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Dài quá trôi hết đề khỏi màn hình: nhìn thấy câu nào giải cấu ấy
Bài 4:
\(A=\frac{\left(x-1\right)+\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2}{\left(x+1\right)\left(x-1\right)}=\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
a) DK x khác +-1
b) \(dk\left(a\right)\Rightarrow A=\frac{2}{\left(x+1\right)}\)
c) x+1 phải thuộc Ước của 2=> x=(-3,-2,0))
1. a) Biểu thức a có nghĩa \(\Leftrightarrow\hept{\begin{cases}x+2\ne0\\x^2-4\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x+2\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)
Vậy vs \(x\ne2,x\ne-2\) thì bt a có nghĩa
b) \(A=\frac{x}{x+2}+\frac{4-2x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{4-2x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-2x+4-2x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-4x+4}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-2}{x+2}\)
c) \(A=0\Leftrightarrow\frac{x-2}{x+2}=0\)
\(\Leftrightarrow x-2=\left(x+2\right).0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)(ko thỏa mãn điều kiện )
=> ko có gía trị nào của x để A=0
Câu 1 :
a) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)
b) Để \(P=1\Leftrightarrow\frac{4x^2+4x}{\left(x+1\right)\left(2x-6\right)}=1\)
\(\Leftrightarrow\frac{4x^2+4x-\left(x+1\right)\left(2x-6\right)}{\left(x+1\right)\left(2x-6\right)}=0\)
\(\Rightarrow4x^2+4x-2x^2+4x+6=0\)
\(\Leftrightarrow2x^2+8x+6=0\)
\(\Leftrightarrow x^2+4x+4-1=0\)
\(\Leftrightarrow\left(x+2-1\right)\left(x+2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=-3\left(TMĐKXĐ\right)\end{cases}}\)
Vậy : \(x=-3\) thì P = 1.
\(a,ĐKXĐ\hept{\begin{cases}x-3\ne0\\x+3\ne0\end{cases}\Leftrightarrow x\ne\pm3}\)
Ta có: \(M=\frac{3}{x-3}-\frac{6x}{9-x^2}+\frac{x}{x+3}\)
\(=\frac{3}{x-3}+\frac{6x}{x^2-9}+\frac{x}{x+3}\)
\(=\frac{3\left(x+3\right)+6x+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x+3}{x-3}\)
\(b,x=\frac{1}{2}\Rightarrow M=\frac{\frac{1}{2}+3}{\frac{1}{2}-3}=-\frac{7}{5}\)
sau khi rút gọn ta được \(P=\frac{x-4}{x-2}\left(x\ne-3;x\ne2;x\ne-2\right)\)
d,ta có \(P=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\left(x\ne-2;x\ne-3;x\ne2\right)\)
để P nguyên mà x nguyên \(\Leftrightarrow x-2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
ta có bảng:
| x-2 | 1 | -1 | 2 | -2 |
| x | 3(tm) | 1(tm) | 4(tm) | 0(tm) |
vậy \(P\in Z\Leftrightarrow x\in\left\{3;1;4;0\right\}\)
e,x2-9=0
\(\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=3\left(tm\right)\\x=-3\left(kotm\right)\end{cases}}\)
thay x=3 vào P đã rút gọn ta có \(P=\frac{3-4}{3-2}=-1\)
vậy với x=3 thì p có giá trị bằng -1
a) ĐKXĐ: x \(\ne\pm3\)
b) = \(\frac{3\left(x-3\right)+x+3+18}{\left(x-3\right)\left(x+3\right)}\)
= \(\frac{4x+12}{\left(x-3\right)\left(x+3\right)}\)= \(\frac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{4}{x-3}\)
c) P = 4 hay \(\frac{4}{x-3}=4\)=> x - 3 = 1 <=> x = 4 (TM)
Vậy ...
