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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) 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
a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)
\(A=\frac{x-1}{x+2}-\frac{x+2}{x-2}-\frac{x^2+12}{4-x^2}=\frac{\left(x-1\right).\left(x-2\right)}{x^2-4}-\frac{\left(x+2\right)^2}{x^2-4}+\frac{x^2+12}{x^2-4}\)
\(=\frac{x^2-3x+2}{x^2-4}-\frac{x^2+4x+4}{x^2-4}+\frac{x^2+12}{x^2-4}=\frac{x^2-7x+10}{x^2-4}=\frac{\left(x-2\right).\left(x-5\right)}{\left(x-2\right).\left(x+2\right)}=\frac{x-5}{x+2}\)
\(A=\frac{x-1}{x+2}-\frac{x+2}{x-2}-\)\(\frac{x^2+12}{4-x^2}\)\(ĐKXĐ\): \(x\ne\pm2\)
\(=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)\(-\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)\(+\frac{x^2+12}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2-2x-x+2-x^2-4x-4+x^2+12}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-7x+10}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-2x-5x+10}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x\left(x-2\right)-5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{\left(x-5\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-5}{x+2}\)
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) :
\(\frac{3}{x-3}=-2\Rightarrow x=1,5\)
\(\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right)\div\left(1-\frac{x}{x+2}\right)\)
\(=\left(\frac{x}{\left(x+2\right)\left(x-2\right)}+\frac{1\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right)\div\left(1-\frac{x}{x+2}\right)\)
\(=\frac{x+\left(x-2\right)-2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\div\left(1-\frac{x}{x+2}\right)\)
\(=\frac{x-2}{1}\div\left(1-\frac{x}{x+2}\right)\)
\(=\frac{x-2}{1}\div\left(\frac{x+2}{x+2}-\frac{x}{x+2}\right)\)
\(=\frac{x-2}{1}\div\left(\frac{x+2-x}{x+2}\right)=\frac{x-2}{1}\div\frac{2}{x+2}\)
\(=\frac{x-2}{1}\times\frac{x+2}{2}=\frac{\left(x-2\right)\left(x+2\right)}{1.2}=\frac{x^2-2^2}{2}=\frac{x^2-2}{1}=x^2-2\)
(Sai thì thôi)
#Học tốt!!!
~NTTH~
Mơn bạn nha☺☺☺