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M = 1/(x+1).(x+2) + 1/(x+2).(x+3) + 1/(x+3).(x+4) + 1/(x+4).(x+5) + 1/x+5
= 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 + 1/x+3 - 1/x+4 + 1/x+4 - 1/x+5 + 1/x+5 = 1/x+1
k mk nha
a, ĐK : \(x\ne1;2;3;4;5\)
b, \(\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x-1}+\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}\)
\(=\dfrac{1}{x}-\dfrac{1}{x-5}=\dfrac{x-5-x}{x\left(x-5\right)}=\dfrac{-5}{x\left(x-5\right)}\)
a: ĐKXĐ: \(x\notin\left\{0;1;2;3;4;5\right\}\)
b: \(P=\dfrac{1}{\left(x-1\right)\cdot x}+\dfrac{1}{\left(x-2\right)\left(x-1\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}-\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}\)
\(=\dfrac{1}{x-5}-\dfrac{1}{x}=\dfrac{x-x+5}{x\left(x-5\right)}=\dfrac{5}{x\left(x-5\right)}\)
a: ĐKXĐ: \(x\notin\left\{0;1;2;3;4;5\right\}\)
b: \(P=\dfrac{1}{x^2-x}+\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}\)
\(=\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\dfrac{-1}{x}+\dfrac{1}{x-1}-\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}-\dfrac{1}{x-4}+\dfrac{1}{x-5}\)
\(=\dfrac{1}{x-5}-\dfrac{1}{x}\)
\(=\dfrac{x-\left(x-5\right)}{x\left(x-5\right)}=\dfrac{5}{x\left(x-5\right)}\)
c: \(x^3-x^2+2=0\)
=>\(x^3+x^2-2x^2+2=0\)
=>\(x^2\cdot\left(x+1\right)-2\left(x-1\right)\left(x+1\right)=0\)
=>\(\left(x+1\right)\left(x^2-2x+2\right)=0\)
=>x+1=0
=>x=-1
Khi x=-1 thì \(P=\dfrac{5}{\left(-1\right)\left(-1-5\right)}=\dfrac{5}{\left(-1\right)\cdot\left(-6\right)}=\dfrac{5}{6}\)
a: \(M=\dfrac{2x^2-10x-x^2+x+30-x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{x+5}\)
b: Để M là số nguyên thì \(x+5\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)
hay \(x\in\left\{-4;-6;-3;-7;0;-10;-15\right\}\)
Bài 1:
a.\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2\left(x+y\right)\)
b.\(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=4x^2\)
Bài 1 :
a) \(ĐKXĐ:x\ne1\)
\(A=\left(\frac{3}{x^2-1}+\frac{1}{x+1}\right):\frac{1}{x+1}\)
\(\Leftrightarrow A=\frac{3+x-1}{\left(x-1\right)\left(x+1\right)}\cdot\left(x+1\right)\)
\(\Leftrightarrow A=\frac{x+2}{x-1}\)
b) Thay x = \(\frac{2}{5}\)vào A ta được :
\(A=\frac{\frac{2}{5}+2}{\frac{2}{5}-1}=\frac{\frac{12}{5}}{-\frac{3}{5}}=-4\)
c) Để \(A=\frac{5}{4}\)
\(\Leftrightarrow\frac{x+2}{x-1}=\frac{5}{4}\)
\(\Leftrightarrow4x+8=5x-5\)
\(\Leftrightarrow x=13\)
d) Để \(A>\frac{1}{2}\)
\(\Leftrightarrow\frac{x+2}{x-1}>\frac{1}{2}\)
\(\Leftrightarrow\frac{x+2}{x-1}-\frac{1}{2}>0\)
\(\Leftrightarrow2x+4-x+1>0\)
\(\Leftrightarrow x+5>0\)
\(\Leftrightarrow x>-5\)
Bài 2 :
a) \(ĐKXĐ:\hept{\begin{cases}x\ne-1\\x\ne0\end{cases}}\)
\(A=\frac{x^2}{x^2+x}-\frac{1-x}{x+1}\)
\(A=\frac{x}{x+1}+\frac{x-1}{x+1}\)
\(\Leftrightarrow A=\frac{2x-1}{x+1}\)
b) Để \(A=1\)
\(\Leftrightarrow\frac{2x-1}{x+1}=1\)
\(\Leftrightarrow2x-1=x+1\)
\(\Leftrightarrow x=2\)
b) Để \(A< 2\)
\(\Leftrightarrow\frac{2x-1}{x+1}< 2\)
\(\Leftrightarrow\frac{2x-1}{x+1}-2< 0\)
\(\Leftrightarrow2x-1-2x-1< 0\)
\(\Leftrightarrow-2< 0\)(luôn đúng)
Vậy A < 2 <=> mọi x
Áp dụng kết quả của phần a, ta có: