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

Biểu thức này không tồn tại max mà chỉ tồn tại min

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

Bạn coi lại mẫu số

TXĐ: \(\left\{{}\begin{matrix}x\in R\\x\notin\left\{0;-1\right\}\end{matrix}\right.\)

a: ĐKXĐ:\(x\notin\left\{2;0\right\}\)

b: \(C=\left(\dfrac{x\left(2-x\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{2-x^2+x}{x^2}\right)\)

\(=\dfrac{-x^3+4x^2-4x-4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{x^2}\)

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

c: Thay x=2017 vào C, ta được:

\(C=\dfrac{2017+1}{2\cdot2017}=\dfrac{1009}{2017}\)

1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)

\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)

\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)

\(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}\)

\(=-8\sqrt{3}\)

2) \(A=\sqrt{12-4x}\) có nghĩa khi:

\(12-4x\ge0\)

\(\Leftrightarrow4x\le12\)

\(\Leftrightarrow x\le\dfrac{12}{4}\)

\(\Leftrightarrow x\le3\)

3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)

\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}\right)^2-1^2}\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)

TXĐ: \(\left\{{}\begin{matrix}x\in R\\x\notin\left\{0;2;-2\right\}\end{matrix}\right.\)

Ta có: \(\left(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

\(=\left(\dfrac{x^2}{x\left(x-2\right)\left(x+2\right)}-\dfrac{6\left(x+2\right)}{3\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\dfrac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\right)\)

\(=\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{x^2-4+10-x^2}{x+2}\)

\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}\)

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{6}\)

\(=\dfrac{-1}{x-2}\)

a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}

Thay x = 2, ta có B không tồn tại

Thay x = -1, ta có B = \(\dfrac{1}{3}\)

b)ĐKXĐ:x ≠ 2,-2

Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x

Do đó không tồn tại x thỏa mãn đề bài

Đặt bthuc = A nhé

ĐKXĐ : \(2x\ne3y\)

\(A=\left[\dfrac{2x\left(4x^2+6xy+9y^2\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{27y^3+36xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{24xy\left(2x-3y\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{2x\left(2x-3y\right)}{\left(2x-3y\right)}+\dfrac{9y^2+12xy}{\left(2x-3y\right)}\right]\)\(=\left[\dfrac{8x^3+12x^2y+18xy^2-27y^3-36xy^2-48x^2y+72xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{4x^2-6xy+9y^2+12xy}{\left(2x-3y\right)}\right]\)

\(=\dfrac{8x^3-36x^2y+36xy^2-27y^3}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\cdot\dfrac{4x^2+6xy+9y^2}{2x-3y}\)

\(=\dfrac{\left(2x-3y\right)^3}{\left(2x-3y\right)^2}=2x-3y\)

Với x = 1/3 ; y = -2 (tmđk) thay vào A ta được : A = 2.1/3 - 3.(-2) = 20/3