Bài 1:

                    \(x^2-8x+y^2+6y+25=0\)

\(\Leftrightarrow\)\(\left(x^2-8x+16\right)+\left(y^2+6y+9\right)=0\)

\(\Leftrightarrow\)\(\left(x-4\right)^2+\left(y+3\right)^2=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x-4=0\\y+3=0\end{cases}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x=4\\y=-3\end{cases}}\)

Vậy...

Bài 2: 

Phương trình có nghiệm duy nhất là    x = -2/3    nên ta có:

          \(\left(4+a\right).\frac{-2}{3}=a-2\)

\(\Leftrightarrow\)\(-\frac{8}{3}-\frac{2}{3}a=a-2\)

\(\Leftrightarrow\)\(a+\frac{2}{3}a=2-\frac{8}{3}\)

\(\Leftrightarrow\)\(\frac{5}{3}a=-\frac{2}{3}\)

\(\Leftrightarrow\)\(a=-\frac{2}{5}\)

Bài 3:

\(A=a^4-2a^3+3a^2-4a+5\)

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

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

\(=\left(a-1\right)\left[a^2\left(a-1\right)+2\left(a-1\right)\right]+3\)

\(=\left(a-1\right)^2\left(a^2+2\right)+3\ge3\)

\(\text{Vậy Min A=3. Dấu "=" xảy ra khi và chỉ khi }a-1=0\Leftrightarrow a=1\)

Bài 4:

\(xy-3x+2y=13\)

\(\Leftrightarrow x\left(y-3\right)+2\left(y-3\right)=7\)

\(\Leftrightarrow\left(x+2\right)\left(y-3\right)=7=1.7=7.1=-1.-7=-7.-1\)

Vậy...

Bài 5:

\(xy-x-3y=2\)

\(\Leftrightarrow x\left(y-1\right)-3\left(y-1\right)=5\)

\(\Leftrightarrow\left(x-3\right)\left(y-1\right)=5=1.5=5.1=-1.-5=-5.-1\)

Vậy....

a, \(Đkxđ:x\ne-3;x\ne2\)

b,\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)

\(=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\)

\(=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)\(=\dfrac{x-4}{x-2}\)

c,\(A=-\dfrac{3}{4}\) khi \(\dfrac{x-4}{x-2}=-\dfrac{3}{4}\)

\(\Leftrightarrow\left(x-4\right).4=-3\left(x-2\right)\)

\(\Leftrightarrow4x-16=-3x+6\)

\(\Leftrightarrow7x=22\)

\(\Leftrightarrow x=\dfrac{22}{7}\)

Vậy khi \(x=\dfrac{22}{7}\) thì \(A=-\dfrac{3}{4}\)

a) ĐKXĐ : \(\left\{{}\begin{matrix}x+3\ne0\\2-x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)

b) \(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}-\dfrac{1}{x-2}\)

\(A=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)

\(A=\dfrac{-x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\)

c) Để \(A=\dfrac{-3}{4}\) thì :

\(A=\dfrac{x-4}{x-2}=\dfrac{-3}{4}\)

\(\Rightarrow\dfrac{x-4}{x-2}+\dfrac{3}{4}=0\)

\(\Rightarrow\dfrac{4\left(x-4\right)}{4\left(x-2\right)}+\dfrac{3\left(x-2\right)}{4\left(x-2\right)}=0\)

\(\Rightarrow4x-16+3x-6=0\)

\(\Rightarrow7x+22=0\)

\(\Rightarrow x=\dfrac{-22}{7}\)

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

Vì \(1\in Z\) để \(A\in Z\) thì \(\dfrac{2}{x-2}\in Z\)

\(\Rightarrow x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Có : \(\left\{{}\begin{matrix}x-2=1=>x=3\\x-2=-1=>x=1\\x-2=2=>x=4\\x-2=-2=>0\end{matrix}\right.\)

Vậy để A nhận gt nguyên thì x \(\in\left\{3;1;4;0\right\}\)

e) \(x^2-9=0\)

\(\Rightarrow\left(x+3\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=3\end{matrix}\right.\)

Thay vào A ta có :

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

d, \(A\) nguyên thì \(\dfrac{x-4}{x-2}\) nguyên.

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

Mà \(1\) nguyên, để \(A\) nguyên thì \(\dfrac{2}{x-2}\in Z\)

\(\Rightarrow x-2\inƯ\left(2\right)=\left\{2;-2;1;-1\right\}\)

Ta có bảng sau:

Vậy A nguyên khi \(x\in\left\{4;0;3;1\right\}\)

e, \(x^2-9=0\)

\(\Leftrightarrow x^2-3^2=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-3\left(ktm\right)\end{matrix}\right.\)

Khi \(x=3\) thì \(A=\dfrac{3-4}{3-2}=-1\)

a/ la 4

b/ la 100

cho rõ lời giải hộ tớ được không và cho cả giá trị x,y nữa

1, Ta có: \(|x-100|+\left(x-y\right)^2+100\ge0+0+100=100\)

                Dấu "=" xảy ra khi \(\hept{\begin{cases}x-100=0\\x-y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=100\\y=100\end{cases}}}\)

2, Áp dụng BĐT \(|a|\ge a\) với \(\forall a\). Dấu "=" xảy ra khi \(a\ge0\)

Áp dụng vào bài toán ta có: \(|x+20|+|47-x|+3^3\ge x+20+47-x+9=76\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x+20\ge0\\47-x\ge0\end{cases}\Leftrightarrow-20\le x\le47}\)