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\(D=x+1-\sqrt{x}=x-\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}>0\)
1/
Xét hiệu $(x+1)^2-4x^2=(x+1)^2-(2x)^2=(x+1-2x)(x+1+2x)$
$=(1-x)(3x+1)$
Do $x\in (0;1)$ nên $1-x>0; 3x+1>0$
$\Rightarrow (x+1)^2-4x^2>0\Rightarrow (x+1)^2> 4x^2$
2/
Xét hiệu:
$(1+x+y)^2-4(x^2+y^2)=x^2+y^2+1+2x+2y+2xy-4x^2-4y^2$
$=1+2x+2y+2xy-3x^2-3y^2$
$=2x(1-x)+2y(1-y)+1+2xy-x^2-y^2$
Vì $x,y\in (0;1)$ nên:
$2x(1-x)>0$
$2y(1-y)>0$
$(x-1)(y-1)>0\Rightarrow xy+1> x+y=x.1+y.1> x^2+y^2$
$\Rightarrow 1+xy-x^2-y^2>0$
$\Rightarrow 1+2xy-x^2-y^2>0$
Suy ra: $2x(1-x)+2y(1-y)+1+2xy-x^2-y^2>0$
$\Rightarrow (1+x+y)^2> 4(x^2+y^2)$
\(a,A=\left(\dfrac{x+14\sqrt{x}-5}{x-25}+\dfrac{\sqrt{x}}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)
\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{x+14\sqrt{x}-5+x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{2x+10\sqrt{x}-\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}+5\right)-\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}\)