\(M=\dfrac{\sqrt{x}+5}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2+7}{\sqrt{x}-2}=1+\dfrac{7}{\sqrt{x}-2}\)

Để M nguyên \(\Leftrightarrow\text{ }7\text{ }⋮\text{ }\left(\sqrt{x}-2\right)\)

=> \(\sqrt{x}-2\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{1;3;9\right\}\)

\(\Rightarrow x\in\left\{1;9;81\right\}\)

Tham Khảo

tích nha

M xác định

\(\Leftrightarrow\hept{\begin{cases}x-1\ne0\\x^2-x\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\left(x-1\right)\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne0;x\ne1\end{cases}}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\ne0\end{cases}}\)

Vậy ĐKXĐ của M là \(\hept{\begin{cases}x\ne1\\x\ne0\end{cases}}\)

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

Thay x=5 ta có: 

\(M=\frac{3.5+1}{5\left(5-1\right)}=\frac{15+1}{5.4}=\frac{16}{20}=\frac{4}{5}\)

Vậy \(M=5\)tại  x=5

\(M=0\)

\(\Leftrightarrow\frac{3x+1}{x\left(x-1\right)}=0\Leftrightarrow3x+1=0\Leftrightarrow x=-\frac{1}{3}\)( thỏa mãn đkxđ)

Vậy với \(x=-\frac{1}{3}\)thì \(M=0\)

\(M=-1\)

\(\Leftrightarrow\frac{3x+1}{x\left(x-1\right)}=-1\Leftrightarrow3x+1=-x^2+x\Leftrightarrow x^2+2x+1=0\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Vậy với \(x=-1\)thì \(M=-1\)

\(4x^2+4x+11\)

\(=\left(2x\right)^2+2.2x.1+1^2+10\)

\(=\left(2x+1\right)^2+10\)

Ta có: \(\left(2x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(2x+1\right)^2+10\ge10\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow\left(2x+1\right)^2=0\Leftrightarrow2x+1=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy Min \(4x^2+4x+11=10\Leftrightarrow x=-\frac{1}{2}\)

voi x =1 

=>M= (1-5)² +2017

M= (-4)² +2017

M= 16 + 2017

M=2033

Vậy M = 2033

b) gia tri nho nhat cua bieu thuc là 2017 

bấm nhầm lớp 6

1: \(D=\dfrac{1}{x+4}+\dfrac{x}{x-4}+\dfrac{24-x^2}{x^2-16}\)

\(=\dfrac{1}{x+4}+\dfrac{x}{x-4}+\dfrac{24-x^2}{\left(x+4\right)\left(x-4\right)}\)

\(=\dfrac{x-4+x\left(x+4\right)+24-x^2}{\left(x+4\right)\left(x-4\right)}\)

\(=\dfrac{-x^2+x+20+x^2+4x}{\left(x+4\right)\left(x-4\right)}=\dfrac{5x+20}{\left(x+4\right)\left(x-4\right)}\)

\(=\dfrac{5\left(x+4\right)}{\left(x+4\right)\left(x-4\right)}=\dfrac{5}{x-4}\)

2: Khi x=10 thì \(D=\dfrac{5}{10-4}=\dfrac{5}{6}\)

3: \(M=\left(x-2\right)\cdot D=\dfrac{5\left(x-2\right)}{x-4}\)

Để M là số nguyên thì \(5\cdot\left(x-2\right)⋮x-4\)

=>\(5\left(x-4+2\right)⋮x-4\)

=>\(5\left(x-4\right)+10⋮x-4\)

=>\(10⋮x-4\)

=>\(x-4\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

=>\(x\in\left\{5;3;6;2;9;-1;14;-6\right\}\)

a ĐK \(a>0\)và \(a\ne1\)

. \(M=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right).\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)

\(=\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{\sqrt{a}-1}{\sqrt{a}}\)

b. Ta có \(M-1=\frac{\sqrt{a}-1}{\sqrt{a}}-1=\frac{\sqrt{a}-1-\sqrt{a}}{\sqrt{a}}=\frac{-1}{\sqrt{a}}< 0\)

Vậy \(M< 1\)

Ta có: \(\left(x-2\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

Th1: x = 2

\(\Rightarrow\left(4.2-5\right)\left(2-7\right)=3.\left(-5\right)=-15\)

Th2: x = -3

\(\Rightarrow\left[4.\left(-3\right)-5\right].\left[\left(-3\right)-7\right]=-17.\left(-10\right)=170\)

vậy ..

hok tốt!!

a, \(M=\left(x-2\right)^2-22\)

Có: \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-2\right)^2-22\ge-22\forall x\)

hay GTNN của M là -22 

Dấu "=" xảy ra tại  \(\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy GTNN của M là -22 tại x=2.

b, \(N=9-|x+3|\)

Có: \(|x+3|\ge0\forall x\)

\(\Rightarrow9-|x+3|\le9\forall x\)

hay GTLN của N là 9

Dấu "=" xảy ra tại \(|x+3|=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

Vậy GTLN của N là 9 tại x = -3.

giúp mình voi , mọi ng oi

\(\dfrac{x-2}{x+5}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+5< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-5\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -5\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -5\end{matrix}\right.\)