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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: \(\begin{cases}x\ne0\\x+5\ne0\end{cases}\Leftrightarrow\begin{cases}x\ne0\\x\ne-5\end{cases}\)
b)\(A=\frac{x^2+2x}{2x+10}+\frac{x+5}{x}-\frac{50-5x}{2x\left(x+5\right)}=\frac{x^2+2x}{2.\left(x+5\right)}+\frac{x+5}{x}-\frac{50-5x}{2x\left(x+5\right)}\)
\(=\frac{x^2+2x}{2x.\left(x+5\right)}+\frac{2\left(x+5\right)^2}{2x\left(x+5\right)}-\frac{50-5x}{2x\left(x+5\right)}\)
\(=\frac{x^2+2x+2x^2+20x+50-50+5x}{2x\left(x+5\right)}=\frac{3x^2+27x}{2x\left(x+5\right)}=\frac{3x.\left(x+9\right)}{2x\left(x+5\right)}=\frac{3x+27}{2x+10}\)
c)Để A=1 thì: \(\frac{3x+27}{2x+10}=1\Rightarrow3x+27=2x+10\Leftrightarrow x=-17\)(nhận)
Vậy x=-17 thì A=1
Câu 3 :
\(a,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\) ĐKXđ : \(x\ne\pm1\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{5\left(x-1\right)}\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{10}{x+1}\)
\(B=\left(\frac{x}{3x-9}+\frac{2x-3}{3x-x^2}\right).\frac{3x^2-9x}{x^2-6x+9}.\)
ĐKXđ : \(x\ne0;x\ne3\)
\(B=\left(\frac{x}{3\left(x-3\right)}+\frac{2x-3}{x\left(3-x\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\left(\frac{x^2}{3x\left(x-3\right)}+\frac{9-6x}{3x\left(x-3\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\frac{x^2-6x+9}{3x\left(x-3\right)}.\frac{3x\left(x-3\right)}{x^2-6x+9}=1\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)
\(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}\)
\(\Leftrightarrow A=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2\left(x+2\right)}{x-3}\)
\(\Leftrightarrow A=\frac{2x-9-\left(x-3\right)\left(x+3\right)+2\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{\left(x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{x+4}{x-3}\)
b) Để \(A\inℤ\)
\(\Leftrightarrow\frac{x+4}{x-3}\inℤ\)
\(\Leftrightarrow1+\frac{7}{x-3}\inℤ\)
\(\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Leftrightarrow x\in\left\{2;4;-4;10\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{2;4;-4;10\right\}\)
c) Để \(A=\frac{3}{5}\)
\(\Leftrightarrow\frac{x+4}{x-3}=\frac{3}{5}\)
\(\Leftrightarrow5x+20=3x-9\)
\(\Leftrightarrow2x+29=0\)
\(\Leftrightarrow x=-\frac{29}{2}\)
d) Để \(A< 0\)
\(\Leftrightarrow\frac{x+4}{x-3}< 0\)
\(\Leftrightarrow1+\frac{7}{x-3}< 0\)
\(\Leftrightarrow\frac{-7}{x-3}< 1\)
\(\Leftrightarrow-7< x-3\)
\(\Leftrightarrow x>-4\)
e) Để \(A>0\)
\(\Leftrightarrow\frac{x+4}{x-3}>0\)
\(\Leftrightarrow1+\frac{7}{x-3}>0\)
\(\Leftrightarrow\frac{-7}{x-3}>1\)
\(\Leftrightarrow-7>x-3\)
\(\Leftrightarrow x< -4\)
\(a)A=(\frac{x}{(x+6)(x+6)}-\frac{x-6}{x(x+6)})\cdot\frac{x(x+6)}{2x-6}+\frac{x}{x-6}\)
\(A=\frac{x^2-(x-6)^2}{x(x+6)(x-6)}\cdot\frac{x(x+6)}{2x-6}-\frac{x}{x-6}=\frac{(x-x+6)(x+x-6)}{(x-6)(2x-6)}-\frac{x}{x-6}\)
