cho x>y>0 vaf x^5+y^5=x-y. Chung minh rang x^4+y^4< 1
Những câu hỏi liên quan
cho x+y+z=0 chung minh rang:
2x^4+2y^4+2z^4=(x^2+y^2+z^2)^2
cho x = a/m , y = b/m vaf x < y .hay chung to rang x<z<y voi z= a+b/2m
cho x, y,z >0 chung minh rang\(\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}< hoac=\frac{3}{ }4\)3/4
Chung minh rang neu 2(x+y) = 5(y+z) = 3(z+x) thi \(\frac{x-y}{4}\) \(\frac{y-z}{5}\)
\(2\left(x+y\right)=5\left(y+z\right)=3\left(z+x\right)\)
\(\Leftrightarrow\frac{x+y}{15}=\frac{y+z}{6}=\frac{z+x}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{z+x}{10}=\frac{y+z}{6}=\frac{\left(z+x\right)-\left(y+z\right)}{10-6}=\frac{x-y}{4}\)
\(\frac{x+y}{15}=\frac{z+x}{10}=\frac{\left(x+y\right)-\left(z+x\right)}{15-10}=\frac{y-z}{5}\)
Suy ra đpcm.
cho x, y , z la cac so nguyen thoa man x . y - x. z + y.z - z^2 +1 =0 chung minh rang x+ y =0
cho x, y duong thoa man:x+y=1.Chung minh rang \(8\left(x^4+y^4\right)+\frac{1}{xy}\ge5\)
\(A=8\left(x^4+y^4\right)+\frac{1}{4xy}+\frac{1}{4xy}+\frac{1}{2xy}\ge8\left(x^4+y^4\right)+\frac{1}{2\left(x^2+y^2\right)}+\frac{1}{2\left(x^2+y^2\right)}+\frac{1}{2xy}\)
\(\Rightarrow A\ge8\left(x^4+y^4\right)+\frac{1}{2\sqrt{2\left(x^4+y^4\right)}}+\frac{1}{2\sqrt{2\left(x^4+y^4\right)}}+\frac{1}{2\left(\frac{x+y}{2}\right)^2}\)
\(\Rightarrow A\ge3\sqrt[3]{8\left(x^4+y^4\right)\cdot\frac{1}{2\sqrt{2\left(x^4+y^4\right)}}\cdot\frac{1}{2\sqrt{2\left(x^4+y^4\right)}}}+\frac{1}{2\cdot\frac{1}{4}}=3+2=5\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)
Cho x,y duong thoa man: x+y=3. Chung minh rang x2y <= 4
Chung minh rang x^10+y^10 chia het cho x^4+x^3y+x^2y^2+xy^3+y^4
cho x,y,z la 3 so thuc tuy y thoa man x+y+z=0 va -1< x<1,-1<y<1,-1<z<1.chung minh rang da thuc x^2+y^4+z^6 co gia tri khong lon hon 2