1+2+3+...........+x=45. tìm x
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a, \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{44}{45}\)
=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{44}{45}\)
=> \(1-\frac{1}{x+1}=\frac{44}{45}\)
=> \(\frac{x}{x+1}=\frac{44}{45}\)
=> x = 44
b, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
.................
\(\frac{1}{45^2}< \frac{1}{44.45}=\frac{1}{44}-\frac{1}{45}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{44}-\frac{1}{45}=1-\frac{1}{45}< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1\)
a) 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=1-1/2+1/2-1/3+1/3-1/4+....+1/x-1/(x+1)=1-1/(x+1)=x/(x+1)=44/45
=> x=44
b/ 1/22 < 1/1.2; 1/32 < 1/2.3; ....; 1/452 < 1/44.45
=> A < 1/1.2+1/2.3+...+1/44.45=1-1/45=44/45 < 1
=> A < 1

\(5x+\left(1+2+3+4\right)=45\)
\(5x+10=45\)
\(5x=35\)
\(x=7\)
Vậy \(x=7\)



b: ĐKXĐ: x∉{-3;1;-1}
Ta có: \(\frac{1}{x^2+2x-3}=\frac{1}{\left(x+1\right)^2}+\frac{1}{48}\)
=>\(\frac{1}{\left(x+3\right)\left(x-1\right)}-\frac{1}{\left(x+1\right)^2}=\frac{1}{48}\)
=>\(\frac{\left(x+1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x+1\right)^2\cdot\left(x-1\right)\left(x+3\right)}=\frac{1}{48}\)
=>\(\frac{x^2+2x+1-\left(x^2+2x-3\right)}{\left(x+1\right)^2\cdot\left(x-1\right)\left(x+3\right)}=\frac{1}{48}\)
=>\(\frac{4}{\left(x+1\right)^2\cdot\left(x-1\right)\left(x+3\right)}=\frac{1}{48}\)
=>\(\left(x+1\right)^2\cdot\left(x^2+2x-3\right)=192\)
=>\(\left(x^2+2x+1\right)\left(x^2+2x-3\right)=192\)
=>\(\left(x^2+2x\right)^2-2\left(x^2+2x\right)-195=0\)
=>\(\left(x^2+2x-15\right)\left(x^2+2x+13\right)=0\)
mà \(x^2+2x+13=\left(x+1\right)^2+12\ge12>0\forall x\)
nên \(x^2+2x-15=0\)
=>(x+5)(x-3)=0
=>\(\left[\begin{array}{l}x+5=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\left(nhận\right)\\ x=3\left(nhận\right)\end{array}\right.\)

x+1 + x+2 + x+3 + x +4 + x+5 = 45
x + x + x + x + x = 45 -1 -2 -3 -4 -5
5x = 30
x =6
9 nha
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