cho x, y > 0, x2-y= y2-x. Tìm A= 2xy -x2-y2+3x-3y+4
tks mọi người nhiều
cho x,y khac nhau biet x2-y=y2-x. tinh gia tri cua bieu thuc A=x2+2xy+y2-3x-3y
Theo bài ra, ta có: \(x^2-y=y^2-x\Leftrightarrow x^2-y^2=-x+y\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)=-\left(x-y\right)\)
\(\Leftrightarrow\left(x+y\right)=-1\)
Ta lại có: \(A=x^2+2xy+y^2-3x-3y=\left(x+y\right)^2-3\left(x+y\right)\)
Thay x+y=-1 vào biểu thức A, ta được: \(A=\left(-1\right)^2-3.\left(-1\right)=1+3=4\)
Vậy A=4
1.
a.(-xy)(-2x2y+3xy-7x)
b.(1/6x2y2)(-0,3x2y-0,4xy+1)
c.(x+y)(x2+2xy+y2)
d.(x-y)(x2-2xy+y2)
2.
a.(x-y)(x2+xy+y2)
b.(x+y)(x2-xy+y2)
c.(4x-1)(6y+1)-3x(8y+4/3)
1.
\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)
\(=2x^3y^2-3x^2y^2+7x^2y\)
\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)
\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)
\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x+y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3\)
\(d,\left(x-y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)^3\)
\(=x^3-3x^2y+3xy^2-y^3\)
2.
\(a,\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3-y^3\)
\(b,\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^3+y^3\)
\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)
\(=24xy+4x-6y-1-24xy-4x\)
\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)
\(=-6y-1\)
#Toru
Phân tích đa thức thành nhân tử:
a) 3x-3y-x2+2xy-y2
b) x2-4x2y2+y2+2xy
c) (x+y)3-(x-y)3
d) x2-5x-14
\(a,=3\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(3-x+y\right)\\ b,=\left(x+y\right)^2-4x^2y^2=\left(x-2xy+y\right)\left(x+2xy+y\right)\\ c,=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\\ =2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\\ =2y\left(3x^2+y^2\right)\\ d,=x^2+2x-7x-14=\left(x+2\right)\left(x-7\right)\)
Bài 1: Rút gọn các biểu thức:
a. (2x - 1)2 - 2(2x - 3)2 + 4
b. (3x + 2)2 + 2(2 + 3x)(1 - 2y) + (2y - 1)2
c. (x2 + 2xy)2 + 2(x2 + 2xy)y2 + y4
d. (x - 1)3 + 3x(x - 1)2 + 3x2(x -1) + x3
e. (2x + 3y)(4x2 - 6xy + 9y2)
f. (x - y)(x2 + xy + y2) - (x + y)(x2 - xy + y2)
g. (x2 - 2y)(x4 + 2x2y + 4y2) - x3(x – y)(x2 + xy + y2) + 8y3
a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)
a) 3x-3y+x2-y2
b) (2xy+1)^2-(2x+y)^2
c)(x2+y2-5)^2-4(x2y2+4xy+4) d) (x2+y2-z2)^2-4x2y2
e) 9x2 +90
x+225-(x-7)^2
bn viết rõ đề đi bn
Vd:x2 là 2.x hay x\(^2\)
Có nhiều chỗ vậy lắm bn ạ,bn viết lại đề đi rồi tụi mk giúp cho.
a) \(3x-3y+x^2-y^2\)
\(=3\left(x-y\right)+\left(x+y\right)\left(x-y\right)\)
\(=\left(3+x+y\right)\left(x-y\right)\)
b) \(\left(2xy+1\right)^2-\left(2x+y\right)^2\)
\(=\left[\left(2xy+1\right)-\left(2x+y\right)\right]\left[\left(2xy+1\right)+\left(2x+y\right)\right]\)
\(=\left(2xy+1-2x-y\right)\left(2xy+1+2x+y\right)\)
\(=\left(y+1\right)\left(2x+1\right)\left(y-1\right)\left(2x-1\right)\)
c) \(\left(x^2+y^2-5\right)^2-4\left(x^2y^2+4xy+4\right)\)
↓
\(=\left(x^2-y^2-2y-1\right)\left(x^2-2xy+y^2-9\right)\)
\(=\left[x^2-\left(y^2+2y+1\right)\right]\left(x^2-2xy+y^2-9\right)\)
\(=\left[x^2-\left(y+1\right)^2\right]\left[\left(x-y\right)^2-3^2\right]\)
\(=\left[x^2-\left(-y-1\right)^2\right]\left(x-y+3\right)\left(x-y-3\right)\)
\(=\left(x+y+1\right)\left(x-y-1\right)\left(x-y+3\right)\left(x-y-3\right)\)
d) \(\left(x^2+y^2-z^2\right)^2-4x^2y^2\)
\(=\left(x^2+y^2-z^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+y^2-z^2-2xy\right)\left(x^2+y^2-z^2+2xy\right)\)
\(=\left[\left(x-y\right)^2-z^2\right]\left[\left(x+y\right)^2-z^2\right]\)
\(=\left(x-y-z\right)\left(x-y+z\right)\left(x+y-z\right)\left(x+y+z\right)\)
e)
- \(9x^2+90=9\left(x+10\right)\)
- \(x+225-\left(x-7\right)^2\)
\(=x+225-\left(x^2-14x+49\right)\)
\(=x+225-x^2+14x-49\)
\(=-x^2+15x+176\)
\(=-\left(x^2-15x-176\right)\)
CM rằng
a) x2+2xy+y2+1>0 với mọi x
b) x2+y2+1≥xy+x+y
c) x2-x+1>0 với mọi số thực x
em mong mọi người giúp đỡ em cảm ơn ạ
a) \(x^2+2xy+y^2+1\\ =\left(x+y\right)^2+1\\Do\left(x+y\right)^2>0\forall x\in R\\ \Rightarrow\left(x+y\right)^2+1>0\forall\in R\)
Bài 1: Rút gọn các biểu thức:
a. (2x - 1)2 - 2 (2x - 3)2 + 4
b. (3x + 2)2 + 2 (2 + 3x) (1 - 2y) + (2y - 1)2
c. (x2 + 2xy)2 + 2 (x2 + 2xy) y2 + y4
d. (x - 1)3 + 3x (x - 1)2 + 3x2 (x -1) + x3
e. (2x + 3y) (4x2 - 6xy + 9y2)
f. (x - y) (x2 + xy + y2) - (x + y) (x2 - xy + y2)
g. (x2 - 2y) (x4 + 2x2y + 4y2) - x3 (x – y) (x2 + xy + y2) + 8y3
a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
Mọi người ơi,giúp mình câu này với ạh,cảm ơn mn
Phân tích đa thức thành nhân tử:
1)(x + 2)(x + 3)(x + 4)(x + 5) - 24
2)x2 - 2xy + y2 + 3x - 3y -1
mọi người giúp mình với nhé,cảm ơn nhiều lắm!💜🥺
1)
\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)
Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:
\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)
2) Bạn xem lại đề!
c) 3x + 3y – x2 – 2xy – y2 d) x3 – x + 3x2y + 3xy2 – y + y3
c) \(3x+3y-x^2-2xy-y^2=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)d) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
\(c,=3\left(x+y\right)-\left(x+y\right)^2=\left(3-x-y\right)\left(x+y\right)\\ d,=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
c) \(=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)
d) \(\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)