Bài làm:

Ta có: \(xy=5\)\(\Rightarrow x=\frac{5}{y}\)

Thay vào ta được:

\(x^2+y^2=26\)

\(\Leftrightarrow\frac{25}{y^2}+y^2=26\)

\(\Leftrightarrow\frac{25+y^4}{y^2}=26\)

\(\Leftrightarrow y^4-26y^2+25=0\)

\(\Leftrightarrow\left(y^4-y^2\right)-\left(25y^2-25\right)=0\)

\(\Leftrightarrow\left(y^2-1\right)\left(y^2-25\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y^2-1=0\\y^2-25=0\end{cases}}\Rightarrow\orbr{\begin{cases}y=\pm1\\y=\pm5\end{cases}}\Rightarrow\orbr{\begin{cases}x=\pm5\\x=\pm1\end{cases}}\)

Vậy ta có các cặp số (x;y) thỏa mãn: \(\left(1;5\right);\left(-1;-5\right);\left(5;1\right);\left(-5;-1\right)\)

Ta có :

\(x^2+y^2=26\Rightarrow x^2+y^2+2xy=26+2.5\)

\(\Rightarrow\left(x+y\right)^2=36\Leftrightarrow x+y=6\left(1\right)\)

\(x^2+y^2=26\Rightarrow x^2+y^2-2xy=26-2.5\)

\(\Rightarrow\left(x-y\right)^2=16\Leftrightarrow x-y=4\left(2\right)\)

Từ ( 1 ) và ( 2 ) \(\Rightarrow x=\frac{6+4}{2}=5\)

\(\Rightarrow y=5-4=1\)

Vậy x = 5 ; y = 1

\(=\frac{\left(2x+1\right)\left(x+1\right)+8-\left(x-1\right)^2}{x^2-1}.\frac{x^2-1}{5}=\)

\(=\frac{2x^2+3x+1+8-x^2+2x-1}{5}=\frac{x^2+5x+8}{5}\)

\(\left(\frac{2x+1}{x-1}+\frac{8}{x^2-1}-\frac{x-1}{x+1}\right)\cdot\frac{x^2-1}{5}\left(x\ne\pm1\right)\)

\(=\left(\frac{2x+1}{x-1}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x+1}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\left(\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\left(\frac{2x^2+3x+1}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\frac{2x^2+3x+1+8-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\frac{\left(x^2+5x+8\right)\cdot\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)5}=\frac{x^2+5x+8}{5}\)

Theo đề, ta có : a= 20 + b 

=> b . (20 + b) = 3 

<=> b² +20b -3 = 0

<=>  \(\orbr{\begin{cases}b=-10+\sqrt{103}\\b=-10-\sqrt{103}\end{cases}\Rightarrow\orbr{\begin{cases}a=10+\sqrt{103}\\a=10-\sqrt{103}\end{cases}}}\)

Nếu đề là rút gọn biểu thức thì...

đkxđ: \(x\ne\pm1\)

Ta có: \(\frac{x+1}{2x-2}+\frac{-2x^2+3}{3x^2-3}\)

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

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

\(=\frac{3x^2+6x+3+6-4x^2}{6\left(x-1\right)\left(x+1\right)}\)

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

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

\(=\frac{\left(x+1\right)\left(x^2-3\right)-\left(2x^2+3\right)\left(2x-2\right)}{\left(x^2-3\right)\left(2x-2\right)}=\frac{x^3-3x+x^2-3-\left(4x^3-4x^2+6x-6\right)}{\left(x^2-3\right)\left(2x-2\right)}\)

\(=\frac{x^3-3x+x^2-3-4x^3+4x^2-6x+6}{\left(x^2-3\right)\left(2x-2\right)}=\frac{-3x^3-9x+5x^2+3}{\left(x^2-3\right)\left(2x-2\right)}\)

\(\frac{x+y}{7x+y}-\frac{6x}{-7x}=\frac{x+y}{7x+y}+\frac{6x}{7x}\)

\(=\frac{49x+13y}{7\left(7x+y\right)}=\frac{49x+13y}{49x+7y}\)

\(=1+\frac{6y}{49x+7y}\)

A = (2x - 3)(x² + 4x) - 2(x³ + 2x + 6)

   = 2x(x² + 4x) - 3(x² + 4x) - 2x³ - 4x - 12

  = 2x³ + 8x² - 3x² - 12x - 2x³ - 4x - 12

  = 5x² - 16x - 12

Thay x = 4 vào biểu thức trên ta có : 5.4² - 16.4 - 12 = 4

B = x(x² + 7x) - (x + 9)(x² + 17)

   = x³ + 7x² - x(x² + 17) - 9(x² + 17)

  = x³ + 7x² - x³ - 17x - 9x² - 153

  = -2x² - 17x - 153 

Thay x = 5 vào biểu thức trên ta có : -2.5² - 17.5  - 153 = -50 - 85 - 153 = -288

A = ( 2x - 3 )( x² + 4x ) - 2( x³ + 2x + 6 )

A = 2x³ + 8x² - 3x² - 12x - 2x³ - 4x - 12

A = 5x² - 16x - 12

Thế A = 4 ta được :

A = 5.4² - 16.4 - 12 = 4

B = x( x² + 7x ) - ( x + 9 )( x² + 17 )

B = x³ + 7x² - ( x³ + 17x + 9x² + 153 )

B = x³ + 7x² - x³ - 17x - 9x² - 153

B = -2x² - 17x - 153

Thế x = 5 ta được :

B = -2.5² - 17.5 - 153 = -288

Dài quá ! Nên vẫn phải làm ^_^.

Bài 1: 

+) \(A=x^2-2x+6=x^2-2x+1+5=\left(x-1\right)^2+5\ge5\)

Min A = 5 \(\Leftrightarrow x=1\)

+) \(B=x^2+6x+12=x^2+6x+9+3=\left(x+3\right)^2+3\ge3\)

Min B = 3 \(\Leftrightarrow x=-3\)

+) \(C=4-x^2+2x=-\left(x^2-2x+4\right)=-\left[\left(x-1\right)^2+3\right]=-\left(x-1\right)^2-3\le-3\)

Max C = -3 \(\Leftrightarrow x=1\)

+) \(D=-x^2+6x=-\left(x^2-6x+9-9\right)=-\left(x-3\right)^2+9\le9\)

Max D = 9 \(\Leftrightarrow x=3\)

Bài 2 :

a) \(x^2-x-3x+3=0\)

\(\Leftrightarrow x^2-4x+4-1=0\)

\(\Leftrightarrow\left(x-2\right)^2-1=0\)

\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

b) \(\left(x-3\right)^2-4=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

c) Xem lại đề hộ mình nha 

d) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

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

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

\(\Leftrightarrow x\in\left\{0;-3;2\right\}\)