1 | + | 1 | + … + | 1 | = | |
3 x 5 | 5 x 7 | 121 x 123 |
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x-\frac{10}{3}=\frac{7}{15}.\frac{3}{5}\)
\(x-\frac{10}{3}=\frac{7}{25}\)
\(x=\frac{7}{25}+\frac{10}{3}=\frac{271}{75}\)
Vậy x = 271/75
\(x-\frac{10}{3}\)\(=\)\(\frac{7}{15}\)\(.\)\(\frac{3}{5}\)
\(x-\frac{10}{3}\)\(=\) \(\frac{7}{25}\)
\(x\) \(=\)\(\frac{7}{25}\)\(+\)\(\frac{10}{3}\)
\(x\) \(=\) \(\frac{271}{75}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tính nhanh:
1+2+3+4+5+6+7+8+9+10
=(1+9)+(2+8)+(3+7)+(4+6)+10+5
= 10 + 10 + 10 + 10 + 10+5
= 50 + 5
= 55
123*45+123*50+123*5
= 123*(45+50+5)
= 123 * 100
= 12300
K mk nha bn
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(x-\dfrac{10}{3}=\dfrac{7}{15}\cdot\dfrac{3}{5}\)
=>\(x-\dfrac{10}{3}=\dfrac{21}{75}=\dfrac{7}{25}\)
=>\(x=\dfrac{7}{25}+\dfrac{10}{3}=\dfrac{21+250}{75}=\dfrac{271}{75}\)
b: \(x+\dfrac{3}{22}=\dfrac{27}{121}\cdot\dfrac{9}{11}\)
=>\(x+\dfrac{3}{22}=\dfrac{243}{1331}\)
=>\(x=\dfrac{243}{1331}-\dfrac{3}{22}=\dfrac{123}{2662}\)
c: \(\dfrac{8}{23}\cdot\dfrac{46}{24}-x=\dfrac{1}{3}\)
=>\(\dfrac{8}{24}\cdot\dfrac{46}{23}-x=\dfrac{1}{3}\)
=>\(\dfrac{2}{3}-x=\dfrac{1}{3}\)
=>\(x=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
d: \(1-x=\dfrac{49}{65}\cdot\dfrac{5}{7}\)
=>\(1-x=\dfrac{49}{7}\cdot\dfrac{5}{65}=\dfrac{7}{13}\)
=>\(x=1-\dfrac{7}{13}=\dfrac{6}{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{3}.x=\frac{1}{3}\)1) x-\(\frac{10}{3}\)=\(\frac{7}{15}.\frac{3}{5}\)
x-10/3=7/25
x=7/25+10/3
x=\(\frac{271}{75}\)
2)\(\frac{8}{23}.\frac{46}{24}.x=\frac{1}{3}\)
2/3.x=1/3
x=1/3:2/3
x=1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x+1}{124}+1+\dfrac{x+2}{123}+1=\dfrac{x+3}{122}+1+\dfrac{x+4}{121}+1\)
\(\Leftrightarrow\dfrac{x+125}{124}+\dfrac{x+125}{123}=\dfrac{x+125}{122}+\dfrac{x+125}{121}\)
\(\Leftrightarrow\left(x+125\right)\left(\dfrac{1}{124}+\dfrac{1}{123}-\dfrac{1}{122}-\dfrac{1}{121}\ne0\right)=0\Leftrightarrow x=-125\)
<=>\(\dfrac{x+1}{124}+\dfrac{x+2}{123}-\dfrac{x+3}{122}-\dfrac{x+4}{121}=0\)
<=>\(\left(\dfrac{x+1}{124}+1\right)+\left(\dfrac{x+2}{123}+1\right)-\left(\dfrac{x+3}{122}+1\right)-\left(\dfrac{x+4}{121}+1\right)=0\)
<=>\(\dfrac{x+125}{124}+\dfrac{x+125}{123}-\dfrac{x+125}{122}-\dfrac{x+125}{121}=0\)
