Ta có:

\(5x\left(x-1\right)=x-1\)

\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=1\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{5};1\right\}\)

Ta có:

\(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=\left(a-c\right)\left[\left(a-b\right)^2-\left(a-b\right)\left(b-c\right)+\left(b-c\right)^2\right]+\left(c-a\right)^3\) \(=\left(a-c\right)\left[\left(a-b\right)^2-\left(a-b\right)\left(b-c\right)+\left(b-c\right)^2-\left(a-c\right)^2\right]\)

\(=\left(a-c\right)\left[\left(a-b\right)\left(a-2b+c\right)+\left(b-a\right)\left(a+b-2c\right)\right]\)

\(=\left(a-c\right)\left(a-b\right)\left[\left(a-2b+c\right)-\left(a+b-2c\right)\right]\)

\(=\left(a-c\right)\left(a-b\right)\left(3c-3b\right)\)

\(=3\left(a-b\right)\left(b-c\right)\left(c-a\right)=3k\)

Vì \(3k⋮3k\) nên \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3⋮3k\)

Ta có:

\(3n^3-2n+17=3n^3+6n^2-6n^2-12n+10n+20-3\)

\(=3n^2\left(n+2\right)-6n\left(n+2\right)+10\left(n+2\right)-3\)

\(=\left(3n^2-6n+10\right)\left(n+2\right)-3\)

Để \(3n^3-2n+17⋮\left(n+2\right)\)

Hay \(\left(3n^2-6n+10\right)\left(n+2\right)-3⋮\left(n+2\right)\)

Mà \(\left(3n^2-6n+10\right)\left(n+2\right)⋮\left(n+2\right)\)

Suy ra \(3⋮\left(n+2\right)\)

Vì \(n\in Z\), suy ra \(n+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

TH1: \(n+2=\pm1\Leftrightarrow\left[{}\begin{matrix}n+2=1\\n+2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=-1\\n=-3\end{matrix}\right.\)

TH2:\(n+2=\pm3\Leftrightarrow\left[{}\begin{matrix}n+2=3\\n+2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=1\\n=-5\end{matrix}\right.\)

Vậy \(n\in\left\{-5;-3;-1;1\right\}\)

a) Đề sai thì phải.Phải là CM: \(x^2-x+1>0\) với mọi x

Ta có:

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

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

Vì \(\left(x-\frac{1}{2}\right)^2\ge0\) nên \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)

Vậy \(x^2-x+1>0\) với mọi \(x\in R\)

b)Ta có:

\(-x^2+2x-4=-\left(x^2-2x+1\right)-3\)

\(=-\left(x-1\right)^2-3\)

Vì \(-\left(x-1\right)^2\le0\) với mọi x nên \(-\left(x-1\right)^2-3< 0\)

Vậy \(-x^2+2x-4< 0\) với mọi \(x\in R\)

Ta có:

\(A=2x^2+x+1\)

\(=2\left(x^2+\frac{1}{2}x+\frac{1}{16}\right)+\frac{7}{8}\)

\(=2\left(x+\frac{1}{4}\right)^2+\frac{7}{8}\)

Vì \(\left(x+\frac{1}{4}\right)^2\ge0\) với mọi x nên \(2\left(x+\frac{1}{4}\right)^2\ge0\)

Suy ra \(A=2\left(x+\frac{1}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\)

Vậy A đạt GTNN là \(\frac{7}{8}\) khi \(\left(x+\frac{1}{4}\right)^2=0\)

\(\Leftrightarrow x+\frac{1}{4}=0\)

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

KL: Vậy A đạt GTNN là \(\frac{7}{8}\) khi \(x=-\frac{1}{4}\)

a) A xác định khi \(\left\{{}\begin{matrix}x>0\\\sqrt{x}-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\\sqrt{x}\ne3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne9\end{matrix}\right.\)

b)Với \(x>0;x\ne9\), ta có:

\(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

Để A đạt giá trị nguyên thì \(\frac{4}{\sqrt{x}-3}\) đạt giá trị nguyên

Hay\(4⋮\left(\sqrt{x}-3\right)\)

Suy ra \(\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

TH1: \(\sqrt{x}-3=\pm1\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=1\\\sqrt{x}-3=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=16\\x=4\end{matrix}\right.\)

TH2: \(\sqrt{x}-3=\pm2\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=2\\\sqrt{x}-3=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=5\\\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=25\\x=1\end{matrix}\right.\)

TH3: \(\sqrt{x}-3=\pm4\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=4\\\sqrt{x}-3=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=7\\\sqrt{x}=-1\left(Loại\right)\end{matrix}\right.\Rightarrow x=49\)

Vậy \(x\in\left\{1;4;16;25;49\right\}\)

Bài 1: Ta có: \(3\sqrt{12}=\sqrt{9}.\sqrt{12}=\sqrt{108}\)

và \(2\sqrt{26}=\sqrt{4}.\sqrt{26}=\sqrt{104}\)

Vì \(108>104\Rightarrow\sqrt{108}>\sqrt{104}\)

Hay \(3\sqrt{12}>2\sqrt{26}\)

Bài 2:

\(\frac{5}{4}\sqrt{12x}-\sqrt{12x}-3=\frac{1}{6}\sqrt{12x}\)

\(\Leftrightarrow\frac{5}{4}\sqrt{12x}-\sqrt{12x}-\frac{1}{6}\sqrt{26}=3\)

