a) A xác định \(\Leftrightarrow\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}}\)

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

\(A=\left[\frac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\frac{2\cdot3x}{3x\left(x+1\right)}-\frac{3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\right]\cdot\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}\cdot\frac{x+1}{2\cdot\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{\left(-8x^2+2\right)\left(x+1\right)}{3x\left(x+1\right)2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{2\left(1-4x^2\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{2\left(1-2x\right)\left(1-2x\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{1+2x}{3x}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{2x+1-3x-1+x^2}{3x}\)

\(A=\frac{x^2-x}{3x}\)

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

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

b) Thay x = 4 ta có :

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

c) Để A thuộc Z thì \(x-1⋮3\)

\(\Rightarrow x-1\in B\left(3\right)=\left\{0;3;6;...\right\}\)

\(\Rightarrow x\in\left\{1;4;7;...\right\}\)

Vậy.....

Cho Bt 

a,Tìm điều kiện xác định và rút gọn bt A

b,Tính giá trị bt A tại x=4

c,tìm x thuộc Z để a thuộc Z

\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\)

\(=\frac{x}{z}+\frac{y}{z}+\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}\)

\(=\left(\frac{x}{z}+\frac{z}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{y}+\frac{y}{x}\right)\)

Áp dụng BĐT AM-GM ta có:

\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\ge2.\sqrt{\frac{x}{z}.\frac{z}{x}}+2.\sqrt{\frac{x}{y}.\frac{y}{x}}+2.\sqrt{\frac{y}{z}.\frac{z}{y}}=2+2+2=6\)

                                                                                                                           đpcm

Svac-xơ 

\(VT=\left(\frac{x+y}{z}+1\right)+\left(\frac{y+z}{x}+1\right)+\left(\frac{z+x}{y}+1\right)-3\)

\(VT=\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{z}-3=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3\)

\(\ge\left(x+y+z\right).\frac{\left(1+1+1\right)^2}{x+y+z}-3=9-3=6\)

a) \(A=\frac{97^3+83^3}{180}-97\cdot83\)

\(A=\frac{\left(97+83\right)\left(97^2-97\cdot83+83^2\right)}{180}-97\cdot83\)

\(A=\frac{180\cdot\left(97^2-97\cdot83+83^2\right)}{180}-97\cdot83\)

\(A=97^2-97\cdot83+83^2-97\cdot83\)

\(A=9409-2\cdot8051+6889\)

\(A=196\)

b) \(B=\left(50^2+48^2+...+2^2\right)-\left(49^2+47^2+...+1^2\right)\)

\(B=50^2+48^2+...+2^2-49^2-47^2-...-1^2\)

\(B=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)

\(B=\left(50+49\right)\left(50-49\right)+\left(48+47\right)\left(48-47\right)+...+\left(2+1\right)\left(2-1\right)\)

\(B=50+49+48+47+...+2+1\)

Số số hạng là : \(\left(50-1\right):1+1=50\)( số )

Tổng B là : \(\left(50+1\right)\cdot50:2=1275\)

Vậy....

a,\(6x-8y=9\)

\(\Rightarrow x=\frac{9+8y}{6}\)

\(y=\frac{6x-9}{8}\)

Vậy....

\(b,11x+18y=120\)

\(\Rightarrow x=\frac{120-18y}{11}\)

\(y=\frac{120-11x}{18}\)

100 nha

______________
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^_^

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

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

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

Dấu "='' khi a = 1

Vậy ..........