Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(A=\dfrac{x^2-8x+16-x^2+16}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-8\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-4x}{\left(x+4\right)\left(x-1\right)}\)
a: Thay x=-4 vào B, ta được:
\(B=\dfrac{-4+3}{-4}=\dfrac{-1}{-4}=\dfrac{1}{4}\)
b: \(P=A\cdot B=\dfrac{x^2-3x+2x-9+3x+9}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}\)
\(=\dfrac{x^2+2x}{\left(x-3\right)}\cdot\dfrac{1}{x}=\dfrac{x+2}{x-3}\)
c: Để P nguyên thì \(x-3\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{4;2;8;-2\right\}\)
a: \(A=\left(\frac{x}{x+2}+\frac{2}{x-2}+\frac{4x}{4-x^2}\right):\frac{2x+1}{8x+16}\)
\(=\frac{x\left(x-2\right)+2\left(x+2\right)-4x}{\left(x-2\right)\left(x+2\right)}\cdot\frac{8\left(x+2\right)}{2x+1}\)
\(=\frac{x^2-2x+2x+4-4x}{x-2}\cdot\frac{8}{2x+1}=\frac{\left(x-2\right)^2}{\left(x-2\right)}\cdot\frac{8}{2x+1}\)
\(=\frac{8\left(x-2\right)}{2x+1}=\frac{8x-16}{2x+1}\)
b: Thay \(x=-2\frac12=-2,5\) vào A, ta được:
\(A=\frac{8\cdot\left(-2,5\right)-16}{2\cdot\left(-2,5\right)+1}=\frac{-20-16}{-5+1}=\frac{-36}{-4}=9\)
c: Để A nguyên thì 8x-16⋮2x+1
=>8x+4-20⋮2x+1
=>-20⋮2x+1
=>2x+1∈{1;-1;5;-5}
=>2x∈{0;-2;4;-6}
=>x∈{0;-1;2;-3}
Kết hợp ĐKXĐ, ta được: x∈{0;-1;-3}
a: \(A=\left(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\right):\left(\frac{3}{x^2-2x^3}\right)\)
\(=\frac{\left(1-2x\right)\left(2x-1\right)+2x\cdot2x-1}{2x\left(2x-1\right)}:\frac{3}{x^2\left(1-2x\right)}\)
\(=\frac{-\left(4x^2-4x+1\right)+4x^2-1}{2x\left(2x-1\right)}\cdot\frac{-x^2\left(2x-1\right)}{3}\)
\(=\frac{-4x^2+4x-1+4x^2-1}{2}\cdot\frac{-x}{3}=\frac{4x-2}{2}\cdot\frac{-x}{3}=-\frac{x\left(2x-1\right)}{3}\)
b: \(A=-\frac{x\left(2x-1\right)}{3}\)
\(=-\frac13\left(2x^2-x\right)\)
\(=-\frac23\left(x^2-\frac12x\right)\)
\(=-\frac23\left(x^2-2\cdot x\cdot\frac14+\frac{1}{16}-\frac{1}{16}\right)=-\frac23\left(x-\frac14\right)^2+\frac23\cdot\frac{1}{16}\)
\(=-\frac23\left(x-\frac14\right)^2+\frac{1}{24}\le\frac{1}{24}\forall x\)
Dấu '=' xảy ra khi x-1/4=0
=>x=1/4(nhận)
A = \(\frac{2x^2+4x}{x^3-4x}+\frac{x^2-4}{x^2+2x}+\frac{2}{2-x}\)
= \(\frac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)}-\frac{2}{x-2}\)
= \(\frac{2}{x-2}+\frac{x-2}{x}-\frac{2}{x-2}\)
= \(\frac{x-2}{x}\)
= \(1-\frac{2}{x}\)
Để A có giá trị nguyên thì \(1-\frac{2}{x}\) nguyên, để \(1-\frac{2}{x}\) nguyên thì \(\frac{2}{x}\) nguyên
\(\frac{2}{x}\) có giá trị nguyên ⇒ \(x=1,-1,2,-2\)