chứng minh rằng: \(\dfrac{2x^2+3xy+y^2}{2x^3+x^2y-2xy^2-y^3}=\dfrac{1}{x-y}\)
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5 tháng 12 2021
\(VT=\dfrac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}=\dfrac{\left(x+y\right)\left(x+2y\right)}{\left(x+2y\right)\left(x-y\right)\left(x+y\right)}=\dfrac{1}{x-y}\)
18 tháng 10 2021
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
5 tháng 11 2017
\(\text{1) }\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\\ =\dfrac{\left(x^7+x^6\right)+\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6+x^4+x^2+1}{x-1}\)
\(\text{3) }\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\\ =\dfrac{\left(x^2-2xy+y^2\right)+\left(2xz-2yz\right)+z^2}{\left(x^2-2xy+y^2\right)-z^2}\\ =\dfrac{\left(x-y\right)^2+2\left(x-y\right)z+z^2}{\left(x-y\right)^2-z^2}\\ =\dfrac{\left(x-y+z\right)^2}{\left(x-y+z\right)\left(x-y-z\right)}\\ =\dfrac{x-y+z}{x-y-z}\)
Z
23 tháng 12 2020
B) Ta có: 2x-2y-x2+2xy-y2
⇔ 2(x-y)-(x2-2xy+y2)
⇔ 2(x-y)-(x-y)2
⇔ (x-y)(2-x+y)
Đúng thì tick nhé
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 12 2021
\(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}=\dfrac{x^4}{xy+2xz}+\dfrac{y^4}{yz+2xy}+\dfrac{z^4}{xz+2yz}\)
\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)
19 tháng 11 2021
\(a,VT=\dfrac{x^2+2xy+4-3x^2-3xy}{\left(x+y\right)\left(x+2y\right)}=\dfrac{-2x^2-xy+4}{\left(x+y\right)\left(x-2y\right)}=VP\\ b,VP=\dfrac{\left(x+y\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{x-y}=VT\)
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 4 2021
\(VT\le\dfrac{x}{2x+2y+2}+\dfrac{y}{2yz+2z+2}+\dfrac{z}{2z+2x+2}\)
Nên ta chỉ cần chứng minh: \(\dfrac{x}{x+y+1}+\dfrac{y}{y+z+1}+\dfrac{z}{z+x+1}\le1\)
\(\Leftrightarrow\dfrac{y+1}{x+y+1}+\dfrac{z+1}{y+z+1}+\dfrac{x+1}{z+x+1}\ge2\)
Thật vậy, ta có:
\(VT=\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(z+x+1\right)}+\dfrac{\left(y+1\right)^2}{\left(y+1\right)\left(x+y+1\right)}+\dfrac{\left(z+1\right)^2}{\left(z+1\right)\left(y+z+1\right)}\)
\(VT\ge\dfrac{\left(x+y+z+3\right)^2}{\left(x^2+y^2+z^2\right)+3\left(x+y+z\right)+xy+yz+zx+3}\)
\(VT\ge\dfrac{6\left(x+y+z\right)+2\left(xy+yz+zx\right)+12}{3\left(x+y+z\right)+xy+yz+zx+6}=2\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 9 2021
\(VT^2\le3\left(\dfrac{1}{2x^2+y^2+3}+\dfrac{1}{2y^2+z^2+3}+\dfrac{1}{2z^2+x^2+3}\right)\)
Mặt khác:
\(\dfrac{1}{2\left(x^2+1\right)+y^2+1}\le\dfrac{1}{4x+2y}=\dfrac{1}{2}\left(\dfrac{1}{x+x+y}\right)\le\dfrac{1}{18}\left(\dfrac{2}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow VT^2\le\dfrac{1}{6}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{3}{2}\)
\(\Rightarrow VT\le\dfrac{\sqrt{6}}{2}\)
\(\dfrac{2x^2+3xy+y^2}{2x^3+x^2y-2xy^2-y^3}=\dfrac{1}{x-y}\)
\(VT=\dfrac{2x^2+3xy+y^2}{2x^3+x^2y-2xy^2-y^3}\)
\(=\dfrac{2x^2+2xy+xy+y^2}{\left(2x^3+x^2y\right)+\left(-2xy^2-y^3\right)}\)
\(=\dfrac{\left(2x^2+2xy\right)+\left(xy+y^2\right)}{x^2\left(2x+y\right)-y^2\left(2x+y\right)}\)
\(=\dfrac{2x\left(x+y\right)+y\left(x+y\right)}{\left(x^2-y^2\right)\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)\left(x+y\right)}{\left(x^2-y^2\right)\left(2x+y\right)}\)
\(=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{1}{x-y}=VP\left(đpcm\right)\)