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Bài 1:
a) \(\dfrac{3x^2-5}{x^2-5x}+\dfrac{5-15x}{5x-25}\)
\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{5\left(1-3x\right)}{5\left(x-5\right)}\)
\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{1-3x}{x-5}\)
\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{x\left(1-3x\right)}{x\left(x-5\right)}\)
\(=\dfrac{3x^2-5+x\left(1-3x\right)}{x\left(x-5\right)}\)
\(=\dfrac{3x^2-5+x-3x^2}{x\left(x-5\right)}\)
\(=\dfrac{-5+x}{x\left(x-5\right)}\)
\(=\dfrac{x-5}{x\left(x-5\right)}\)
\(=\dfrac{1}{x}\)
b) \(\dfrac{4+x^3}{x-3}-\dfrac{2x+2x^2}{x-3}+\dfrac{2x-13}{x-3}\)
\(=\dfrac{\left(4+x^3\right)-\left(2x+2x^2\right)+\left(2x-13\right)}{x-3}\)
\(=\dfrac{4+x^3-2x-2x^2+2x-13}{x-3}\)
\(=\dfrac{x^3-2x^2-9}{x-3}\)
\(=\dfrac{x^3-3x^2+x^2-9}{x-3}\)
\(=\dfrac{x^2\left(x-3\right)+\left(x-3\right)\left(x+3\right)}{x-3}\)
\(=\dfrac{\left(x-3\right)\left(x^2+x+3\right)}{x-3}\)
\(=x^2+x+3\)
c) \(\dfrac{2}{x-5}+\dfrac{x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}+\dfrac{x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{2\left(x+5\right)+x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{2x+10+x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3}{x+5}\)
d) Đề sai?
Bài 2:
\(A=2\left(x+1\right)+\left(3x+2\right)\left(3x-2\right)-9x^2\)
\(A=2x+2+9x^2-4-9x^2\)
\(A=2x-2\)
\(A=2\left(x-1\right)\)
Thay x = 15 vào A ta được:
\(A=2\left(15-1\right)\)
\(A=2.14=28\)
\(a,\dfrac{x^2-2x}{x^2-4}=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)
b) \(\dfrac{x^2+5x+4}{x^2-1}=\dfrac{x^2+x+4x+4}{x^2-1}=\dfrac{\left(x+1\right)\left(x+4\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+4}{x-1}\)
c) \(\dfrac{x^4+4}{x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1}\)
\(=\dfrac{x^4+4x^2-4x^2+4}{x^3+2x-2x^2-x^2+2x-1-1}\)
\(=\dfrac{\left(x^2+2\right)^2-4x^2}{\left(x^3+2x-2x^2\right)-\left(x^2-2x+2\right)}\)
\(=\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x\left(x^2+2-2x\right)-\left(x^2+2-2x\right)}\)
\(=\dfrac{x^2+2+2x}{x-1}\)
Bài 2:
a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
\(=\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{10}{2x+1}\)
b) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\dfrac{1-2x+x^2}{x\left(x+1\right)}:\dfrac{1+x^2-2x}{x}\)
\(=\dfrac{1}{x+1}\)
c) Trong ngoặc giữa hai phân số là dấu gì vậy ?
