`(1-5x)/(x-1)>=1`

`<=>(1-5x)/(x-1)-1>=0`

`<=>(1-5x-x+1)/(x-1)>=0`

`<=>(2-6x)/(x-1)>=0(x ne 1)`

`<=>(6x-2)/(x-1)<=0`

`<=>(x-2/6)/(x-1)<=0`

`TH1:x-2/6>=0,x-1<0` 

`<=>2/6<=x<1(TM)`

`TH2:x-2/6<=0,x-1>0`

`<=>1<x<=2/6`(vô lý)

Vậy `2/6<=x<1`

Ta có: \(\dfrac{1-5x}{x-1}\ge1\)

\(\Leftrightarrow\dfrac{1-5x-x+1}{x-1}\ge0\)

\(\Leftrightarrow\dfrac{-6x+2}{x-1}\ge0\)

Trường hợp 1: \(\dfrac{-6x+2}{x-1}=0\)

\(\Leftrightarrow-6x+2=0\)

\(\Leftrightarrow-6x=-2\)

hay \(x=\dfrac{1}{3}\)

Trường hợp 2: \(\dfrac{-6x+2}{x-1}>0\)

\(\Leftrightarrow\dfrac{1}{3}< x< 1\)

Vậy: \(\dfrac{1}{3}\le x< 1\)

Ta có: \(x\ge1;y\ge1\)

\(\Rightarrow\left\{{}\begin{matrix}x\left(y-1\right)\ge0\\y\left(x-1\right)\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}xy\ge x^2\\xy\ge y^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}1+xy\ge1+x^2\\1+xy\ge1+y^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{1+x^2}\ge\dfrac{1}{1+xy}\left(1\right)\\\dfrac{1}{1+y^2}\ge\dfrac{1}{1+xy}\left(2\right)\end{matrix}\right.\)

Cộng vế với vế (1) và (2) ta được : {cái bđt ở đầu bài chép xuống đây}

a) (x≤3)





 

a: Ta có: \(\sqrt{x-1}=2\)

\(\Leftrightarrow x-1=4\)

hay x=5

b: Ta có: \(\sqrt{3-x}=4\)

\(\Leftrightarrow3-x=16\)

hay x=-13

c: Ta có: \(2\cdot\sqrt{3-2x}=\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{3-2x}=\dfrac{1}{4}\)

\(\Leftrightarrow-2x+3=\dfrac{1}{16}\)

\(\Leftrightarrow-2x=-\dfrac{47}{16}\)

hay \(x=\dfrac{47}{32}\)

d: Ta có: \(4-\sqrt{x-1}=\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{2}\)

\(\Leftrightarrow x-1=\dfrac{49}{4}\)

hay \(x=\dfrac{53}{4}\)

e: Ta có: \(\sqrt{x-1}-3=1\)

\(\Leftrightarrow\sqrt{x-1}=4\)

\(\Leftrightarrow x-1=16\)

hay x=17

f:Ta có: \(\dfrac{1}{2}-2\cdot\sqrt{x+2}=\dfrac{1}{4}\)

\(\Leftrightarrow2\cdot\sqrt{x+2}=\dfrac{1}{4}\)

\(\Leftrightarrow\sqrt{x+2}=\dfrac{1}{8}\)

\(\Leftrightarrow x+2=\dfrac{1}{64}\)

hay \(x=-\dfrac{127}{64}\)

\(VT\ge\dfrac{1}{\left(a^2+1\right)-1}+\dfrac{1}{\left(b^2+1\right)-1}+\dfrac{1}{\left(c^2+1\right)-1}+4-\dfrac{4}{ab+1}+4-\dfrac{4}{bc+1}+4-\dfrac{4}{ca+1}\)

\(VT\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{4}{ab+1}-\dfrac{4}{bc+1}-\dfrac{4}{ca+1}+12\)

Mặt khác \(a;b;c\ge1\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\Rightarrow ab+1\ge a+b\) (và tương tự...)

\(\Rightarrow VT\ge\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}-\dfrac{4}{a+b}-\dfrac{4}{b+c}-\dfrac{4}{c+a}+12\)

\(VT\ge\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}-\dfrac{4}{a+b}-\dfrac{4}{b+c}-\dfrac{4}{c+a}+1+1+1+9\)

\(VT\ge\left(\dfrac{2}{a+b}-1\right)^2+\left(\dfrac{2}{b+c}-1\right)^2+\left(\dfrac{2}{c+a}-1\right)^2+9\ge9\)

\(\dfrac{x^3}{2y+1}+\dfrac{2y+1}{9}+\dfrac{1}{3}\ge3\sqrt[3]{\dfrac{x^3\left(2y+1\right)}{27\left(2y+1\right)}}=x\)

Tương tự: \(\dfrac{y^3}{2z+1}+\dfrac{2z+1}{9}+\dfrac{1}{3}\ge y\) ; \(\dfrac{z^3}{2x+1}+\dfrac{2x+1}{9}+\dfrac{1}{3}\ge z\)

Cộng vế:

\(VT+\dfrac{2\left(x+y+z\right)+3}{9}+1\ge x+y+z\)

\(\Rightarrow VT\ge\dfrac{7}{9}\left(x+y+z\right)-\dfrac{4}{3}\ge\dfrac{7}{9}.3\sqrt[3]{xyz}-\dfrac{4}{3}=1\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=1\)

\(\dfrac{1}{2x+1}+\dfrac{\left(\dfrac{1}{3}\right)^2}{1}\ge\dfrac{\left(1+\dfrac{1}{3}\right)^2}{2x+1+1}=\dfrac{8}{9}\left(\dfrac{1}{x+1}\right)\)

Tương tự: \(\dfrac{1}{2y+1}+\dfrac{1}{9}\ge\dfrac{8}{9}.\dfrac{1}{y+1}\) ; \(\dfrac{1}{2z+1}+\dfrac{1}{9}\ge\dfrac{8}{9}.\dfrac{1}{z+1}\)

Cộng vế:

\(VT+\dfrac{1}{3}\ge\dfrac{8}{9}\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right)\ge\dfrac{4}{3}\)

\(\Rightarrow VT\ge1\)

\(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{8}-1>=0\)

=>\(\dfrac{16\sqrt{x}-x-2\sqrt{x}-1-8\sqrt{x}-8}{8\left(\sqrt{x}+1\right)}>=0\)

=>-x+6căn x-9>=0

=>x=3

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

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

=0