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\(K=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{2017bc}{abc+2017bc+2017b}\)
\(K=\frac{a}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{2017bc}{2017\left(bc+b+1\right)}\)
\(K=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}=\frac{bc+b+1}{bc+b+1}=1\)
\(H=\frac{4x^2+4-4x^2-4x-1}{x^2+1}=\frac{4\left(x^2+1\right)}{x^2+1}-\frac{\left(2x+1\right)^2}{x^2+1}=4-\frac{\left(2x+1\right)^2}{x^2+1}\le4\)
\(\Rightarrow H_{max}=4\) khi \(x=-\frac{1}{2}\)
Thay abc = 2017 vào A ta có:
\(A=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{abc^2}{ac+abc^2+abc}\)
\(=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{bc}{1+bc+b}=1\)
những dạng có cho tích hoặc tổng bằng một số nào đó và trog đa thức cần tính có tích hoặc tổng hoặc số đó thj kiểu j cx p thay vào bn ak.
hỳ mik tự rút đc kinh nghiệm đó mờ
a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)
a) Ta thấy x=-2 thỏa mãn ĐKXĐ của B.
Thay x=-2 và B ta có :
\(B=\frac{2\cdot\left(-2\right)+1}{\left(-2\right)^2-1}=\frac{-3}{3}=-1\)
b) Rút gọn :
\(A=\frac{3x+1}{x^2-1}-\frac{x}{x-1}\)
\(=\frac{3x+1-x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{-x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)
Xấu nhỉ ??
\(a,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) (x khác -3; khác 0)
\(=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x}{2x.\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x-x+6}{2x.\left(x+3\right)}=\frac{2x+6}{x.\left(2x+6\right)}=\frac{1}{x}\)
\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\) (x khác 0 , khác 1/2 khác -1/2 )
\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\left(\frac{4x^2+4x+1}{\left(2x-1\right)\left(2x+1\right)}-\frac{4x^2-4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5.\left(2x-1\right)}{4x}=\frac{10}{2x+1}\)
a) \(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6}{1-x}\)
\(=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(x-1\right)\left(2x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{4x^2-3x+17+2x^2-x-2x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{-12x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=-\frac{12}{x^2+x+1}\)
b) \(\frac{1}{x^2-x+1}-\frac{x^2+2}{x^3+1}+1=\frac{x+1-x^2-2+x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x-x^2+x^3}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x}{x+1}\)
c) \(N=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{a}{a\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{1+b}{b+1+bc}+\frac{abc^2}{ac+abc^2+abc}\)
\(N=\frac{1+b}{b+1+bc}+\frac{abc^2}{ac\left(1+bc+b\right)}\)
\(N=\frac{1+b}{b+1+bc}+\frac{bc}{1+bc+b}\)
\(N=\frac{1+b+bc}{b+1+bc}\)
\(N=1.\)