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a) ĐKXĐ: \(\hept{\begin{cases}x+3\ne0\\3-x\ne0\\x^2-9\ne0\end{cases}}\) <=> \(\hept{\begin{cases}x\ne-3\\x\ne3\\x\ne\pm3\end{cases}}\)
Ta có: A = \(\frac{x+1}{x+3}-\frac{x-1}{3-x}+\frac{2x-2x^2}{x^2-9}\)
A = \(\frac{\left(x+1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x+3\right)\left(x-1\right)}{\left(x+3\right)\left(x-3\right)}+\frac{2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
A = \(\frac{x^2-2x-3+x^2+2x-3+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
A = \(\frac{2x-6}{\left(x-3\right)\left(x+3\right)}\)
A = \(\frac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
A = \(\frac{2}{x+3}\)
b) Để A nhận giá trị dương <=> 2 \(⋮\)x + 3
<=> x + 3 \(\in\)Ư(2) = {1; 2}
Lập bảng:
| x + 3 | 1 | 2 |
| x | -2 | -1 |
Vậy ....
a) \(ĐKXĐ:x^2-3x\ne0\)\(\Leftrightarrow x\left(x-3\right)\ne0\)\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x-3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne3\end{cases}}\)
\(A=\frac{x^3-3x^2-x+3}{x^2-3x}=\frac{\left(x^3-3x^2\right)-\left(x-3\right)}{x\left(x-3\right)}=\frac{x^2\left(x-3\right)-\left(x-3\right)}{x\left(x-3\right)}\)
\(=\frac{\left(x^2-1\right)\left(x-3\right)}{x\left(x-3\right)}=\frac{\left(x-1\right)\left(x+1\right)}{x}\)
b) Với \(x=2\)( thoả mãn ĐKXĐ ) \(\Rightarrow A=\frac{\left(2-1\right)\left(2+1\right)}{2}=\frac{3}{2}\)
a) \(\left(2x+3\right)^2-3\left(x-4\right)\left(x+4\right)=\left(x-2\right)^2+1\)
\(\Leftrightarrow4x^2+12x+9-3\left(x^2-16\right)=x^2-4x+4+1\)
\(\Leftrightarrow4x^2+12x+9-3x^2+48=x^2-4x+5\)
\(\Leftrightarrow x^2+12x+57=x^2-4x+5\)
\(\Leftrightarrow16x+52=0\)
\(\Leftrightarrow x=-\frac{13}{4}\)
b) \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\)Xem lại đề !
c) \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)
\(\Leftrightarrow x^2-x-x^2-x+12=5x\)
\(\Leftrightarrow-2x+12=5x\)
\(\Leftrightarrow7x-12=0\)
\(\Leftrightarrow x=\frac{12}{7}\)
d) \(\left(2x+1\right)\left(2x-1\right)=4x\left(x-7\right)-3x\)
\(\Leftrightarrow4x^2-1=4x^2-28x-3x\)
\(\Leftrightarrow28x+3x-1=0\)
\(\Leftrightarrow31x-1=0\)
\(\Leftrightarrow x=\frac{1}{31}\)
a,P=\(\frac{x^2\left(x-3\right)+3\left(x-3\right)}{(x-3)^2}\)
=\(\frac{x^2+3}{x-3}\)
a) Điều kiện xác định: \(x^2-6x+9=\left(x-3\right)^2\ne0\)
\(\Rightarrow x\ne3\)
ĐKXĐ: \(x\ne3\)
\(P=\frac{x^3-3x^2+3x-9}{x^2-6x+9}\)
\(P=\frac{\left(x-3\right)\left(x^2+3\right)}{\left(x-3\right)\left(x-3\right)}\)
\(P=\frac{x^2+3}{x-3}\)
b) +) x = 2
\(P=\frac{2^2+3}{2-3}=-7\)
+) x = -3
\(P=\frac{\left(-3\right)^2+3}{-3-3}=1\)
Câu 1a : tự kết luận nhé
\(2\left(x+3\right)=5x-4\Leftrightarrow2x+6=5x-4\Leftrightarrow-3x=-10\Leftrightarrow x=\frac{10}{3}\)
Câu 1b : \(\frac{1}{x-3}-\frac{2}{x+3}=\frac{5-2x}{x^2-9}\)ĐK : \(x\ne\pm3\)
\(\Leftrightarrow x+3-2x+6=5-2x\Leftrightarrow-x+9=5-2x\Leftrightarrow x=-4\)
c, \(\frac{x+1}{2}\ge\frac{2x-2}{3}\Leftrightarrow\frac{x+1}{2}-\frac{2x-2}{3}\ge0\)
\(\Leftrightarrow\frac{3x+3-4x+8}{6}\ge0\Rightarrow-x+11\ge0\Leftrightarrow x\le11\)vì 6 >= 0
1) 2(x + 3) = 5x - 4
<=> 2x + 6 = 5x - 4
<=> 3x = 10
<=> x = 10/3
Vậy x = 10/3 là nghiệm phương trình
b) ĐKXĐ : \(x\ne\pm3\)
\(\frac{1}{x-3}-\frac{2}{x+3}=\frac{5-2x}{x^2-9}\)
=> \(\frac{x+3-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{5-2x}{\left(x-3\right)\left(x+3\right)}\)
=> x + 3 - 2(x - 3) = 5 - 2x
<=> -x + 9 = 5 - 2x
<=> x = -4 (tm)
Vậy x = -4 là nghiệm phương trình
c) \(\frac{x+1}{2}\ge\frac{2x-2}{3}\)
<=> \(6.\frac{x+1}{2}\ge6.\frac{2x-2}{3}\)
<=> 3(x + 1) \(\ge\)2(2x - 2)
<=> 3x + 3 \(\ge\)4x - 4
<=> 7 \(\ge\)x
<=> x \(\le7\)
Vậy x \(\le\)7 là nghiệm của bất phương trình
Biểu diễn
-----------------------|-----------]|-/-/-/-/-/-/>
0 7
B
B