\(\Rightarrow\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}-\frac{a+b+c}{a+b+c}=0\)

\(\Rightarrow\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}\right)=0\)

xét: \(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\left(\text{vì a+b+c khác 0}\right)\)

\(\text{ta có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)

\(\Rightarrow\frac{ab+bc+ac}{abc}-\frac{1}{a+b+c}=0\)

\(\Rightarrow\frac{\left(ab+bc+ac\right).\left(a+b+c\right)-abc}{abc.\left(a+b+c\right)}=0\)

\(\Rightarrow\left(ab+bc+ac\right).\left(a+b+c\right)-abc=0\)

\(\Rightarrow\left(b+a\right).\left(c+a\right).\left(c+b\right)=0\)

\(\Rightarrow\hept{\begin{cases}b=-a\\a=-c\\c=-b\end{cases}}\)

\(M=\left(-b^{101}+b^{101}\right).\left(-c^{2017}+c^{2017}\right).\left(b^{2019}+-b^{2019}\right)=0\)

p/s: dài nhỉ =) 

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

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

Để \(N=\frac{3}{x+1}\)là số nguyên 

\(\Rightarrow3⋮x+1\Rightarrow x+1\in U\left(3\right)=\left\{\pm1;\pm3\right\}\)

bn tự lập bảng ha ~ 

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

ĐKXĐ: x khác 1

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

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

đặt \(m=\frac{1}{x-1}\Rightarrow A=1+-m+2m^2=2.\left(m^2-\frac{m.1}{2}+\frac{1}{16}\right)+\frac{7}{8}\)

\(A=2.\left(m-\frac{1}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\)

dấu = xảy ra khi \(m-\frac{1}{4}=0\)

\(\Rightarrow m=\frac{1}{4}=\frac{1}{x-1}\Rightarrow x=5\)

p/s: ko chắc lắm, 60% thôi >:

Ai cho mình đề thì bất cứ khi nào gặp câu trả lời mình cũng k đúng hết .

Ý của mình là k đúng cho bạn nào cho mình đề á

\(a,A=4x+2+2x-2-5x-6x^2-44x+2+2x-2-5x-6x^2-4\)

\(=\left(4x+2x-5x-44x+2x-5x\right)-\left(6x^2+6x^2\right)+\left(2-2+2-2-4\right)\)

\(=-46x-12x^2-4\)

Thay \(x=7373\) vào biểu thức A, ta có :

\(-46.7373-12.7373^2-4\)

\(-652672710\)

 ta có:   (3x-1)^2=(x-1)^2

           =>9x²-6x+1=x²-2x+1

           =>9x²-6x+1-x²+2x-1=0

            =>8x²-4x=0

            =>4x(2x-1)=0

           =>4x=0 ; 2x-1=0

           =>x=0 ; x=1/2

         vậy x = (0,1/2)

5x - xy + y² - 5y

= ( 5x - 5y ) - ( xy - y² )

= 5( x - y ) - y( x - y )

= ( 5 - y )( x - y )

\(5x-xy+y^2-5y\)

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

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

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

Để \(2x^3-5x^2+6x+m⋮2x-5\)  thì :

\(2x^3-5x^2+6x+m=\left(2x-5\right)\cdot Q\)

Đặt \(x=\frac{5}{2}\)ta có :

\(2\left(\frac{5}{2}\right)^3-5\left(\frac{5}{2}\right)^2+6\cdot\frac{5}{2}+m=\left(2\cdot\frac{5}{2}-5\right)\cdot Q\)

\(15+m=0\)

\(m=-15\)

Vậy........

Bài làm chỉ mang t/c tham khảo,chưa biết đúng hay sai.

Ta có: \(\frac{2x^3-5x^2+6x+m}{2x-5}=\frac{2x^3-5x^2+2x-5+4x+5+m}{2x-5}\)

\(=1+\frac{2x^3-5x^2+4x+5+m}{2x-5}=1+\frac{2x^3-5x^2+2x-5+2x+10+m}{2x-5}\)

\(=2+\frac{2x^3-5x^2+2x+10+m}{2x-5}=3+\frac{2x^3-5x^2+15+m}{2x-5}\)

\(=104+\frac{1}{15}m\).

Để \(2x^3-5x^2+6x+m⋮2x-5\) thì \(\frac{1}{15}m\) là số nguyên hay \(\frac{m}{15}\) nguyên hay \(m\in B\left(15\right)\)