x+y-2\right)-y.\left(x+y-2\right)+\left(x+y-2\right)+1\)

\(M=x^2.0+y.0+0+1\)

\(M=1\)

\(N=x^3+x^2y-2x^2-xy^2+x^2y+2xy+2y+2x-2\)

\(N=x^3+x^2y-2x^2-xy^2+x^2y+2xy+2y+2x-\left(-4+2\right)\)

\(N=\left(x^3+x^2y-2x^2\right)-\left(x^2y+xy^2-2xy\right)+\left(2x+2y-4\right)+2\)

\(N=\left(x^2x+x^2y-2x^2\right)-\left(xyx+xyy-2xy\right)+\left(2x+2y-4\right)+2\)

\(N=x^2\left(x+y-2\right)-xy\left(x+y-2\right)+2\left(x+y-2\right)+2\)

\(N=x^2.0-xy.0+2.0+2\)

\(N=2\)

\(P=x^4+2x^3y-2x^3+x^2y^2-2x^2y-x\left(x+y\right)+2x+3\)

\(P=\left(x^4+x^3y-2x^3\right)+\left(x^3y+x^2y^2-2x^2y\right)-\left(x^2+xy-2x\right)+3\)\(P=\left(x^3x+x^3y-2x^3\right)+\left(x^2y.x+x^2yy-2x^2y\right)-\left(xx+xy-2x\right)+3\)

\(P=x^3\left(x+y-2\right)+x^2y\left(x+y-2\right)-x\left(x+y-2\right)+3\)

\(P=x^3.0+x^2y.0-x.0+3\)

\(P=3\)

Tích mình nha!

Mà bài này hình như học ở lớp 7 rồi!

??? đề bài

Tìm min mn  ạ

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

a: =(xy-2x)-(y^2-2y)

=x(y-2)-y(y-2)

=(x-y)(y-2)

b: =(x^2-2xy+y^2)-(x-y)

=(x-y)^2-(x-y)

=(x-y)(x-y-1)

c: =(x^2-1)-(2xy-2y)

=(x-1)(x+1)-2y(x-1)

=(x-1)(x+1-2y)

d: =(x+3)(x+3-2x+5)

=(x+3)(8-x)

\(a,xy-2x-y^2+2y\)

\(=x\left(y-2\right)-y\left(y-2\right)\)

\(=\left(x-y\right)\left(y-2\right)\)

\(b,x^2-2xy+y^2-x+y\)

\(=\left(x-y\right)^2-\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y-1\right)\)

\(c,x^2-1-2xy+2y\)

\(=\left(x-1\right)\left(x+1\right)-2y\left(x-1\right)\)

\(=\left(x-1\right)\left(x+1-2y\right)\)

\(d,\left(x+3\right)^2-\left(2x-5\right)\left(x+3\right)\)

\(=\left(x+3\right)\left(x+3-2x+5\right)\)

\(=\left(x+3\right)\left(-x+8\right)\)

#Urushi

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

mấy bn ơi, giúp mk nhanh vs nha!!!!!!!!!!!

a/ A = 2x² + y² - 2xy - 2x + 3

= (x² - 2xy + y²) + (x² - 2x + 1) + 2

= (x - y)² + (x - 1)² + 2\(\ge2\)