a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}

Thay x = 2, ta có B không tồn tại

Thay x = -1, ta có B = \(\dfrac{1}{3}\)

b)ĐKXĐ:x ≠ 2,-2

Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x

Do đó không tồn tại x thỏa mãn đề bài

Phân thức có nghĩa khi a;b;c không đồng thời bằng 0

Khi đó:

\(\dfrac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2+2ab+2bc+2ca\right)+\left(ab+bc+ca\right)^2}{a^2+b^2+c^2+ab+bc+ca}\)

\(=\dfrac{\left(a^2+b^2+c^2\right)^2+2\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)+\left(ab+bc+ca\right)^2}{a^2+b^2+c^2+ab+bc+ca}\)

\(=\dfrac{\left(a^2+b^2+c^2+ab+bc+ca\right)^2}{a^2+b^2+c^2+ab+bc+ca}\)

\(=a^2+b^2+c^2+ab+bc+ca\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\Rightarrow\left\{{}\begin{matrix}bc=-ab-ac\\ab=-bc-ac\\ac=-ab-bc\end{matrix}\right.\)

\(M=\dfrac{1}{a^2+bc-ab-ac}+\dfrac{1}{b^2+ac-ab-bc}+\dfrac{1}{c^2+ab-bc-ac}\)

\(=\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-c\right)-a\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)-b\left(c-a\right)}\)

\(=\dfrac{1}{\left(a-b\right)\left(a-c\right)}-\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{b-c-\left(a-c\right)+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)

a) Đặt \(A=\frac{\left(a+b\right)^2-c^2}{a+b+c}=\frac{\left(a+b\right)^2}{a+b}-\frac{c^2}{c}=a+b-c\)

b)Đặt \(B=\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{a+b-c}{a+c-b}\)

Auto giải thích thêm câu b) (để tránh bị các thành phần spammer bắt bẻ)

\(\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{a+b-c}{a+c-b}\) vì:

\(\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left[\left(a+b\right)-c\right]\left[\left(a+b\right)+c\right]}{\left[\left(a+c\right)-b\right]\left[\left(a+c\right)+b\right]}=\frac{a+b-c}{a+c-b}\)

a: ĐKXĐ:\(x\notin\left\{2;0\right\}\)

b: \(C=\left(\dfrac{x\left(2-x\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{2-x^2+x}{x^2}\right)\)

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

\(=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}=\dfrac{x+1}{2x}\)

c: Thay x=2017 vào C, ta được:

\(C=\dfrac{2017+1}{2\cdot2017}=\dfrac{1009}{2017}\)

Lời giải:

Xét tử :

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

\(=\frac{a^2}{a(a-b)-c(a-b)}+\frac{b^2}{b(b-c)-a(b-c)}+\frac{c^2}{c(c-a)-b(c-a)}\)

\(=\frac{a^2}{(a-c)(a-b)}+\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)}\)

\(=\frac{a^2(c-b)+b^2(a-c)+c^2(b-a)}{(a-b)(b-c)(c-a)}\)

\(=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}=1\)

Xét mẫu (tương tự bên tử)

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

\(=\frac{bc(c-b)+ac(a-c)+ab(b-a)}{(a-b)(b-c)(c-a)}=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(a-b)(b-c)(c-a)}\)

\(=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}=1\)

Do đó:

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

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