câu c) mang tính mua vui hay gì hả bn

mếu thật thì x=0,x=số nào cx đc(câu trả lời này mang tính mua vui thôi nhé)

b) (4√x + 4)/(x + 2√x + 5) ≥ 1

⇔ (4√x + 4)/(x + 2√x + 5) - 1 ≤ 0

Do x ≥ 0 ⇒ x + 2√x + 5 > 0

⇒ (4√x + 4)/(x + 2√x + 5) - 1 ≤ 0

⇔ (4√x + 4) - (x + 2√x + 5) ≤ 0

⇔ 4√x + 4 - x - 2√x - 5 ≤ 0

⇔ -x + 2√x - 1 ≤ 0

⇔ -(x - 2√x + 1) ≤ 0

⇔ -(√x - 1)² ≤ 0 (luôn đúng)

Vậy (4√x + 4)/(x + 2√x + 5) ≤ 1 với mọi x ≥ 0

a: \(P=\dfrac{x+8\sqrt{x}+8-x-4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}+2\right)}:\dfrac{x+\sqrt{x}+3+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)}\)

\(=\dfrac{4\left(\sqrt{x}+1\right)}{x+2\sqrt{x}+5}\)

b: 4(căn x+1)>=4

x+2căn x+5>=5

=>P<=4/5<1

bn đăng bên toán nhé

\(A=\dfrac{x-2014}{\dfrac{x^2-4x+4-x^2-2x-1}{\left(x+1\right)\left(x-2\right)}:\dfrac{x^2-4x+4+x^2+2x+1}{\left(x+1\right)\left(x-2\right)}}\)

\(=\dfrac{x-2014}{\dfrac{-6x+3}{\left(x+1\right)\left(x-2\right)}\cdot\dfrac{\left(x+1\right)\left(x-2\right)}{2x^2-2x+5}}\)

\(=\left(x-2014\right)\cdot\dfrac{2x^2-2x+5}{-6x+3}\)

Để A>=0 thì \(\left(x-2014\right)\left(-6x+3\right)>=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-2014\right)< =0\)

=>1/2<x<=2014

Ta có \(B\ge\dfrac{\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)^2}{2}\) \(=\dfrac{\left(1+\dfrac{1}{xy}\right)^2}{2}\)

Lại có \(xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{1}{4}\)

\(\Rightarrow B\ge\dfrac{\left(1+4\right)^2}{2}=\dfrac{25}{2}\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

Vậy GTNN của B là \(\dfrac{25}{2}\) khi \(x=y=\dfrac{1}{2}\)

Hàm xác định trên R

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

Hàm đã cho là hàm lẻ

\(a,\left(12x^2y^2-6xy^2\right):3xy+2y=6xy^2\left(2x-1\right):3xy+2y=2y\left(2x-1\right)+2y=4xy-2y+2y=4xy\)

\(b,\dfrac{4}{x+1} + \dfrac{8}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{4\left(x-1\right)+8}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4x-4+8}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4x+4}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{4}{x-1}\)

\(c,\dfrac{1 }{x+1}- \dfrac{1}{x-1} +\dfrac{ 2x}{x^2-1} \)

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

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

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

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

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

\(a,=4xy-2y+2y=4xy\\ b,\dfrac{4}{x+1}+\dfrac{8}{\left(x+1\right)\left(x-1\right)}=\dfrac{4x-4+8}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4x+4}{\left(x+1\right)\left(x-1\right)}=\dfrac{4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x-1}\\ c,\dfrac{1}{x+1}-\dfrac{1}{x-1}+\dfrac{2x}{x^2-1}=\dfrac{x-1-x-1+2x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{2x-2}{\left(x-1\right)\left(x+1\right)}=\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x+1}\)

4) | x-1/3| -1/3=1/3

\(C=\dfrac{-\left(x+1\right)+2\left(x-1\right)+5-x}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

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

\(C=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)

\(\Rightarrow C=\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1+x\right)\left(1-x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

\(\Rightarrow C=\dfrac{1+x+2\left(1-x\right)-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

\(\Rightarrow C=\dfrac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{-\left(1-x\right)\left(x+1\right)}{1-2x}\)

\(\Rightarrow C=-2.\dfrac{-1}{1-2x}\)

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