<=>\(\left(x+125\right)\left(\dfrac{1}{124}+\dfrac{1}{123}-\dfrac{1}{122}-\dfrac{1}{121}\right)=0\)
<=>x+125=0
<=>x=-125
![](https://rs.olm.vn/images/avt/0.png?1311)
1. So sánh
a) \(25^{50}\) và \(2^{300}\)
\(25^{50}=25^{1.50}=\left(25^1\right)^{50}=25^{50}\)
\(2^{300}=2^{6.50}=\left(2^6\right)^{50}=64^{50}\)
Vì \(25< 64\) nên \(25^{50}< 64^{50}\)
Vậy \(25^{50}< 2^{300}\)
b) \(625^{15}\) và \(12^{45}\)
\(625^{15}=625^{1.15}=\left(625^1\right)^{15}=625^{15}\)
\(12^{45}=12^{3.15}=\left(12^3\right)^{15}=1728^{15}\)
Vì \(625< 1728\) nên \(625^{15}< 1728^{15}\)
Vậy \(625^{15}< 12^{45}\)
1.So sánh
a)\(25^{50}\) và \(2^{300}\)
Ta có : \(2^{300}=\left(2^6\right)^{50}=64^{50}\)
Vì \(25^{50}< 64^{50}\) nên \(25^{50}< 2^{300}\)
b)\(625^{15}\) và \(12^{45}\)
Ta có : \(12^{45}=\left(12^3\right)^{15}=1728^{15}\)
Vì \(625^{15}< 1728^{15}\) nên \(625^{15}< 12^{45}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Rightarrow x+35=103\\ \Rightarrow x=68\\ b,\Rightarrow x+7=42\\ \Rightarrow x=35\\ c,\Rightarrow x-15=9\\ \Rightarrow x=24\\ d,\Rightarrow x+23=31\\ \Rightarrow x=8\\ e,\Rightarrow18-x=6\\ \Rightarrow x=12\\ f,\Rightarrow x+14=34\\ \Rightarrow x=20\\ g,\Rightarrow2x+1=21\\ \Rightarrow x=11\\ h,\Rightarrow3\left(x+1\right)=99\\ \Rightarrow x+1=33\Rightarrow x=32\\ i,5\left(x-3\right)=25\\ \Rightarrow x-3=5\\ \Rightarrow x=8\\ j,\Rightarrow8\left(2x+7\right)=88\\ \Rightarrow2x+7=11\Rightarrow x=2\\ k,\Rightarrow5\left(x+4\right)=85\\ \Rightarrow x+4=17\\ \Rightarrow x=13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. $121-3(x-5)=6$
$3(x-5)=121-6=115$
$x-5=115:3=\frac{115}{3}$
$x=\frac{115}{3}+5=\frac{130}{3}$
b.
$2x-138=2^3.3^2=72$
$2x=72+138=210$
$x=210:2=105$
c.
$x-3\vdots 7$
$\Rightarrow x-3\in\left\{0;7;14;21;28;35;42;49; 56;...\right\}$
Mà $10< x< 50$ nên $x\in\left\{14;21;28;35;42;49\right\}$
d.
$27\vdots x+1$
$\Rightarrow x+1\in\left\{\pm 1; \pm 3; \pm 9; \pm 27\right\}$
$\Rightarrow x\in\left\{0; -2; -4; 2; 8; -10; 26; -28\right\}$
a ) 121-3.(x - 5 ) = 6
3.(x-5) = 121 -6
3. (x-5)=115
x-5 = 115:3
x-5=35
x=35+5
x = 40
b) 2x - 138 = 2'3. 3'2
2x -138=8.9
2x-138=72
2x=72+138
2x=210
x=210:2
x=105
c) theo bài ra : x-3 ∈ B(7)
ta có B(7)=(0,7,14,21,28,35,49,56,...)
=) x-3 ∈ ( 0,7,14,21,28,35,49,56,...)
=) x ∈( 3 , 10,17,24,31,38,42,58,..)
mà 10 <x<50 nên x ∈ ( 17 , 24 ,31,38,42 )
vậy x ∈(17,24,31,38,42)