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a, ĐKXĐ của biểu thức là :
x\(\ne2\) và x\(\ne-3\)
b, Rút gọn A :
\(\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}-\dfrac{1}{x-2}=\dfrac{x+2}{x+3}-\dfrac{5}{x^2-2x+3x-6}-\dfrac{1}{x-2}=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x^2-2x\right)+\left(3x-6\right)}-\dfrac{1}{x-2}=\dfrac{x+2}{x+3}-\dfrac{5}{x\left(x-2\right)+3\left(x-2\right)}-\dfrac{1}{x-2}=\dfrac{x+2}{x+3}-\dfrac{2}{\left(x-2\right)\left(x+3\right)}-\dfrac{1}{x-2}=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{1\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\dfrac{\left(x+2\right)\left(x-2\right)-5-x+3}{\left(x-2\right)\left(x+3\right)}=\dfrac{1}{x+3}\)
c, Ta có
A=\(\dfrac{1}{x+3}=-\dfrac{3}{4}\Leftrightarrow\dfrac{1.4}{\left(x+3\right)4}=\dfrac{-3\left(x+3\right)}{\left(x+3\right).4}\Leftrightarrow4=-3\left(x+3\right)\Leftrightarrow4=-3x-9\Leftrightarrow3x=-9-4\Leftrightarrow3x=-13\Rightarrow x=-\dfrac{13}{3}\left(TM\text{Đ}K\text{X}\text{Đ}\right)\)
Cho mk sữa lại nhé Đỗ Linh Chi
Câu B :
rút gon ta đc A=x+3
Tại A=\(-\dfrac{3}{4}\)
ta có
x+3=\(-\dfrac{3}{4}\Leftrightarrow\dfrac{4\left(x+3\right)}{4}=-\dfrac{3}{4}\Leftrightarrow4\left(x+3\right)=-3\Leftrightarrow4x+12=-3\Leftrightarrow4x=-15\rightarrow x=\dfrac{15}{4}\)
\(A=\left(\dfrac{x+y}{y}+\dfrac{2y}{x-y}\right)\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\left(\dfrac{2x^2+2-2x^2+x}{2\left(2x-1\right)}\right)\cdot\dfrac{1-2x}{x+2}\)
\(=\dfrac{x^2-y^2+2y^2}{y\left(x-y\right)}\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\dfrac{x+2}{2\left(2x-1\right)}\cdot\dfrac{-\left(2x-1\right)}{x+2}\)
\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)
\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
a) ĐKXĐ:
\(\begin{cases} x+3\ne 0\\ x^2+x-6 \ne 0 \Rightarrow (x+3)(x-2) \ne 0\\ 2-x\ne 0 \end{cases} \\\Leftrightarrow \begin{cases} x\ne -3\\ x\ne 2 \end{cases} \)
b) Với \(x\ne-3;x\ne2\) ta có:
\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{1}{x-2}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x^2-4-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x-4}{x-2}\)
a: DKXĐ: x<>1; x<>-1
b: \(A=\dfrac{x^2+2x+1+6-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{x^2+2x+7-x^2+x-3x+3}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)
Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
a: ĐKXĐ: x<>-3; x<>2
b: \(A=\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\dfrac{x-4}{x-2}\)
c: Để \(A=\dfrac{-3}{4}\) thì \(\dfrac{x-4}{x-2}=\dfrac{-3}{4}\)
=>-3x+6=4x-16
=>-7x=-22
=>x=22/7