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Tìm điều kiện của biến x để giá trị của biểu thức P được xác định.
a) ĐK:\(\begin{cases} x + 2≠0\\ x - 2≠0 \end{cases}\)⇔\(\begin{cases} x ≠ -2\\ x≠ 2 \end{cases}\)
Vậy biểu thức P xác định khi x≠ -2 và x≠ 2
b) P= \(\dfrac{3}{x+2}\)-\(\dfrac{2}{2-x}\)-\(\dfrac{8}{x^2-4}\)
P=\(\dfrac{3}{x+2}\)+\(\dfrac{2}{x-2}\)-\(\dfrac{8}{(x-2)(x+2)}\)
P= \(\dfrac{3(x-2)}{(x-2)(x+2)}\)+\(\dfrac{2(x+2)}{(x-2)(x+2)}\)-\(\dfrac{8}{(x-2)(x+2)}\)
P= \(\dfrac{3x-6+2x+4-8}{(x-2)(x+2)}\)
P=\(\dfrac{5x-10}{(x-2)(x+2)}\)
P=\(\dfrac{5(x-2)}{(x-2)(x+2)}\)
P=\(\dfrac{5}{x+2}\)
Vậy P=\(\dfrac{5}{x+2}\)
a: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{4}{x-2}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)
\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{6\left(x+1\right)\cdot x\left(x+2\right)}{3\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x}{x-2}\)
Câu 2:
a: ĐKXĐ: \(x\notin\left\{0;2\right\}\)
b: Sửa đề: \(A=\left(\dfrac{2x-x^2}{2x^2+8}-\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\left(\dfrac{2}{x^2}-\dfrac{x-1}{x}\right)\)
\(=\left(\dfrac{2x-x^2}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{2-x\left(x-1\right)}{x^2}\)
\(=\left(\dfrac{\left(2x-x^2\right)\left(x-2\right)-4x^2}{2\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{2-x^2+x}{x^2}\)
\(=\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x^2+4\right)\left(x-2\right)}\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{x^3-2x^2-2x^2+4x+4x^2}{2\left(x^2+4\right)\left(x-2\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3+4x}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)\left(x+1\right)}{2\left(x^2+4\right)\cdot x^2}=\dfrac{x+1}{2x}\)
c: Khi x=2024 thì \(A=\dfrac{2024+1}{2\cdot2024}=\dfrac{2025}{4048}\)
Câu 1:
a: \(25x^2\left(x-3y\right)-15\left(3y-x\right)\)
\(=25x^2\left(x-3y\right)+15\left(x-3y\right)\)
\(=\left(x-3y\right)\left(25x^2+15\right)\)
\(=\left(x-3y\right)\cdot5\cdot\left(5x^2+3\right)\)
b: \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=\left(x^4-x^2\right)-\left(4x^2-4\right)\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
1. ĐKXĐ: \(x\ne\pm1\)
2. \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x+3}{x+1}\right)\cdot\dfrac{x+1}{2}\)
\(=\dfrac{\left(x+1\right)^2-\left(x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{x^2+2x+1-x^2+4x-3}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{6x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{2\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x-3}{x-1}\)
3. Tại x = 5, A có giá trị là:
\(\dfrac{5-3}{5-1}=\dfrac{1}{2}\)
4. \(A=\dfrac{x-3}{x-1}\) \(=\dfrac{x-1-3}{x-1}=1-\dfrac{3}{x-1}\)
Để A nguyên => \(3⋮\left(x-1\right)\) hay \(\left(x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(tmđk\right)\\x=0\left(tmđk\right)\\x=4\left(tmđk\right)\\x=-2\left(tmđk\right)\end{matrix}\right.\)
Vậy: A nguyên khi \(x=\left\{2;0;4;-2\right\}\)
ĐK: \(x\ne\pm2\)
\(A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right).\dfrac{x+2}{2}\)
\(=\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right].\dfrac{x+2}{2}\)
\(=\dfrac{x-2x-2+x-2}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{2}\)
\(=\dfrac{2}{2-x}\)