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\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{13.15}\right).x=\dfrac{-26}{45}\\ \Leftrightarrow\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{13.15}\right).x=\dfrac{-52}{45}\\ \Leftrightarrow\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}\right).x=\dfrac{-52}{45}\\ \Leftrightarrow\left(1-\dfrac{1}{15}\right).x=\dfrac{-52}{45}\\ \Leftrightarrow\dfrac{14}{15}.x=\dfrac{-52}{45}\\ \Leftrightarrow x=-\dfrac{26}{21}\)
(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
\(a,x+2010=2021\\ \Rightarrow x=2021-2010\\ \Rightarrow x=11\\ b,x^5=32\\ \Rightarrow x^5=2^5\\ \Rightarrow x=2\\ c,2^x=16\\ \Rightarrow2^x=2^4\\ \Rightarrow x=4\\ d,4.2^x-3=125\\ \Rightarrow4.2^x=128\\ \Rightarrow2^x=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5\\ e,3^x+3^{x+1}=324\\ \Rightarrow3^x+3^x.3=324\\ \Rightarrow\left(1+3\right).3^x=324\\ \Rightarrow4.3^x=324\\ \Rightarrow3^x=81\\ \Rightarrow3^x=3^4\\ \Rightarrow x=4\)
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
\(2x+7⋮x+2\)
\(\Leftrightarrow x+2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{-1;-3;1;-5\right\}\)
\(\Rightarrow2\left(x+2\right)+3⋮x+2\\ \Rightarrow x+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Rightarrow x\in\left\{-5;-3;-1;1\right\}\)
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
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