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\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
\(=\frac{1}{x}\)
ta có: \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
=\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
= \(\frac{1}{x}\)
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)=1/3
<=>1/x-1/x+1+1/x+1-1/x+2+1/x+2-1/x+3+1/x+3-1/x+4=1/3
<=>1/x-1/x+4=1/3
<=>x+4/x(x+4)-x/x(x+4) ( quy dong mau ) =1/3
<=>4/x(x+4)=1/3
<=> 4.3=x(x+4) ( nhan cheo )
<=> x(x+4)=12
<=> x^2+4x-12=0
<=>x^2-2x+6x-12=0
<=>x(x-2) + 6(x-2) =0
<=> (x-2)(x+6)=0
<=> x-2 =0 hoac x +6=0
<=>x=2 hoac x= -6
Vay x thuoc ( 2,-6 )
K mk nha !!
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x\text{+}2\right)}\text{+}\frac{1}{\left(x\text{+}2\right)\left(x\text{+}3\right)}+\frac{1}{\left(x\text{+}3\right)\left(x\text{+}4\right)}=\frac{1}{3}\)
\(\Rightarrow\frac{1}{x}-\frac{1}{x\text{+}1}\text{+}\frac{1}{x\text{+}1}-\frac{1}{x\text{+}2}\text{+}.....\text{+}\frac{1}{x\text{+}3}-\frac{1}{x\text{+}4}=\frac{1}{3}\)
\(\Rightarrow\)\(\frac{1}{x}-\frac{1}{x\text{+}4}=\frac{1}{3}\)
\(\Rightarrow\frac{x\text{+}4}{x\left(x\text{+}4\right)}-\frac{x}{x\left(x\text{+}4\right)}=\frac{1}{3}\)
\(\Rightarrow\frac{4}{x\left(x\text{+}4\right)}=\frac{1}{3}\)
\(\Rightarrow\frac{4}{x\left(x\text{+}4\right)}=\frac{4}{12}\)
\(\Rightarrow x\left(x\text{+}4\right)=12\)
mà x và x+4 cách nhau 4 đơn vị \(\Rightarrow x=2\)và x+4\(=\)6
Vậy \(x=2\)
d) \(\frac{1}{\left(x+2\right)}+\frac{1}{\left(x+3\right)}=\frac{1}{\left(x+2\right)\left(x+3\right)}\)
ĐKXĐ : \(x\ne-2;x\ne-3\)
\(\Leftrightarrow x+3+x+2=1\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-2\) (không nhận)
Vậy : \(S=\varnothing\)
Giai phương trình sau :
a) \(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}\)
ĐKXĐ : \(x\ne1;x\ne-5\)
Với điều kiện trên ta có :
\(\Leftrightarrow\)\(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{-3}{x-1}=\frac{5}{x+5}\)
\(\Leftrightarrow10-3\left(x+5\right)=5\left(x-1\right)\)
\(\Leftrightarrow10-3x-15=5x-5\)
\(\Leftrightarrow-8x=0\)
\(\Leftrightarrow x=0\) (nhận)
Vậy : \(S=\left\{0\right\}\)
quá dễ tách ra thành 1\x-1\x+1+1\x+1-1\x+2+1\x+2-1\x+3+1\x+3-1\x+4+...+1\x+5-1\x+6
=1\x-1\x+6
=6\x(x+6)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}\)
\(=\frac{1}{x}-\frac{1}{x+6}\)\(=\frac{6}{x\left(x+6\right)}\)
ta có
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\)
\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+....+\frac{1}{x+6}\)
\(=\frac{1}{x}-\frac{1}{x+6}\)
ĐK: \(x\in R\backslash\left\{-4,-3,-2,-1\right\}\)
PT ban đầu
\(\Leftrightarrow\frac{x+2-x-1}{\left(x+1\right)\left(x+2\right)}+\frac{x+3-x-2}{\left(x+2\right)\left(x+3\right)}+\frac{x+4-x-3}{\left(x+3\right)\left(x+4\right)}+\frac{x+5-x-4}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+5}=403\\ \Leftrightarrow x+5=\frac{1}{403}\Leftrightarrow x=\frac{-2014}{403}\)
Chúc bạn học tốt nha
.
Sr bạn nha, nhưng điều kiện là \(x\in R\backslash\left\{-5,-4,-3,-2,-1\right\}\). (Xét thiếu :>)
Chúc bạn học tốt nha.
a,\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
b,Áp dụng câu a:
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
\(=\frac{1}{x}\)
a)
\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
b) S =\(\frac{1}{x}-\frac{1}{x+5}+\frac{1}{x+5}=\frac{1}{x}\)
\(\left(\frac{x+1}{39}+1\right)+\left(\frac{x+2}{38}+1\right)=\left(\frac{x+3}{37}+1\right)+\left(\frac{x+4}{36}+1\right)\)
\(\Leftrightarrow\frac{x+40}{39}+\frac{x+40}{38}-\frac{x+40}{37}-\frac{x+40}{36}=0\)
\(\Leftrightarrow\left(x+40\right)\left(\frac{1}{39}+\frac{1}{38}-\frac{1}{37}-\frac{1}{36}\right)=0\)
<=> x+40=0 (vì \(\frac{1}{39}+\frac{1}{38}-\frac{1}{37}-\frac{1}{36}\ne\)0)
<=> x=-40
Vậy x=-40