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b)
\(P=A-B=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x^2-9}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-9-x^2+9}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x\left(2-x\right)}{\left(x-3\right)\left(x-2\right)}\\ =-\dfrac{x\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =-\dfrac{x}{x-3}\)
c)
Để \(P\le1\) thì:
\(-\dfrac{x}{x-3}\le1\)
\(\Leftrightarrow\dfrac{x}{x-3}\ge1\\ \Leftrightarrow x-3-x\ge1\\ \Leftrightarrow-3\ge1\left(vô.lý\right)\)
Vậy không tồn tại giá trị x để \(P\le1\)
`HaNa♬D`
Làm lại nha cái này đúng, kia sai nha=)
b)
Với \(\left\{{}\begin{matrix}x\ne3\\x\ne2\end{matrix}\right.\)
\(P=A-B=(\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)})+\dfrac{2x-1}{x-3}\\ =\left(\dfrac{2x-9-x^2-9}{\left(x-3\right)\left(x-2\right)}\right)+\dfrac{\left(2x-1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2}{\left(x-3\right)\left(x-2\right)}+\dfrac{2x^2-4x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2+2x^2-4x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x^2-3x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x^2-2x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x\left(x-2\right)-\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}=\dfrac{x-1}{x-3}\)
c)
Để P\(\ge1\) thì:
\(\dfrac{x-1}{x-3}\ge1\\ \Leftrightarrow x-3-x+1-1\ge0\\ \Leftrightarrow-3\ge0\left(vô.lý\right)\)
Vậy không tồn tại giá trị x để \(P\ge1\)
`HaNa☘D`
f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)
g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)
h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)
n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)
p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)
k: \(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-6}{x^2-4}\)
m: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
`a,` Với `x=3`
\(B=\dfrac{x^2-x}{2x+1}\\ \Rightarrow\dfrac{3^2-3}{2\cdot3+1}\\ =\dfrac{9-3}{6+1}\\ =\dfrac{6}{7}\)
`b,` Ta có `M=A*B`
\(M=\left(\dfrac{1}{x-1}+\dfrac{x}{x^2-1}\right)\cdot\dfrac{x^2-x}{2x+1}\\ =\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x\left(x-1\right)}{2x+\text{ }1}\\ =\left(\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x\left(x-1\right)}{2x+1}\\ =\dfrac{x+1+x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x\left(x-1\right)}{2x+1}\\ =\dfrac{2x+1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x\left(x-1\right)}{2x+1}\\ =\dfrac{x}{x+1}\)
`c,` Để `M=1/2`
`=> x/(x+1)=1/3`
`<=> (3x)/(3(x+1))= (x+1)/(3(x+1))`
`<=> 3x=x+1`
`<=>3x-x=1`
`<=>2x=1`
`<=>x=1/2`
a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}
Thay x = 2, ta có B không tồn tại
Thay x = -1, ta có B = \(\dfrac{1}{3}\)
b)ĐKXĐ:x ≠ 2,-2
Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x
Do đó không tồn tại x thỏa mãn đề bài
1,
\(A=\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x^2+x-2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2-4}{\left(x-2\right)\left(x+2\right)}\)
\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)
2.
\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
3.
Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)
4.
\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)
\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)
5.
\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)
\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)
\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)
\(a,\)Với \(x\ne-3,x\ne2\) ta có :
\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}-\dfrac{1}{x-2}\)
\(=\dfrac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x-4}{x-2}\)
\(b,\) \(A=-3\Leftrightarrow\dfrac{x-4}{x-2}=-3\)
\(\Leftrightarrow x-4=-3\left(x-2\right)\)
\(\Leftrightarrow x-4+3x-6=0\)
\(\Leftrightarrow4x=10\Rightarrow x=\dfrac{10}{4}=\dfrac{5}{2}\)