minh cung lop 8

x^3-xy^2-6xy^2+6y^3

=x(x^2-y^2)-6y^2(x-y)

=x(x-y)(x+y)-6y^2(x-y)

=(x-y)(x(x+y)-6y^2)

ban cu hoc thuoc HẰNG ĐẲNG THỨC LÀ RA NGAY CO J CỨ HỎI MINH MINH CHUYEN TOAN , ANH , HOA

ta có: \(\dfrac{x^2+y^2}{xy}=\dfrac{2}{3}\Rightarrow2xy=3x^2+3y^2\\ \Rightarrow6xy=9x^2+9y^2\)

thay vào M, ta được:

\(M=\dfrac{x^2+9x^2+9y^2+y^2}{x^2-9x^2-9y^2+y^2}=\dfrac{10x^2+10y^2}{-8x^2-8y^2}\\ M=\dfrac{10\left(x^2+y^2\right)}{-8\left(x^2+y^2\right)}=\dfrac{10}{-8}=-\dfrac{5}{4}\)

\(A=\dfrac{x^2+y^2}{xy}+\dfrac{xy}{x^2+y^2}=\dfrac{x^2+y^2}{4xy}+\dfrac{xy}{x^2+y^2}+\dfrac{3\left(x^2+y^2\right)}{4xy}\)

\(A\ge2\sqrt{\dfrac{\left(x^2+y^2\right)xy}{4xy\left(x^2+y^2\right)}}+\dfrac{3.2xy}{4xy}=\dfrac{5}{2}\)

Dấu "=" xảy ra khi \(x=y\)

\(C=\dfrac{\left(x+y\right)^2-4xy}{xy}+\dfrac{6xy}{\left(x+y\right)^2}=\dfrac{\left(x+y\right)^2}{xy}+\dfrac{6xy}{\left(x+y\right)^2}-4\)

\(C=\dfrac{3\left(x+y\right)^2}{8xy}+\dfrac{6xy}{\left(x+y\right)^2}+\dfrac{5\left(x+y\right)^2}{8xy}-4\)

\(C\ge2\sqrt{\dfrac{18xy\left(x+y\right)^2}{8xy\left(x+y\right)^2}}+\dfrac{5.4xy}{8xy}-4=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(x=y\)

\(C=\dfrac{\left(x+y\right)^2-4xy}{xy}+\dfrac{6xy}{\left(x+y\right)^2}=\dfrac{\left(x+y\right)^2}{xy}+\dfrac{6xy}{\left(x+y\right)^2}-4\)

\(C=\dfrac{3\left(x+y\right)^2}{8xy}+\dfrac{6xy}{\left(x+y\right)^2}+\dfrac{5\left(x+y\right)^2}{8xy}-4\)

\(C\ge2\sqrt{\dfrac{18xy\left(x+y\right)^2}{8xy\left(x+y\right)^2}}+\dfrac{5.4xy}{8xy}-4=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(x=y\)

\(a,=12x^3y-4x^2y^2+3xy^3\\ b,=x^3+3x^2-5x+3x^2+9x-15-x^3-4x^2+4x\\ =2x^2+8x-15\)

b: Ta có: \(\left(x+3\right)\left(x^2+3x-5\right)-x\left(x-2\right)^2\)

\(=x^3+3x^2-5x+3x^2+9x-15-x^3+4x^2-4x\)

\(=10x^2-15\)

a/ Cho x, y ≥ 1. Chứng minh: 1/(1 + x^2) + 1/(1 + y^2) ≥ 2/(1 + xy)

b/ Đề:...Tìm GTLN

Có:

\(\dfrac{1}{4x^2-4x+2}=\dfrac{1}{\left(2x-1\right)^2+1}\le\dfrac{1}{2}\forall x\ge1\)

\(\dfrac{1}{9y^2+6y+2}=\dfrac{1}{\left(3y+1\right)^2+1}\le\dfrac{1}{2}\forall y\ge0\)

\(\Rightarrow A=\dfrac{1}{4x^2-4x+2}+\dfrac{1}{9y^2+6y+2}\le\dfrac{1}{2}+\dfrac{1}{2}=1\)

Vậy MAX_A = 1 khi \(\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

\(B=x+y+\dfrac{6}{x}+\dfrac{24}{y}=\left(\dfrac{3x}{2}+\dfrac{6}{x}\right)+\left(\dfrac{3y}{2}+\dfrac{24}{y}\right)-\dfrac{3}{2}\left(x+y\right)\)

\(B\ge2\sqrt{\dfrac{18x}{2x}}+2\sqrt{\dfrac{72y}{2y}}-\dfrac{3}{2}.6=15\)

\(B_{min}=15\) khi \(\left(x;y\right)=\left(2;4\right)\)

\(\dfrac{x^2-9y^2}{x^2+xy-6y^2}=\dfrac{\left(x-3y\right)\left(x+3y\right)}{\left(x-2y\right)\left(x+3y\right)}=\dfrac{x-3y}{x-2y}\)

Giải:

\(\left(3x^2y-\dfrac{1}{2}x^2+\dfrac{1}{5}xy\right)6xy^3\)

\(=3x^2y.6xy^3-\dfrac{1}{2}x^2.6xy^3+\dfrac{1}{5}xy.6xy^3\)

\(=18x^3y^4-3x^3y^3+\dfrac{6}{5}x^2y^4\)

Vậy ...