\(\Leftrightarrow\frac{1}{12}\sqrt{12x}=3\)

\(\Leftrightarrow\sqrt{\frac{1}{12^2}}.\sqrt{12x}=3\)

\(\Leftrightarrow\sqrt{\frac{x}{12}}=3\)

\(\Leftrightarrow\frac{x}{12}=9\)

\(\Leftrightarrow x=108\)

Bài 3: Với \(x>0;y>0\), ta có:

\(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}+\sqrt{y}}=\frac{\sqrt{x^2}.\sqrt{y}-\sqrt{y^2}.\sqrt{x}}{\sqrt{xy}}.\left(\sqrt{x}+\sqrt{y}\right)\)

\(=\frac{\sqrt{x^2y}-\sqrt{xy^2}}{\sqrt{xy}}.\left(\sqrt{x}+\sqrt{y}\right)\)

\(=\frac{\sqrt{xy}\left(\sqrt{y}-\sqrt{x}\right)}{\sqrt{xy}}.\left(\sqrt{y}+\sqrt{x}\right)\)

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

\(=y-x\)

Ta có: \(13⋮\left(x-3\right)\)

\(\Rightarrow x-3\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

TH1: \(x-3=\pm1\Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

TH2: \(x-3=\pm13\Leftrightarrow\left[{}\begin{matrix}x-3=13\\x-3=-13\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=16\\x=-10\end{matrix}\right.\)

Vậy \(x\in\left\{-10;2;4;16\right\}\)

a) \(3,6-\left|x-0,4\right|=0\)

\(\Leftrightarrow\left|x-0,4\right|=3,6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-0,4=3,6\\x-0,4=-3,6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3,2\end{matrix}\right.\)

Vậy \(x\in\left\{4;-3,2\right\}\)

b) Ta có:

\(\frac{x}{2}=y=\frac{z}{3}=\frac{2y}{2}=\frac{x-2y+z}{2-2+3}=\frac{210}{3}=70\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{2}=70\\y=70\\\frac{z}{3}=70\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=140\\y=70\\z=210\end{matrix}\right.\)

Vậy \(x=140\); \(y=70\); \(z=210\)

c)\(\left|x+0,25\right|-4=\frac{1}{4}\)

\(\Leftrightarrow\left|x+\frac{1}{4}\right|=\frac{17}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{4}=\frac{17}{4}\\x+\frac{1}{4}=\frac{-17}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{-9}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{4;\frac{-9}{2}\right\}\)

d) \(x:\left(0,25\right)^4=\left(0,5\right)^2\)

\(\Leftrightarrow x=\left(0,25\right)^4.\left(0,5\right)^2\)

\(\Leftrightarrow x=\left(0,5\right)^8.\left(0,5\right)^2\)

\(\Leftrightarrow x=\left(0,5\right)^{10}=\left(\frac{1}{2}\right)^{10}=\frac{1}{2^{10}}=\frac{1}{1024}\)

Vậy \(x=\frac{1}{1024}\)

e) \(3^{x-1}+5.3^{x-1}=162\)

\(\Leftrightarrow6.3^{x-1}=162\)

\(\Leftrightarrow3^{x-1}=27\)

\(\Leftrightarrow3^{x-1}=3^3\)

\(\Leftrightarrow x-1=3\)

\(\Leftrightarrow x=4\)

f) \(\frac{x}{-25}=\frac{2}{5}\)

\(\Leftrightarrow x=\left(-25\right).\frac{2}{5}=-10\)

Vậy \(x=-10\)

g) \(\left|x+\frac{3}{4}\right|-\frac{3}{4}=\sqrt{\frac{1}{9}}\)

\(\Leftrightarrow\left|x+\frac{3}{4}\right|-\frac{3}{4}=\frac{1}{3}\)

\(\Leftrightarrow\left|x+\frac{3}{4}\right|=\frac{13}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{13}{12}\\x+\frac{3}{4}=-\frac{13}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=-\frac{11}{6}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{3};-\frac{11}{6}\right\}\)

a) Với \(x\in\left\{0;3\right\}\), ta có:

\(B=\left(\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\right):\frac{2x-2}{x}\)

\(=\left[\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x\left(x-3\right)}\right]:\frac{2\left(x-1\right)}{x}\)

\(=\frac{\left(x-3\right)^2-x^2+9}{x\left(x-3\right)}.\frac{x}{2\left(x-1\right)}\)

\(=\frac{\left(x-3\right)^2-\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}.\frac{x}{2\left(x-1\right)}\)

\(=\frac{\left(x-3\right)\left[\left(x-3\right)-\left(x+3\right)\right]}{x\left(x-3\right)}.\frac{x}{2\left(x-1\right)}\)

\(=\frac{-6\left(x-3\right)}{x\left(x-3\right)}.\frac{x}{2\left(x-1\right)}\)

\(=\frac{-6x\left(x-3\right)}{2x\left(x-3\right)\left(x-1\right)}=\frac{-3}{x-1}\)

b) Để B luôn nhận giá trị nguyên thì \(-3⋮\left(x-1\right)\) với \(x\notin\left\{0;3\right\}\)

Hay \(3⋮\left(x-1\right)\)

\(\Rightarrow x-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

TH1: \(x-1=\pm1\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

TH2: \(x-1=\pm3\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

Ta lấy \(x\in\left\{-2;2;4\right\}\) (vì \(x\ne0\))

Vậy \(x\in\left\{-2;2;4\right\}\)