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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=\(-\dfrac{3}{4}\Leftrightarrow\dfrac{4\left(x+3\right)}{4}=-\dfrac{3}{4}\Leftrightarrow4\left(x+3\right)=-3\Leftrightarrow4x+12=-3\Leftrightarrow4x=-15\rightarrow x=\dfrac{15}{4}\)
để A xác định
\(\Rightarrow\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x^2\ne4\end{cases}}\Rightarrow x\ne\pm2\)
\(A=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{x^2-4}\)
\(A=\frac{4.x-8}{\left(x+2\right).\left(x-2\right)}+\frac{3.x+6}{\left(x-2\right).\left(x+2\right)}-\frac{5x-6}{\left(x-2\right).\left(x+2\right)}\)
\(A=\frac{4x-8+3x+6-5x+6}{\left(x+2\right).\left(x-2\right)}=\frac{2.\left(x+2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{2}{x-2}\)
\(\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{x^2-4}=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{4x-8}{\left(x+2\right)\left(x-2\right)}+\frac{3x+4}{\left(x-2\right)\left(x+2\right)}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}=\frac{4x-8+3x+4-5x+6}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{2x+2}{\left(x+2\right)\left(x-2\right)}=\frac{2x+2}{x^2-4}\)
C, \(x=4\Rightarrow A=\frac{2x+2}{x^2-4}=\frac{-6}{12}=\frac{-1}{2}\)
d, \(A\inℤ\Leftrightarrow2x+2⋮x^2-4\Leftrightarrow2x^2+2x-2x^2+8⋮x^2-4\Leftrightarrow2x+8⋮x^2-4\)
\(\Leftrightarrow2x^2+8x⋮x^2-4\Leftrightarrow16⋮x^2-4\)
\(x^2-4\inℕ\)
\(\Rightarrow x^2\in\left\{0;4;12\right\}\)
Thử lại thì 12 ko là số chính phương vậy x=0 hoặc x=2 thỏa mãn
mk học lớp 6 mong mn thông cảm nếu có sai sót
\(A=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}+\dfrac{1}{2-x}\)
a ) ĐKXĐ :\(x\ne2\) và \(x\ne-3\).
Rút gọn : \(A=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}+\dfrac{1}{2-x}\)
\(\Leftrightarrow A=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{1}{x-2}\)
\(\Leftrightarrow A=\dfrac{\left(x+2\right)\left(x-2\right)-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow A=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\)
b ) Khi \(A=-\dfrac{3}{4},\) thì :
\(\dfrac{x-4}{x-2}=-\dfrac{3}{4}\)
\(\Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\)
\(\Leftrightarrow4x-16=-3x+6\)
\(\Leftrightarrow x=\dfrac{22}{7}\).
c ) Ta có : \(\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)
Vậy để A nguyên thi \(x-2⋮2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Thay vào từng cái sẽ ra nha :**
d ) Ta có : \(x^2-9=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
+ ) Khi x = 3 , thì :
\(A=\dfrac{3-4}{3-2}=\dfrac{-1}{1}=-1\)
+ ) Khi x = -3, thì :
\(A=\dfrac{-3-4}{-3-2}=\dfrac{-7}{-5}=\dfrac{7}{5}.\)
Vậy ........
\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
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,ĐK: \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)
c, Với x = 4 thỏa mãn ĐKXĐ thì
\(A=\frac{-3}{4-3}=-3\)
d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)
Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
\(\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{5x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
a ) ĐKXĐ : \(x\ne0,x\ne-5\)
b ) Rút gọn trước cái đã
\(\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{5x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+10x^2+50x-10x-50+50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+12x^2+35x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x+7\right)}{2x\left(x+5\right)}=\dfrac{x+7}{2x}\)
Khi \(A=1\), thì :
\(\dfrac{x+7}{2x}=1\Leftrightarrow x=7\)
Khi A = 3, thì :
\(\dfrac{x+7}{2x}=3\Leftrightarrow x=-1.\)
Bài 2 :
a ) ĐKXĐ : x\(\ne-3;2\)
b ) \(\dfrac{x-2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}+\dfrac{1}{2-x}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\)
c ) Khi \(A=-\dfrac{3}{4}\), thì :
\(\dfrac{x-4}{x-2}=-\dfrac{3}{4}\)
\(\Leftrightarrow4x-16=-3x+6\)
\(\Leftrightarrow x=\dfrac{22}{7}\)
d ) Ta có :
\(A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)
Để A nguyên thi \(x-2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Thay vào rồi tìm ra nếu x có trong đkxđ thì loại .
e ) \(x^2-9=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Thay từng x vào A là tìm ra
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