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a) thay x = -3 vào biểu thức, ta có:
\(A=\frac{\left(-3\right)^2+2.\left(-3\right)}{\left(-3\right)+1}=-\frac{3}{2}\)
b) M = A.B
\(M=\left(-\frac{3}{2}\right)\left(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{4-x^2}\right)\)
\(M=-\frac{3\left(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{4-x^2}\right)}{2}\)
\(M=-\frac{3.\frac{8}{x+2}}{2}\)
\(M=-\frac{\frac{24}{x+2}}{2}\)
\(M=-\frac{24}{2\left(x+2\right)}\)
\(M=-\frac{12}{x+2}\)
2) a) Ta có B = \(\frac{x+2}{x-2}-\frac{x-2}{x+2}-\frac{16}{4-x^2}=\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}=\frac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{8}{x-2}\)
Khi |x - 1| = 2
=> \(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Khi x = 3 (thỏa mãn) => A = \(\frac{3^2-2.3}{3+1}=\frac{3}{4}\)
Khi x = - 1 (không thỏa mãn) => Không tìm được A
b) Ta có P = \(A.B=\frac{x^2-2x}{x+1}.\frac{8}{x-2}=\frac{8x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{8x}{x+1}\)
Đẻ P < 8
=> \(\frac{8x}{x+1}< 8\Leftrightarrow\frac{x}{x+1}< 1\)
=> \(\orbr{\begin{cases}x< x+1\left(x>-1\right)\\x>x+1\left(x< -1\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}0x< 1\left(tm\right)\\0x>1\left(\text{loại}\right)\end{cases}}\)
Vậy x > - 1 thì P < 8
a) Ta có: A = \(\left(\frac{x}{x-1}+\frac{x}{x^2-1}\right):\left(\frac{2}{x^2}-\frac{2-x^2}{x^3+x^2}\right)\)
A = \(\left(\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{2\left(x+1\right)}{x^2\left(x+1\right)}-\frac{2-x^2}{x^2\left(x+1\right)}\right)\)
A = \(\left(\frac{x^2+x+x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{2x+2-2+x^2}{x^2\left(x+1\right)}\right)\)
A = \(\left(\frac{x^2+2x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x^2+2x}{x^2\left(x+1\right)}\right)\)
A = \(\frac{x\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x^2\left(x+1\right)}{x\left(x+2\right)}\)
A = \(\frac{x^2}{x+1}\)
b) ĐKXĐ: x \(\ne\)\(\pm\)1; x \(\ne\)0; x \(\ne\)-2
Ta có: A = 4
<=> \(\frac{x^2}{x+1}=4\)
<=> x2 = 4(x + 1)
<=> x2 - 4x - 4 = 0
<=>(x2 - 4x + 4) - 8 = 0
<=> (x - 2)2 = 8
<=> \(\orbr{\begin{cases}x-2=\sqrt{8}\\x-2=-\sqrt{8}\end{cases}}\)
<=> \(\orbr{\begin{cases}x=2\sqrt{2}+2\\x=2-2\sqrt{2}\end{cases}}\)(tm)
Vậy ...
c) Ta có: A < 0
<=> \(\frac{x^2}{x+1}< 0\)
Do x2 \(\ge\)0 => x + 1 < 0
=> x < -1
Vậy để A < 0 thì x < -1 và x khác -2
1. Ta có:
\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)
\(=\frac{2}{x}-\frac{1}{x+2014}\)
\(=\frac{2\left(x+2014\right)}{x\left(x+2014\right)}-\frac{x}{x\left(x+2014\right)}\)
\(=\frac{2x+4028-x}{x\left(x+2014\right)}=\frac{x+4028}{x\left(x+2014\right)}\)
2a) ĐKXĐ: x \(\ne\)1 và x \(\ne\)-1
b) Ta có: A = \(\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
A = \(\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)
A = \(x-1+x+1-3\)
A = \(2x-3\)
c) Với x = 3 => A = 2.3 - 3 = 3
c) Ta có: A = -2
=> 2x - 3 = -2
=> 2x = -2 + 3 = 1
=> x= 1/2
a. A=\(1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)
\(=1+\left(\frac{x+1+x+1-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right).\frac{x\left(x^2-x+1\right)}{x^2\left(x-2\right)}\)
\(=1+\frac{-2x^2+4x}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1+\frac{-2x\left(x-2\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1-\frac{2}{x+1}=\frac{x-1}{x+1}\)
b.\(\left|x-\frac{3}{4}\right|=\frac{5}{4}\Rightarrow\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}\)
Với \(x=2\Rightarrow A=\frac{2-1}{2+1}=\frac{1}{3}\)
Với \(x=-\frac{1}{2}\Rightarrow A=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3\)
Câu a bài 2 hình như sai đề thf phải :(( gửi bạn câu b ở đây nhé
\(\frac{x-18}{3-2x}+\frac{11x}{3-2x}=\frac{12x-18}{3-2x}=\frac{6\left(2x-3\right)}{3-2x}=-6\)