Đại số lớp 8

Gọi \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b},\)ta có :

\(M.\frac{c}{a-b}=1+\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)=1+\frac{c}{a-b}.\frac{b^2-bc+ac-a^2}{ab}\)

\(=1+\frac{c}{a-b}.\frac{\left(a-b\right)\left(c-a-b\right)}{ab}=1+\frac{2c^2}{ab}=1+\frac{2c^3}{abc}\)

Tương tự : \(M.\frac{a}{b-c}=1+\frac{2a^3}{abc},M.\frac{b}{c-a}=1+\frac{2b^3}{abc}.\)

Vậy \(A=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=9\)

\(3x^2-27x=0\)

\(3x\left(x-9\right)=0\)

\(\left[\begin{array}{nghiempt}x=0\\x-9=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=0\\x=9\end{array}\right.\)

\(\frac{2}{3}x\left(x^2-4\right)=0\)

\(\frac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)

\(\left[\begin{array}{nghiempt}x=0\\x-2=0\\x+2=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=0\\x=2\\x=-2\end{array}\right.\)

a)\(3x^2-27x=0\)

\(3x\left(x-9\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}3x=0\\x-9=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=9\end{array}\right.\)

b) \(\frac{2}{3}x\left(x^2-4\right)=0\)

\(\frac{2}{3}x\left(x+2\right)\left(x-2\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}\frac{2}{3}x=0\\x+2=0\\x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=-2\\x=2\end{array}\right.\)

\(25x^2-10x+1-y^2\)

\(=\left(5x-1\right)^2-y^2\)

\(=\left(5x-y-1\right)\left(5x+y-1\right)\)

\(4y^2-4x^2-4y+1\)

\(=\left(2y-1\right)^2-\left(2x\right)^2\)

\(=\left(2y+2x-1\right)\left(2y-2x-1\right)\)

\(-y^2+6y-9+x^2\)

\(=x^2-\left(y-3\right)^2\)

\(=\left(x+y-3\right)\left(x-y+3\right)\)

\(A=\frac{2016a}{ab+2016a+2016}+\frac{b}{bc+b+2016}+\frac{c}{ac+c+1}\)

\(A=\frac{2016a}{ab+2016a+abc}+\frac{b}{bc+b+2016}+\frac{bc}{abc+bc+b}\)

\(A=\frac{2016a}{a\left(b+2016+bc\right)}+\frac{b}{bc+b+2016}+\frac{bc}{2016+bc+b}\)

\(A=\frac{2016}{b+2016+bc}+\frac{b}{bc+b+2016}+\frac{bc}{2016+bc+b}\)

\(A=\frac{2016+b+bc}{2016+b+bc}=1\)

Thay : 2016 = abc

ta có :

\(A=\frac{a^2bc}{ab+a^2bc+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)

\(A=\frac{a^2bc}{ab\left(1+ac+c\right)}+\frac{b}{b\left(c+1+ac\right)}+\frac{c}{ac+c+1}\)

\(A=\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}+\frac{c}{ac+c+1}\)

\(A=\frac{ac+c+1}{ac+c+1}\)

\(A=1\)

vậy \(A=\frac{2016.a}{ab+2016.a+2016}+\frac{b}{bc+b+2016}+\frac{c}{ac+c+1}=1\)

Chúc bạn học tốt !

bạn làm được chưa biết chỉ mình vs nhé

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

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

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

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

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

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

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

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

\(=\frac{x+2}{x\left(x+2\right)}\)

\(=\frac{1}{x}\)

\(P=\frac{1}{x^2+2x+3x+6}+\frac{1}{x^2+3x+4x+12}+\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30^{ }}\)

\(P=\frac{1}{\left(x^2+2x\right)+\left(3x+6\right)}+\frac{1}{\left(x^2+3x\right)+\left(4x+12\right)}+\frac{1}{\left(x^2+4x\right)+\left(5x+20\right)}+\frac{1}{\left(x^2+5x\right)+\left(6x+30\right)}\)

\(P=\frac{1}{x\left(x+2\right)+3\left(x+2\right)}+\frac{1}{x\left(x+3\right)+4\left(x+3\right)}+\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}\)

\(P=\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)

\(P=\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}\)

\(P=\frac{1}{x+2}-\frac{1}{x+6}\)

\(P=\frac{x+6}{\left(x+2\right)\left(x+6\right)}-\frac{x+2}{\left(x+2\right)\left(x+6\right)}\)

\(P=\frac{x+6-x+2}{\left(x+2\right)\left(x+6\right)}\)

\(P=\frac{4}{\left(x+2\right)\left(x+6\right)}\)

Chúc cậu học tốt nha !

P= 1/ x²+5x+6 + 1/ x²+ 7x+12 + 1/x²+9x+20 +1/ x²+11x+30
P=1/ x²+2x+3x+6 + 1/ x²+3x+4x+12 + 1/ x ²+4x+5x+20 + 1/ x²+6x+5x+30

P=1/x.(x+2) + 3.(x+2) + 1/ x.(x+3)+4.(x+3) + 1/ x.(x+4) + 5.(x+4) + 1/ x.(x+6)+5.(x +6)
P= 1/ (x+2).(x+3) + 1/(x+3).(x+4)+1/ ( x+4).(x+5)+1/(x+6).(x+5)
P=1.(x+4)(x+5)(x+6) /(x+2)(x+3)(x+4).(x+5)(x+6) + 1.(x+2)(x+5)(x+6)/ ( x+2)(x+3)(x+4)(x+5)(x+6) + 1(x+2)(x+3)(x+6)/ (x+2)(x+3)(x+4)(x+5)(x+6) +1.(x+2)(x+3)(x+4)/(x+2)(x+3)(x+4)(x+5)(x+6)
P = (x+4)(x+5)(x+6)+(x+2)(x+5)(x+6)+(X+2)(x+3)(x+6)+(x+2)(x+3)(x+4)/(x+2)(x+3)(x+4)(x+5)(x+6)

P=4/(x+2)(x+6)

Sorry .mk tính sai kết quả phải là 2/(x+2).(x+6) còn phần trên đúng rồi