Bài 4:

=>(x-5)*3/10=1/5x+5

=>3/10x-3/2=1/5x+5

=>1/10x=5+3/2=6,5

=>0,1x=6,5

=>x=65

Lời giải:

Ta có: \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}\)

\(=\dfrac{\left(n+a\right)-n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}\)

Vậy, \(\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\)

Bài 1:

a, \(\left(x-2\right)^2=9\)

\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)

b, \(\left(3x-1\right)^3=-8\)

\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)

\(\Rightarrow x=-\dfrac{1}{3}\)

c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)

\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)

d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)

\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)

Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)

e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)

Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)

Câu 2:

a: \(\Leftrightarrow12x-60=7x-5\)

=>5x=55

=>x=11

b: \(\Leftrightarrow\left(2x-3\right)^{2010}\left[\left(2x-3\right)^2-1\right]=0\)

=>(2x-3)(2x-2)(2x-4)=0

hay \(x\in\left\{\dfrac{3}{2};1;2\right\}\)

n lớn hơn 2 và ko chia hết cho 3 nên n tồn tại dưới 2 dạng là 3k+1 hoặc 3k+2
Nếu n có dạng 3k + 2
n^2 + 1 = ( 3k + 2 )^2 + 1 = 9k^2 + 12k + 5
n^2 - 1 = 9k^2 + 12k + 3 chia hết cho 3
=> Ko thể đồng thời là số nguyên tố
Nếu n có dạng 3k + 1
n^2 + 1= ( 3k + 1 )^2 + 1 = 9k^2 + 6k + 2
n^2 - 1= ( 3k + 1 )^2 - 1 = 9k^2 + 6k chia hết cho 3
=> Ko thể đồng thời là số nguyên tố
Vậy với n thuộc N , n > 2 và ko chia hết cho 3 thì n^2 + 1 và n^2 - 1 ko thể đồng thời là số nguyên tố

thanks

\(VP=\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}=VT\)

Ta có :

\(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}\)

Vậy \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{a}{n\left(n+a\right)}hay\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\)

a) \(\dfrac{5}{7}-1\dfrac{4}{7}\left(450\%+\dfrac{2}{3}x\right)=\dfrac{-1}{14}\)

\(\dfrac{5}{7}-\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{-1}{14}\)

\(\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{5}{7}+\dfrac{1}{14}\)

\(\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{11}{14}\)

\(\dfrac{9}{2}+\dfrac{2}{3}x=\dfrac{11}{14}:\dfrac{11}{7}=\dfrac{11}{14}.\dfrac{7}{11}\)

\(\dfrac{9}{2}+\dfrac{2}{3}x=\dfrac{1}{2}\)

\(\dfrac{2}{3}x=\dfrac{1}{2}-\dfrac{9}{2}=-4\)

\(x=-4:\dfrac{2}{3}=-4.\dfrac{3}{2}=-6\)

Vậy x = \(-6\)

b) \(100=6.7^{\left|x+2\right|}-194\)

\(100+194=6.7^{\left|x+2\right|}\)

\(294=6.7^{\left|x+2\right|}\)

\(294:6=49=7^{\left|x+2\right|}\)

\(\Rightarrow7^2=7^{\left|x+2\right|}\)

\(\Rightarrow2=\left|x+2\right|\Rightarrow\pm2=x+2\)

+ x + 2 = -2 \(\Rightarrow\) x = - 4

+ x + 2 = 2 \(\Rightarrow\) x = 0

Vậy x = - 4 hoặc 0

a) Để phân số \(\dfrac{3}{n-2}\) là số nguyên thì n - 2 \(⋮\) 3

\(\Rightarrow\) n - 2 \(\in\) Ư(3)

\(\Rightarrow\) n - 2 \(\in\){3; -3; 1;-1}

n \(\in\){5; -1; 3; 2}

c) \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+......+\dfrac{1}{28.29}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{29}-\dfrac{1}{30}\)

\(=\dfrac{1}{3}-\dfrac{1}{30}\)

\(=\dfrac{10}{30}-\dfrac{1}{30}\)

\(=\dfrac{9}{30}\)

=\(\dfrac{3}{10}\)

A=1/2.9+1/9.7+1/7.19+...+1/252.509

=?

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