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\(\Leftrightarrow\dfrac{x+1}{\left(x-3\right)\left(x+2\right)\cdot B}=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}\)
\(\Leftrightarrow B=\dfrac{x-1}{\left(x-3\right)\left(x+2\right)}\)
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
a) \(\dfrac{\left(x+2\right)P}{x-2}=\dfrac{\left(x-1\right)Q}{x^2-4}\)
\(\Leftrightarrow\left(x^2-4\right)\left(x+2\right)P=\left(x-2\right)\left(x-1\right)Q\)
\(\Leftrightarrow\)\(\left(x+2\right)^2\left(x-2\right)P=\left(x-2\right)\left(x-1\right)Q\)
\(\Leftrightarrow\)\(\left(x+2\right)^2P=\left(x-1\right)Q\)
\(\Leftrightarrow P=x-1\)
\(Q=\left(x+2\right)^2=x^2+4x+4\)
b)\(\dfrac{\left(x+2\right)P}{x^2-1}=\dfrac{\left(x-2\right)Q}{x^2-2x+1}\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+2\right)P=\left(x+1\right)\left(x-1\right)\left(x-2\right)Q\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)P=\left(x+1\right)\left(x-2\right)Q\)
\(\Leftrightarrow P=\left(x+1\right)\left(x-2\right)=x^2-x-2\)
\(Q=\left(x-1\right)\left(x+2\right)=x^2+x-2\)
a) A = \(\dfrac{1}{x-1}-\dfrac{4}{x+1}+\dfrac{8x}{\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{x+1-4x+4+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x+5}{\left(x-1\right)\left(x+1\right)}=\dfrac{5}{x-1}\) => đpcm
b) \(\left|x-2\right|=3=>\left[{}\begin{matrix}x-2=3< =>x=5\left(C\right)\\x-2=-3< =>x=-1\left(L\right)\end{matrix}\right.\)
Thay x = 5 vào A, ta có:
A = \(\dfrac{5}{5-1}=\dfrac{5}{4}\)
c) Để A nguyên <=> \(5⋮x-1\)
| x-1 | -5 | -1 | 1 | 5 |
| x | -4(C) | 0(C) | 2(C) | 6(C) |
Ta co:
\(\dfrac{1}{x^2-4}=\dfrac{1}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow\dfrac{1}{\left(x-2\right)\left(x+2\right)}=\dfrac{a}{x-2}+\dfrac{b}{x+2}\)
\(\Rightarrow\dfrac{a\left(x+2\right)+b\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{ax+2a+bx-2b}{\left(x-2\right)\left(x+2\right)}\)
Ta có: \(\dfrac{1}{x^2-4}=\dfrac{a}{x-2}+\dfrac{b}{x+2}\Rightarrow\dfrac{1}{x^2-4}=\dfrac{ax+2a+bx-2b}{x^2-4}\)
\(\Rightarrow ax+2a+bx-2b=1\)
\(\Rightarrow x\left(a+b\right)+\left(2a-2b\right)=0x+1\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=0\\2a-2b=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{4}\\b=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy: \(a=\dfrac{1}{4};b=-\dfrac{1}{4}\).