\(=\frac{6(2x-6)}{(x-6)(2x-6)}-\frac{x}{x-6}=\frac{6}{(x-6)}-\frac{x}{x-6}\cdot\frac{6-x}{x-6}=-1\)
\(b)\text{A luôn = -1 với mọi x}\)
a) ĐKXĐ: x khác +-1
b) \(\frac{x+1}{x-1}+\frac{x-2}{x+1}-\frac{2x^2+x+5}{x^2-1}\)
\(=\frac{x+1}{x-1}+\frac{x-2}{x+1}-\frac{2x^2+x+5}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x^2+x+5}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{\left(x+1\right)^2+\left(x-2\right)\left(x-1\right)-\left(2x^2+x+5\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=-\frac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=-\frac{2}{x-1}\)
a. tìm điều kiện xác định của P
ĐKXĐ: \(x\ne0;x\ne\pm1\)
\(P=\left(\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{2\left(x+1\right)}\right):\frac{x+1}{2x}\)
\(P=\frac{4x+\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\times\frac{2x}{x+1}\)
\(P=\frac{4x+x^2-2x+1}{2\left(x-1\right)\left(x+1\right)}\times\frac{2x}{x+1}\)
\(P=\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\times\frac{x}{x+1}\)
\(P=\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\times\frac{x}{x+1}\)
\(P=\frac{x}{x-1}\)
b. tìm x
Với P = 2 ta có:
\(\frac{x}{x-1}=2\)
=> x = 2(x-1)
=> x = 2x -2
=> 2x - x = 2
=> x = 2
Vậy với x = 2 thì P = 2
c. với 0 < x < 1 . hãy so sánh P với |P|
\(P=\frac{x}{x-1}\)
Với 0< x < 1 thì x -1 <0 ; x>0 => P <0
Suy ra P< |P| ( vì |P| >0)
Câu hỏi tương tự Đọc thêm Báo cáoToán lớp 8A. DE P XAC DINH
<=>X^2-1 KHÁC 0<=>X KHAC -1 VÀ X KHÁC 1
<=>2X+2 KHAC 0 <=>X KHAC-1
<=>2X KHAC 0 <=>X KHAC 0
=> X KHAC O HOAC X KHAC +-1
TACO:( 2X / X^2-1 +X-1/ 2X+2 ) : X+1 / 2X
=[2X . 2 / (X+1)(X-1). 2 + (X-1)(X-1) / 2(X+1)(X-1) ] : X+1/2X
=[4X+(X-1)^2] / 2(X+1)(X-1) :X+1 / 2X
=(4X+X^2-2X+1) / 2(X+1)(X-1) : X+1/2X
=X^2+2X+1 / 2(X-1)(X+1) : X+1 / 2X
=(X+1)^2 / 2(X-1)(X+1) : X+1/2X
=(X+1) / 2(X-1) . 2X/X+1
=X/X-1
B. DE P=2
<=>X/X-1=2
<=>X=2(X-1)=2X-2=X+X-2
TA CÓ: X +X-2 = X+0
=>X-2=0
=>X=2
C .VI 0<X<1
=>X / X-1 = |X/X-1|
=>P=|P|
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) Vì \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\) nên điều kiện xác định của A là \(x^3-1\ne0\)
=> \(x\ne1\)
b) Rút gọn A:
\(A=\frac{5x+1+\left(1-2x\right)\left(x-1\right)+2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{5x+1+x-1-2x^2+2x+2x^2+2x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{10x+2}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{2\left(5x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
c) Vì \(x^2+x+1=\left(x^2+2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
Nên để A > 0 thì \(5x+1\) và \(x-1\) phải cùng dấu.
TH1: \(\hept{\begin{cases}5x+1>0\\x-1>0\end{cases}}\) => \(x>1\)
TH2: \(\hept{\begin{cases}5x+1< 0\\x-1< 0\end{cases}}\) => \(x< -\frac{1}{5}\)
Vậy để A > 0 thì \(x>1\) hoặc \(x< -\frac{1}{5}\)