a: \(x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

b: \(x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

c: \(=x^2-2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x,y\ne0\)

Áp dụng bất đẳng thức AM-GM ta có :

\(x+\frac{1}{x-1}=\left[\left(x-1\right)+\frac{1}{x-1}\right]+1\ge2\sqrt{\left(x-1\right)\cdot\frac{1}{x-1}}+1=2+1=3\left(đpcm\right)\)

Đẳng thức xảy ra <=> x = 2 

Ta có: \(\frac{x-2}{\sqrt{x-1}+1}\)

\(=\frac{x-1-1}{\sqrt{x-1}+1}\)

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

\(=\sqrt{x-1}-1\)

Ta có: \(\sqrt{x-1}\ge0\forall x\) thỏa mãn ĐKXĐ

\(\Leftrightarrow\sqrt{x-1}-1\ge-1\forall x\) thoả mãn ĐKXĐ

\(\Leftrightarrow\frac{x-2}{\sqrt{x-1}+1}\ge-1\forall x\ge1\)(đpcm)

ặt $x+1=t$ thì $t>0$ và  $x=-1+t$. Ta có

           $2x+\genfrac{}{}{0ex}{}{1}{{\left(x+1\right)}^2}=2\left(-1+t\right)+\genfrac{}{}{0ex}{}{1}{t^2}=-2+t+t+\genfrac{}{}{0ex}{}{1}{t^2}$

                                                                       $\ge -2+3\sqrt[3]{t.t.\genfrac{}{}{0ex}{}{1}{t^2}}=-2+3=1$  

1

Có $x^8-x^7+x^5-x^4+x^3-x+1=\frac{x^{10}+x^5+1}{x^2+x+1}$
$x^{10}+x^5+1=(x^5+\frac{1}{2}{)}^2+\frac{3}{4}$
$\Rightarrow x^{10}+x^5+1>0$
$x^2+x+1=(x+\frac{1}{2}{)}^2+\frac{3}{4}>0$
$\Rightarrow x^8-x^7+x^5-x^4+x^3-x+1>0$

ích mk nha bạn

Viết lại câu trả lời được "Copy" trên mạng bởi "Thần hộ vệ ...."

\(x^8-x^7+x^5-x^4+x^3-x+1=\frac{x^{10}+x^5+1}{x^2+x+1}=\frac{\left(x^5+\frac{1}{2}\right)^2+\frac{3}{4}}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}>0\)

Câu 1 :

\(P=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

Câu 2 :

Ta có :

\(\Delta=m^2+16>0\)

\(=>\) phương trình có 2 nghiệm phân biệt .

Theo định lý vi-ét ta có :

\(\left\{{}\begin{matrix}x_1+x_2=m\\x_1.x_2=-4\end{matrix}\right.\)

Thay vào ta được :

\(\dfrac{2m+7}{m^2+8}\ge-\dfrac{1}{8}\)

\(\Leftrightarrow16m+56\ge-m^2-8\)

\(\Leftrightarrow m^2+16m+64\ge0\)

\(\Leftrightarrow\left(m+8\right)^2\ge0\) ( đúng )