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a: Ta có: \(A=\left(\dfrac{4x}{\left(x-2\right)\left(x+2\right)}+\dfrac{2x-4}{x+2}\right)\cdot\dfrac{x+2}{2x}-\dfrac{2}{x-2}\)
\(=\dfrac{4x+2\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{2x}-\dfrac{2}{x-2}\)
\(=\dfrac{4x+2x^2-8x+8}{x-2}\cdot\dfrac{1}{2x}-\dfrac{2}{x-2}\)
\(=\dfrac{2x^2-12x+8}{2x\left(x-2\right)}-\dfrac{2}{x-2}\)
\(=\dfrac{2x^2-12x+8-4x}{2x\left(x-2\right)}=\dfrac{2x^2-16x+8}{2x\left(x-2\right)}\)
\(=\dfrac{x^2-8x+4}{x\left(x-2\right)}\)
b: Thay x=4 vào A, ta được:
\(A=\dfrac{4^2-8\cdot4+4}{4\cdot\left(4-2\right)}=\dfrac{-12}{4\cdot2}=\dfrac{-12}{8}=-\dfrac{3}{2}\)
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 \(x=-\frac{1}{2}\Rightarrow A=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3\)
a) \(A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x^2-3x+2}\)
\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x^2-x-2x+2}\)
\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x\left(x-1\right)-2\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{\left(x-1\right)\left(x-2\right)}\)
\(\Leftrightarrow A=\frac{\left(4x-1\right)\left(x-1\right)-\left(x-3\right)\left(x-2\right)-2x+4}{\left(x-2\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{4x^2-4x-x+1-x^2+2x+3x-6-2x+4}{\left(x-2\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{3x^2-2x-1}{\left(x-2\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{3x^2-3x+\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}\)\(=\frac{3x\left(x-1\right)+\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}\)\(=\frac{\left(x-1\right)\left(3x+1\right)}{\left(x-2\right)\left(x-1\right)}\)\(=\frac{3x+1}{x-2}\)
b)\(\frac{3x+1}{x-2}=\frac{3x-6+7}{x-2}=\frac{3x-6}{x-2}+\frac{7}{x-2}=3+\frac{7}{x-2}\)
Ta có : \(x-2\inƯ_7\left\{-7;-1;1;7\right\}\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-2=-7\\x-2=-1\\x-2=1\\x-2=7\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}\text{x=-5}\\\text{x=1}\\\text{x=3}\\\text{x}=9\end{array}\right.\)
\(\text{x}=1\) (loại)
Vậy giá trị nguyên tập hợp x là:
x=-5;3;9
a,\(A=\frac{6x+12}{\left(x+2\right)\left(2x-6\right)}=\frac{6\left(x+2\right)}{2\left(x+2\right)\left(x-3\right)}=\frac{3}{x-3}\)
b, Giá trị của x để phân thức có giá trị bằng (-2) :
\(\frac{3}{x-3}=-2\Rightarrow x=1,5\)
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
cái này nó hơi khó 1 tí nên chú ý chút khác lên lever :>
a, \(A=\left(\frac{4x}{x^2+2x}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)ĐK : x khác 0 ; 2 ; -2
\(=\left(\frac{4x}{x\left(x+2\right)}+\frac{2}{x-2}-\frac{6-5x}{\left(2-x\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)
\(=\left(\frac{4x\left(x-2\right)}{MTC}+\frac{2x\left(x+2\right)}{MTC}+\frac{\left(6-5x\right)x}{MTC}\right):\frac{x+1}{x-2}\)
\(=\left(\frac{4x^2-8x+2x^2+4x+6x-5x^2}{MTC}\right):\frac{x+1}{x-2}\)
\(=\frac{x^2+2x}{x\left(x+2\right)\left(x-2\right)}.\frac{x-2}{x+1}=\frac{1}{x+1}\)
b, Ta có : \(x^2-2x=8\Leftrightarrow x^2-2x-8=0\)
\(\left(x-4\right)\left(x+2\right)=0\)<=> \(x=4;-2\)
TH1 : Thay x = 4 ta được : \(\frac{1}{4+1}=\frac{1}{5}\)
TH2 : Thay x = -2 ta được : ( ktmđkxđ )
\(A=\left(\frac{4x}{x^2+2x}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right)\div\frac{x+1}{x-2}\)
a)\(=\left(\frac{4x}{x\left(x+2\right)}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\right)\times\frac{x-2}{x+1}\)
\(=\left(\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\right)\times\frac{x-2}{x+1}\)
\(=\left(\frac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}\right)\times\frac{x-2}{x+1}\)
\(=\frac{x+2}{\left(x-2\right)\left(x+2\right)}\times\frac{x-2}{x+1}\)
\(=\frac{1}{x+1}\)
b) x2 - 2x = 8
<=> x2 - 2x - 8 = 0
<=> x2 - 4x + 2x - 8 = 0
<=> x( x - 4 ) + 2( x - 4 ) = 0
<=> ( x - 4 )( x + 2 ) = 0
<=> x = 4 ( tm ) hoặc x = -2 ( ktm )
Với x = 4 ( tm ) => A = 1/5
Với x = -2 ( ktm ) => A không xác định
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
\(M=\left(-\frac{3}{2}\right)\left(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{4-x^2}\right)\)
\(M=-\frac{3\left(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{4-x^2}\right)}{2}\)
\(M=-\frac{3.\frac{8}{x+2}}{2}\)
\(M=-\frac{\frac{24}{x+2}}{2}\)
\(M=-\frac{24}{2\left(x+2\right)}\)
\(M=-\frac{12}{x+2}\)