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b,\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
=>\(\dfrac{bc}{abc}+\dfrac{ac}{bac}+\dfrac{ab}{abc}=0\)
=>\(\dfrac{ab+ac+bc}{abc}=0\)
=>ab+ac+bc=0
=>ab=-ac-bc
ac=-ab-bc
bc=-ab-ac
N=\(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ca}+\dfrac{1}{c^2+2ab}\)
N=\(\dfrac{1}{a^2+bc+bc}+\dfrac{1}{b^2+ca+ca}+\dfrac{1}{c^2+ab+ab}\)
N=\(\dfrac{1}{a^2-ab-ac+bc}+\dfrac{1}{b^2-ab-bc+ca}+\dfrac{1}{c^2-ac-bc+ab}\)
N=\(\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-a\right)-c\left(b-a\right)}+\dfrac{1}{c\left(c-a\right)-b\left(c-a\right)}\)
N=\(\dfrac{1}{\left(a-c\right)\left(a-b\right)}+\dfrac{1}{\left(b-c\right)\left(b-a\right)}+\dfrac{1}{\left(c-b\right)\left(c-a\right)}\)
N=\(\dfrac{b-c}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}-\dfrac{a-c}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}+\dfrac{a-b}{\left(b-c\right)\left(a-c\right)\left(a-b\right)}\)
N=\(\dfrac{b-c-a+c+a-b}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)=0
\(a,\left(\dfrac{1}{x-1}-\dfrac{x}{x-1^2}.\dfrac{x^2+1+x}{x+1}\right):\dfrac{1}{x^2-1}\\ =\left(\dfrac{1}{x-1}-\dfrac{x\left(x^2+1+x\right)}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1}{x^2-1}\\ =\left(\dfrac{1\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x^3+x+x^2}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1}{x^2-1}\)
\(\dfrac{x+1-x^3-x-x^2}{\left(x-1\right)\left(x+1\right)}:\dfrac{1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x+1-x^3-x-x^2\right)\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=1-x^3-x^2\)
b,
thay x=\(\dfrac{1}{2}\) vào bt M ta được:
\(1-\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^2=\dfrac{5}{8}\)
Ta có:(12x^3-7x^2-14x+14): (4x-5)= (3x^2+2x-1)+9: (4x-5). Để (12x^3-7x^2-14x+14)chia hết cho (4x-5) thì 9 phải chia hết cho(4x-5).=>4x-5 thuộc vào ước của 9=+-1;+-3;+-9.xét từng giá trị để tìm x thỏa mãn khi x<0. Sau đó kết luận.
a) ĐKXĐ: \(x\ne\pm2\)
b) \(A=\left(\dfrac{1}{x-2}-\dfrac{1}{x+2}\right)\cdot\dfrac{x^2-4x+4}{4}\)
\(=\dfrac{x+2-x+2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x-2\right)^2}{4}\)
\(=\dfrac{4\left(x-2\right)^2}{4\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x-2}{x+2}\)
c) Với \(x=4\) thoả mãn điều kiện \(x\ne\pm2\), nên thay \(x=4\) vào A, ta có:
\(A=\dfrac{4-2}{4+2}=\dfrac{2}{6}=\dfrac{1}{3}\)
a) A xác định \(\Leftrightarrow\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
b) \(A=\left(\dfrac{1}{x-2}-\dfrac{1}{x+2}\right)\cdot\dfrac{x^2-4x+4}{4}\)
\(A=\dfrac{x+2-x+2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x-2\right)^2}{4}\)
\(A=\dfrac{4\cdot\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)\cdot4}\)
\(A=\dfrac{x-2}{x+2}\)
c) Thay x = 4 ( thỏa mãn ĐKXĐ ), ta có :
\(A=\dfrac{4-2}{4+2}=\dfrac{2}{5}=\dfrac{1}{3}\)