a) \(ĐKXĐ:x\ne\pm3\)
b) \(A=\left(\frac{x}{x+3}+\frac{3-x}{x+3}\cdot\frac{x^2+3x+9}{x^2-9}\right):\frac{3}{x+3}\)
\(\Leftrightarrow A=\left(\frac{x}{x+3}-\frac{\left(x-3\right)\left(x^2+3x+9\right)}{\left(x+3\right)\left(x^2-9\right)}\right):\frac{3}{x+3}\)
\(\Leftrightarrow A=\left(\frac{x}{x+3}-\frac{x^2+3x+9}{\left(x+3\right)^2}\right):\frac{3}{x+3}\)
\(\Leftrightarrow A=\frac{x^2+3x-x^2-3x-9}{\left(x+3\right)^2}:\frac{3}{x+3}\)
\(\Leftrightarrow A=\frac{-9\left(x+3\right)}{3\left(x+3\right)^2}\)
\(\Leftrightarrow A=\frac{-3}{x+3}\)
c) Tại \(x=-\frac{1}{2}\)
\(\Leftrightarrow A=\frac{-3}{-\frac{1}{2}+3}\)
\(\Leftrightarrow A=\frac{-6}{5}\)
d) Để \(A>0\)
\(\Leftrightarrow\frac{-3}{x+3}>0\)
\(\Leftrightarrow x+3< 0\)(Vì -3 < 0)
\(\Leftrightarrow x< -3\)
e) +) Với \(A>\frac{-1}{2}\)
\(\Leftrightarrow\frac{-3}{x+3}>-\frac{1}{2}\)
\(\Leftrightarrow-6>-x-3\)
\(\Leftrightarrow x>3\)(tm)
+) Với \(A< -\frac{1}{2}\)
\(\Leftrightarrow\frac{-3}{x+3}< -\frac{1}{2}\)
\(\Leftrightarrow-6< -x-3\)
\(\Leftrightarrow x< 3\)(chú ý : \(x\ne-3\))
+) Với \(A=-\frac{1}{2}\)
\(\Leftrightarrow-\frac{3}{x+3}=-\frac{1}{2}\)
\(\Leftrightarrow x+3=6\)
\(\Leftrightarrow x=3\)(ktm)
Vậy \(\orbr{\begin{cases}A>-\frac{1}{2}\\A< -\frac{1}{2}\end{cases}}\)
\(\text{a, ĐKXĐ: }\hept{\begin{cases}x+3\ne0\\x-3\ne0\\3x^2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne\mp3\\x\ne0\end{cases}}\)
\(A=\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\left[\frac{\left(3-x\right)\left(x+3\right)^2}{\left(x+3\right)\left(x+3\right)\left(x-3\right)}+\frac{x}{x+3}\right]\cdot\frac{x+3}{3x^2}\)
\(=\frac{x-x-3}{x+3}\cdot\frac{x+3}{3x^2}\)
\(=-\frac{1}{x^2}\)
b, với x=\(-\frac{1}{2}\)ta có:
\(A=-\frac{1}{\left(-\frac{1}{2}\right)^2}=-4\)
c, Để A<0 thì \(-\frac{1}{x^2}< 0\text{ mà }x^2>0\left(\text{vì x khác 0 ĐKXĐ}\right)\)
Với x khác 0 thì thỏa mãn!
