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\(2x+7⋮x+2\)

\(\Leftrightarrow x+2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{-1;-3;1;-5\right\}\)

9 tháng 10 2021

\(\Rightarrow2\left(x+2\right)+3⋮x+2\\ \Rightarrow x+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Rightarrow x\in\left\{-5;-3;-1;1\right\}\)

17 tháng 3 2022

\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{13.15}\right).x=\dfrac{-26}{45}\\ \Leftrightarrow\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{13.15}\right).x=\dfrac{-52}{45}\\ \Leftrightarrow\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}\right).x=\dfrac{-52}{45}\\ \Leftrightarrow\left(1-\dfrac{1}{15}\right).x=\dfrac{-52}{45}\\ \Leftrightarrow\dfrac{14}{15}.x=\dfrac{-52}{45}\\ \Leftrightarrow x=-\dfrac{26}{21}\)

17 tháng 3 2022

(11.3+13.5+...+113.15).x=−2645⇔(21.3+23.5+...+213.15).x=−5245⇔(1−13+13−15+...+113−115).x=−5245⇔(1−115).x=−5245⇔1415.x=−5245⇔x=−2621

15 tháng 5 2022

tách ra ạ

15 tháng 5 2022

hok đăng bài thi nha~

2 tháng 12 2021

\(a,x+2010=2021\\ \Rightarrow x=2021-2010\\ \Rightarrow x=11\\ b,x^5=32\\ \Rightarrow x^5=2^5\\ \Rightarrow x=2\\ c,2^x=16\\ \Rightarrow2^x=2^4\\ \Rightarrow x=4\\ d,4.2^x-3=125\\ \Rightarrow4.2^x=128\\ \Rightarrow2^x=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5\\ e,3^x+3^{x+1}=324\\ \Rightarrow3^x+3^x.3=324\\ \Rightarrow\left(1+3\right).3^x=324\\ \Rightarrow4.3^x=324\\ \Rightarrow3^x=81\\ \Rightarrow3^x=3^4\\ \Rightarrow x=4\)

2 tháng 12 2021

a, <=> x= 11

b, <=> x5 =25

<=> x=2

c, <=> 2x =24

<=> x=4

d, 4.2x -3 =125

<=> 4.2x =128

<=> 2x = 32

<=>2x = 25

<=> x=5

e, <=> 3x +3x.3=324

<=> 4.3x =324

<=> 3x =81

<=> 3x =34

<=> x=4

 

Gọi số học sinh khối 6 là x

Theo đề, ta có: \(x-3\in BC\left(8;12;15\right)\)

\(\Leftrightarrow x-3\in\left\{120;240;360;...\right\}\)

\(\Leftrightarrow x\in\left\{123;243;363\right\}\)

mà 200<=x<=300

nên x=243

3 tháng 10 2021

Gọi số học sinh khối 6 là a

a + 3 \(⋮8;12;15\)

\(\Rightarrow\) \(a+3\in BC\left(8;12;15\right)\)

8 = 2 . 3

12 = 22 . 3

15 = 3 . 5

\(\Rightarrow\) BCNN (8; 12; 15) = 22 . 3 . 5 = 60

Mà 203 < a + 3 < 303 học sinh

\(\Rightarrow\) a + 3 \(\in\) {240; 300}

\(\Rightarrow\) a \(\in\) {237; 207}

Bài 2:

\(10M=\dfrac{10^{12}+10}{10^{12}+1}=1+\dfrac{9}{10^{12}+1}\)

\(10N=\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)

Ta có: \(10^{12}+1>10^{11}+1\)

=>\(\dfrac{9}{10^{12}+1}< \dfrac{9}{10^{11}+1}\)

=>\(\dfrac{9}{10^{12}+1}+1< \dfrac{9}{10^{11}+1}+1\)

=>10M<10N

=>M<N

Bài 1:

\(A=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{900}\right)\)

\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{30}\right)\cdot\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{30}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{29}{30}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{31}{30}\)

\(=\dfrac{1}{30}\cdot\dfrac{31}{2}=\dfrac{31}{60}\)

4: Ta có: x+2x+...+18x=1368

\(\Leftrightarrow153x=1368\)

hay \(x=\dfrac{152}{17}\)

=>x+1/2+x+1/6+...+x+1/90=99,9

=>\(\left(x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)=99.9\)

=>\(9x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=99.9\)

=>9x+(1-0,1)=99,9

=>9x=99,9+0,1-1=100-1=99

=>x=11