cho x+y=1 x,y la so duong
tim gtln cua x:(x4+y2)+y:(y4+x2)
tính giá tri của các bieur thưc sau
a=x2(x+y)-y2(x+y)+x2-y2+2(x+y)+3 biết x+y+1=0
b=x4-xy3+x3y-y4+1 biết x+y=0
Cho x,y la cac so duong thoa man : x+y≤1. Tim GTNN cua:
P=(x4+y4+1)(1/x4+1/y4+1)
g) (x − y)(x + y)(x2+y2)(x4+y4)
h) (x − 3)(x + 3)(x − 4) − (x + 1)3
\(g,=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)=\left(x^4-y^4\right)\left(x^4+y^4\right)=x^8-y^8\)
\(b,=\left(x^2-9\right)\left(x-4\right)-\left(x^3+3x^2+3x+1\right)\\ =x^3-4x^2-9x+36-x^3-3x^2-3x-1\\ =-7x^2-12x+36\)
Cho x,y la cac so duong thoa man : x+y≤1. Tim GTNN cua:
P=(x4+y4+1)(1/x4+1/y4+1)
Can gap mn oi!!!
\(P=\left(x^4+y^4+\dfrac{1}{256}+\dfrac{255}{256}\right)\left(\dfrac{1}{x^4}+\dfrac{1}{y^4}+1\right)\)
\(P=\left(x^4+y^4+\dfrac{1}{256}\right)\left(\dfrac{1}{x^4}+\dfrac{1}{y^4}+1\right)+\dfrac{255}{256}\left(\dfrac{1}{x^4}+\dfrac{1}{y^4}+1\right)\)
\(P\ge\left(\dfrac{x^2}{x^2}+\dfrac{y^2}{y^2}+\dfrac{1}{16}\right)^2+\dfrac{255}{256}\left(\dfrac{1}{2}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)^2+1\right)\)
\(P\ge\left(\dfrac{33}{16}\right)^2+\dfrac{255}{256}\left(\dfrac{1}{2}\left(\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\right)^2+1\right)\)
\(P\ge\left(\dfrac{33}{16}\right)^2+\dfrac{255}{256}\left(\dfrac{1}{8}\left(\dfrac{4}{x+y}\right)^4+1\right)\ge\left(\dfrac{33}{16}\right)^2+\dfrac{255}{256}\left(\dfrac{4^4}{8}+1\right)=\dfrac{297}{8}\)
\(P_{min}=\dfrac{297}{8}\) khi \(x=y=\dfrac{1}{2}\)
Chứng minh: ( x 3 + x 2 y + x y 2 + y 3 )(x - y) = x 4 – y 4
Ta có: VT = ( x 3 + x 2 y + x y 2 + y 3 )(x - y)
= ( x- y). ( x 3 + x 2 y + x y 2 + y 3 ).
= x. ( x 3 + x 2 y + x y 2 + y 3 ) - y( x 3 + x 2 y + x y 2 + y 3 )
= x 4 + x 3 y + x 2 y 2 + x y 3 – x 3 y – x 2 y 2 – x y 3 – y 4
= x 4 – y 4 = VP (đpcm)
Vế trái bằng vế phải nên đẳng thức được chứng minh.
1, Cho biết x+y=15 và xy=50. Tính giá trị của các biểu thức:
a. A=x2+y2
b. B=x4+y4
c. C=x2-y2
2, Cho biết x-y=15 và xy=50. Hãy tính x2+y2 ; x2-y2 rồi từ đó suy ra kết quả của x4-y4.
a: \(A=x^2+y^2=\left(x+y\right)^2-2xy=15^2-2\cdot50=115\)
c: \(x-y=\sqrt{\left(x+y\right)^2-4xy}=\sqrt{15^2-4\cdot50}=5\)
\(C=x^2-y^2=\left(x+y\right)\left(x-y\right)=15\cdot5=75\)
chứng minh giá trị biểu thức sau không phụ thuộc vào giá trị của biến:
a, A = y (x2 - y2) (x2 + y2) - y (x4 - y4)
b, B = (x - 1)3 - (x - 1) (x2 + x + 1) - 3 (1 - x) x
a) \(A=y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)=0\)
b) \(B=\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x=x^3-3x^2+3x-1-x^3-x^2-x+x^2+x+1-3x+3x^2=0\)
a: Ta có: \(A=y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)
\(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)\)
=0
b: Ta có: \(B=\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)
\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)
=0
Rút gọn biểu thức:
x(x + y)(x2+y2)(x4+y4)(x8+y8)(x - y) + xy16
\(x\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\left(x-y\right)+xy^{16}\\ =x\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\\ =x\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\\ =x\left(x^4-y^4\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\\ =x\left(x^8-y^8\right)\left(x^8+y^8\right)+xy^{16}\\ =x\left(x^{16}-y^{16}\right)+xy^{16}\\ =x^{17}-xy^{16}+xy^{16}\\ =x^{17}\)
\(x\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\left(x-y\right)+xy^{16}\)
\(=x\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\)
\(=x\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\)
\(=x\left(x^4-y^4\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\)
\(=x\left(x^8-y^8\right)\left(x^8+y^8\right)+xy^{16}\)
\(=x\left(x^{16}-y^{16}\right)+xy^{16}\)
\(=x^{17}-xy^{16}+xy^{16}\)
\(=x^{17}\)
\(x(x+y)(x^2+y^2)(x^4+y^4)(x^8+y^8)(x-y)+xy^{16}\\=x(x-y)(x+y)(x^2+y^2)(x^4+y^4)(x^8+y^8)+xy^{16}\\=x(x^2-y^2)(x^2+y^2)(x^4+y^4)(x^8+y^8)+xy^{16}\\=x(x^4-y^4)(x^4+y^4)(x^8+y^8)+xy^{16}\\=x(x^8-y^8)(x^8+y^8)+xy^{16}\\=x(x^{16}-y^{16})+xy^{16}\\=x^{17}-xy^{16}+xy^{16}\\=x^{17}\\Toru\)
Chứng minh các bất đẳng thức sau với x, y, z > 0
a) x2 + y2 ≥ (x + y)2/2
b) x3 + y3 ≥ (x + y)3/4
c) x4 + y4 ≥ (x + y)4/8
d) x2 + y2 + z2 ≥ xy + yz + zx
e) x2 + y2 + z2 ≥ (x + y + z)2/3
f) x3 + y3 + z3 ≥ 3xyz
a: Ta có: \(\left(x+y\right)^2\)
\(=x^2+2xy+y^2\)
\(\Leftrightarrow x^2+y^2=\dfrac{\left(x+y\right)^2}{2xy}\ge\dfrac{\left(x+y\right)^2}{2}\forall x,y>0\)