Bài 1:
a: ĐKXĐ: x<>1/3; x<>-1/3
b: \(M=\left(\dfrac{-3x}{3x-1}+\dfrac{2x}{3x+1}\right)\cdot\dfrac{\left(3x-1\right)^2}{2\left(3x^2+5\right)}\)
\(=\dfrac{-9x^2-3x+6x^2-2x}{\left(3x-1\right)\left(3x+1\right)}\cdot\dfrac{\left(3x-1\right)^2}{2\left(3x^2+5\right)}\)
\(=\dfrac{-3x^2-5x}{\left(3x+1\right)}\cdot\dfrac{1}{2\left(3x^2+5\right)}=\dfrac{-3x^2-5x}{2\left(3x+1\right)\left(3x^2+5\right)}\)
Bài 2:
\(P=\left(\dfrac{x}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{x\left(x+5\right)}\right)\cdot\dfrac{x\left(x+5\right)}{2x-5}+\dfrac{x}{5-x}\)
\(=\dfrac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x\left(x+5\right)}{2x-5}+\dfrac{x}{5-x}\)
\(=\dfrac{5}{x-5}+\dfrac{x}{5-x}=-1\)
Câu 1:
a: ĐKXĐ: x<>1/3; x<>-1/3
b: \(M=\dfrac{-9x^2-3x+6x^2-2x}{\left(3x+1\right)\left(3x-1\right)}\cdot\dfrac{\left(3x-1\right)^2}{2\left(3x^2+5\right)}\)
\(=\dfrac{-3x+1}{3x+1}\)
c: x=1/3 thì loại bởi nó không thỏa ĐKXĐ
1) \(\dfrac{A\left(x-5\right)}{\left(x+1\right)\left(x-5\right)}=\dfrac{3x\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}\)
\(\Rightarrow A=3x\)
2) \(\dfrac{\left(x+3\right)\left(x-2\right)}{A\left(x-3\right)}=\dfrac{\left(5x-1\right)\left(x-2\right)}{\left(5x-1\right)\left(x^2+3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+3\right)}{A\left(x-3\right)}=\dfrac{1}{\left(x^2+3\right)}\)
\(\Rightarrow A=\dfrac{\left(x^2+3\right)\left(x+3\right)}{x-3}\)
3) \(\dfrac{\left(x-5\right)\left(x+5\right)}{\left(x+5\right)\left(2x-3\right)}=\dfrac{\left(x-5\right)A}{\left(2x-3\right)\left(x+2\right)}\)
\(\Leftrightarrow1=\dfrac{A}{\left(x+2\right)}\)
\(\Leftrightarrow A=x+2\)
a.\(\dfrac{5\left(x-3\right)}{4\left(x+1\right)}\) : \(\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x+1\right)^2}\)
= \(\dfrac{5\left(x-3\right)}{4\left(x+1\right)}\). \(\dfrac{\left(x+1\right)^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{5\left(x+1\right)}{4\left(x+3\right)}\)
b. \(\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\). \(\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}\)
= \(\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)
c.Tương tự hai câu trên nka!!
d. (\(\dfrac{1}{x\left(x+1\right)}\)-\(\dfrac{2-x}{x+1}\)).(\(\dfrac{x}{x-1}\))
=( \(\dfrac{1}{x\left(x+1\right)}\)-\(\dfrac{2x-x^2}{x\left(x+1\right)}\)). ....
= \(\dfrac{\left(1-x\right)^2}{x\left(x+1\right)}\). ...
= \(\dfrac{x-1}{x+1}\)
\(B=\left(\dfrac{x}{x^2-25}-\dfrac{x-5}{x^2+5x}\right):\dfrac{2x-5}{x^2+5x}+\dfrac{x}{5-x}\) (1).
Đkxđ: \(x\ne\pm5;\)
(1) \(=\left(\dfrac{x}{\left(x+5\right)\left(x-5\right)}-\dfrac{x-5}{x\left(x+5\right)}\right):\dfrac{2x-5}{x\left(x+5\right)}+\dfrac{x}{5-x}\)
\(=\left(\dfrac{x^2-\left(x-5\right)\left(x-5\right)}{x\left(x+5\right)\left(x-5\right)}\right):\dfrac{2x-5}{x\left(x+5\right)}+\dfrac{x}{5-x}\)
\(=\dfrac{25}{x\left(x+5\right)\left(x-5\right)}.\dfrac{x\left(x+5\right)}{2x-5}-\dfrac{x}{x-5}\)
\(=\dfrac{25}{\left(x-5\right)\left(2x-5\right)}-\dfrac{x}{x-5}\)
\(=\dfrac{25-x\left(2x-5\right)}{\left(x-5\right)\left(2x-5\right)}\)
\(=\dfrac{25-2x^2+5x}{\left(x-5\right)\left(2x-5\right)}\)