a) ĐKXĐ: \(x\ne\pm3\)
\(A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\left(\frac{3-x}{x+3}.\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\left(\frac{3-x}{x-3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\frac{\left(3-x\right)\left(x+3\right)+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{3x^2}\)
\(=\frac{3\left(3-x\right)}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{3x^2}\)
\(=-\frac{1}{x^2}\)
Bạn viết biểu thức A ra đi rồi bọn mình mới làm được chứ -.-
Đk : \(x\ne\pm3\)
Để B>A
\(\Leftrightarrow\frac{3}{x+3}>4\)
Rõ ràng: \(x+3>0\)
\(\Rightarrow\frac{3}{x+3}>4\)
\(\Leftrightarrow3>4\left(x+3\right)\)
\(\Leftrightarrow3>4x+12\)
\(\Leftrightarrow-9>4x\)
\(\Leftrightarrow x< \frac{-9}{4}\)
KL: \(x\in Z,x< \frac{-9}{4},x\ne\pm3\)
a) ĐKXĐ: x - 3 \(\ne\)0 x \(\ne\)3
9 - x2 \(\ne\)0 <=> x \(\ne\)\(\pm\)3
x + 3 \(\ne\)0 x \(\ne\)-3
\(\frac{6x-12}{2x^2-18}\) \(\ne\)0 \(6x-12\ne0\) và \(2x^2-18\ne0\)
x \(\ne\)\(\pm\)3
<=> \(x\ne2\) và x \(\ne\)\(\pm\)3
<=> x \(\ne\)\(\pm\)3 và x \(\ne\)2
Ta có: B = \(\left(\frac{x+3}{x-3}+\frac{2x^2-6}{9-x^2}+\frac{x}{x+3}\right):\frac{6x-12}{2x^2-18}\)
B = \(\left(\frac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{2x^2-6}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{6\left(x-2\right)}{2\left(x^2-9\right)}\)
B = \(\left(\frac{x^2+6x+9-2x^2+6+x^2-3x}{\left(x-3\right)\left(x+3\right)}\right):\frac{3\left(x-2\right)}{\left(x-3\right)\left(x+3\right)}\)
B = \(\frac{3x+15}{\left(x+3\right)\left(x-3\right)}\cdot\frac{\left(x-3\right)\left(x+3\right)}{3\left(x-2\right)}\)
B = \(\frac{3\left(x+5\right)}{3\left(x-2\right)}\)
B = \(\frac{x+5}{x-2}\)
b) (sai đề)
c) Ta có: B = \(\frac{x+5}{x-2}=\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)
Để B \(\in\)Z <=> 7 \(⋮\)x - 2 <=> x - 2 \(\in\)Ư(7) = {1; -1; 7; -7}
Lập bảng:
| x - 2 | 1 | -1 | 7 | -7 |
| x | 3 (ktm) | 1 | 9 | -5 |
Vậy ...
a) \(\text{ĐKXĐ:}\hept{\begin{cases}x\ne\pm3\\x\ne2\end{cases}}\)
\(B=\left(\frac{x+3}{x-3}+\frac{2x^2-6}{9-x^2}+\frac{x}{x+3}\right):\frac{6x-12}{2x^2-18}\)
\(B=\left[\frac{x+3}{x-3}+\frac{2x^2-6}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right].\frac{2\left(x^2-9\right)}{6\left(x-2\right)}\)
\(B=\left[\frac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{2x^2-6}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right]\)
\(B=\left[\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}-\frac{2x^2-6}{\left(x-3\right)\left(x+3\right)}+\frac{x^2-3x}{\left(x-3\right)\left(x+3\right)}\right].\frac{2\left(x^2-9\right)}{6\left(x-2\right)}\)
\(B=\frac{x^2+6x+9-\left(2x^2-6\right)+x^2-3}{\left(x-3\right)\left(x+3\right)}.\frac{2\left(x^2-9\right)}{6\left(x-2\right)}\)
\(B=\frac{3\left(x+5\right)}{\left(x-3\right)\left(x+3\right)}.\frac{2\left(x-3\right)\left(x+3\right)}{6\left(x-2\right)}\)
\(B=\frac{x+5}{x-2}\)
b) Ta có: \(\frac{x+5}{x-2}=1+\frac{7}{x-2}\)
Để B nguyên thì: \(7⋮x-2\)
\(\Rightarrow x-2\inƯ\left(7\right)\)
\(\RightarrowƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
Ta có bảng:
| x - 2 | -1 | 1 | -7 | 7 |
| x | 1 | 3 (loại) | -5 | 9 |
Vậy: \(x\in\left\{1;-5;9\right\}\)
b, \(B=\frac{\frac{x}{x+3}-\frac{9}{x^2+6x+9}}{\frac{3}{x+3}}=\frac{\frac{x}{x+3}-\frac{3^2}{x^2+2\cdot3\cdot x+3^2}}{\frac{3}{x+3}}\)
\(=\frac{\frac{x}{x+3}-\left(\frac{3}{x+3}\right)^2}{\frac{3}{x+3}}=1-\frac{3}{x+3}\)
a, Vậy điều kiện là \(x\ne3\)
c, \(B=\frac{1}{3}\Leftrightarrow1-\frac{3}{x+3}=\frac{1}{3}\)
\(\Rightarrow\frac{3}{x+3}=\frac{2}{3}\Leftrightarrow x=\frac{3}{2}\)