a. \(5x\left(2x-7\right)+2x\left(8-5x\right)=5\)

\(\Rightarrow10x^2-35x+16x-10x^2=5\)

\(\Rightarrow-19x=5\)

\(\Rightarrow x=-\frac{5}{19}\)

b. \(x\left(x-\frac{1}{3}\right)-\frac{1}{2}x\left(2x-3\right)=\frac{1}{4}\)

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

\(\Rightarrow\frac{7}{6}x=\frac{1}{4}\)

\(\Rightarrow x=\frac{3}{14}\)

c. \(5\left(x^2-3x+1\right)+x\left(1-5x\right)=x-2\)

\(\Rightarrow5x^2-15x+5+x-5x^2=x-2\)

\(\Rightarrow-14x+5=x-2\)

\(\Rightarrow-14x-x=-2-5\)

\(\Rightarrow-15x=-7\)

\(\Rightarrow x=\frac{7}{15}\)

a, \(5x\left(2x-7\right)+2x\left(8-5x\right)=5\)

\(\Leftrightarrow10x^2-35x+16x-10x^2=5\)

\(\Leftrightarrow-19x=5\Leftrightarrow x=-\frac{5}{19}\)

b, \(x\left(x-\frac{1}{3}\right)-\frac{1}{2}x\left(2x-3\right)=\frac{1}{4}\)

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

\(\Leftrightarrow\frac{7}{6}x=\frac{1}{4}\Leftrightarrow x=\frac{3}{14}\)

c, \(5\left(x^2-3x+1\right)+x\left(1-5x\right)=x-2\)

\(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)

\(\Leftrightarrow-15x+7=0\Leftrightarrow x=\frac{7}{15}\)

\(VP=\frac{6}{\sqrt{\left(3a+bc\right)\left(3b+ca\right)\left(3c+ab\right)}}\)

\(=\frac{6}{\sqrt{\left[\left(a+b+c\right)a+bc\right]\left[\left(a+b+c\right)b+ca\right]\left[\left(a+b+c\right)c+ab\right]}}\)

\(=\frac{6}{\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+1\right)^2}}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)

\(VT=\frac{1}{3a+bc}+\frac{1}{3b+ca}+\frac{1}{3c+ab}\)

\(=\frac{1}{\left(a+b+c\right)a+bc}+\frac{1}{\left(a+b+c\right)b+ac}+\frac{1}{\left(a+b+c\right)c+ab}\)

\(=\frac{\left(b+c\right)+\left(a+c\right)+\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)

Vậy VT = VP, đẳng thức được chứng minh

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

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

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

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

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

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

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

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

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

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

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