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a) \(A=\dfrac{x^2-4x+4}{5x-10}.\) ĐK: \(x\ne2.\)
b) \(A=\dfrac{x^2-4x+4}{5x-10}=\dfrac{\left(x-2\right)^2}{5\left(x-2\right)}=\dfrac{x-2}{5}.\)
c) \(Thay\) \(x=-2018:\) \(\dfrac{-2018-2}{5}=-404.\)
a) ĐK: \(x\ne4,x\ne2;x\ne-2\)
b) \(A=\dfrac{x^3}{x-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)
\(A=\dfrac{x^3}{\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(A=\dfrac{x^3-x^2-2x-2x+4}{\left(x+2\right)\left(x-2\right)}\)
\(A=\dfrac{x^3-x^2-4x+4}{\left(x+2\right)\left(x-2\right)}\)
\(A=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x+2\right)\left(x-2\right)}\)
\(A=\dfrac{\left(x-1\right)\left(x^2-4\right)}{x^2-4}\)
\(A=x-1\)
c) \(A=0\) khi:
\(x-1=0\)
\(\Leftrightarrow x=1\left(tm\right)\)
d) A dương khi: \(A>0\)
\(x-1>0\)
\(\Leftrightarrow x>1\)
Kết hợp với đk:
\(x>1,x\ne4,x\ne2\)
a: ĐKXĐ: x<>-1
b: \(P=\left(1-\dfrac{x+1}{x^2-x+1}\right)\cdot\dfrac{x^2-x+1}{x+1}\)
\(=\dfrac{x^2-x+1-x-1}{x^2-x+1}\cdot\dfrac{x^2-x+1}{x+1}=\dfrac{x^2-2x}{x+1}\)
c: P=2
=>x^2-2x=2x+2
=>x^2-4x-2=0
=>\(x=2\pm\sqrt{6}\)
a: ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};\dfrac{1}{2};-2\right\}\)
b: \(B=\dfrac{4x^2+4x+1-4-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{2x+1}{x+2}\)
\(=\dfrac{8x-4}{2x-1}\cdot\dfrac{1}{x+2}=\dfrac{4}{x+2}\)
a: ĐKXĐ: x<>1/2; x<>-1/2; x<>0
b: \(A=\dfrac{4x^2+4x+1-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{4x}\cdot\dfrac{5}{2x+1}=\dfrac{10}{2x+1}\)
Lời giải:
a.
ĐKXĐ: $x\neq \pm 2$
b.
\(P=\left[\frac{4(x-2)}{(x+2)(x-2)}+\frac{3(x+2)}{(x+2)(x-2)}-\frac{5x+2}{(x-2)(x+2)}\right].\frac{x+2}{2}\)
\(=\frac{4(x-2)+3(x+2)-(5x+2)}{(x-2)(x+2)}.\frac{x+2}{2}=\frac{2(x-2)}{(x-2)(x+2)}.\frac{x+2}{2}=1\)
a: ĐKXĐ: x<>4; x<>-4
b: \(A=\dfrac{\left(x-4\right)\left(x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{x-1}{x+4}\)
c: Để A nguyên thì x+4-5 chia hết cho x+4
=>\(x+4\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-3;-5;1;-9\right\}\)
Sao mà ra đc { -3;-5;1;-9} đc vậy ạ