Gọi cạnh lớn nhất là a 

a + ( a-3 ) + (a - 4 ) + (a-5) = 4a - 12

4a -12 = 80

4a = 92

a = 23 

a - 3 = 23 - 3 = 20 

a - 4 = 20 - 1 = 19

a - 5 = 19 - 1 = 18 

Vậy các cạnh của hình tứ giác là : 23,20,19,18

Gọi cạnh lớn nhất là a 

a + ( a-3 ) + (a - 4 ) + (a-5) = 4a - 12

4a -12 = 80

4a = 92

a = 23 

a - 3 = 23 - 3 = 20 

a - 4 = 20 - 1 = 19

a - 5 = 19 - 1 = 18 

Vậy các cạnh của hình tứ giác là : 23,20,19,18

a) 12⁸ . 9¹² = ( 3 . 2² )⁸ . ( 3² )¹²

= 3⁸ .  2¹⁶ . 3²⁴ = 3³² . 2¹⁶ = ( 3² )¹⁶ . 2¹⁶ = ( 3² . 2 )¹⁶ = 18¹⁶

b) 45¹⁰ . 5³⁰ = 45¹⁰ . ( 5³ )¹⁰

=  ( 45 . 5³ )¹⁰ = 5625¹⁰

Mà 75²⁰ = ( 75² )¹⁰ = 5625¹⁰

Vậy 45¹⁰ . 5³⁰ = 75²⁰

tìm 1 hay ìm 1 vậy bạn 

gọi abc là chu kì của số thập phân vô hạn tuần hoàn đơn ( 0 < abc < 999 ) thì phân số phải tìm là : \(\frac{\overline{abc}}{999}\)

\(\frac{\overline{abc}}{999}=\frac{\overline{abc}}{3^3.37}=\frac{\overline{abc}.37^2}{3^3.37^3}=\frac{\overline{abc}}{\left(3.37\right)^3}\)

ta đặt \(\frac{\overline{abc}}{\left(3.37\right)^3}=\frac{x^3.37^3}{\left(3.37\right)^3}\)với x \(\in\)N*

\(\Rightarrow\)abc . 37² = x³ . 37³

\(\Rightarrow\)abc = 37x³ < 999

\(\Rightarrow\)x³ \(\in\){ 1 ; 8 }

\(\Rightarrow\)x \(\in\){ 1 '; 2 }

\(\Rightarrow\)abc \(\in\){ 037 ; 296 }

vậy phân số cần tìm là : \(\frac{037}{999}=\frac{1}{27};\frac{296}{999}=\frac{8}{27}\)

\(1.77777777778=\frac{177777777778}{100000000000}\)

Đúng 100% chúc bạn zui~^^

\(=\frac{88888888889}{500000}\)

bc = \(\sqrt{ab^2+ac^2}\)= \(\sqrt{5^2+12^2}\)= \(\sqrt{25+144}\)= \(\sqrt{169}\)= 13

xét 2 tam giác abc, adc

ab = ad

ac chung

\(\widehat{bac}\) = \(\widehat{\text{dac }}\)= 90^o

=> tam giác abc = tam giác adc

cảm ơn 

7²⁴=(7²)¹²=49¹²

Vì 49>40 nên 49¹²>40¹²

=> 7²⁴>40¹²

7^24 > 40^12

x+4/2001+x+3/2002=-x+2/2003+x+1/2004

x=...

\(\frac{x+4}{2001}+\frac{x+3}{2002}=\frac{x+2}{2003}+\frac{x+1}{2004}\)

\(\Leftrightarrow\left(\frac{x+4}{2001}+1\right)+\left(\frac{x+3}{2002}+1\right)=\left(\frac{x+2}{2003}+1\right)+\left(\frac{x+1}{2004}+1\right)\)

\(\Leftrightarrow\frac{x+2005}{2001}+\frac{x+2005}{2002}=\frac{x+2005}{2003}+\frac{x+2005}{2004}\)

\(\Leftrightarrow\frac{x+2005}{2001}+\frac{x+2005}{2002}-\frac{x+2005}{2003}-\frac{x+2005}{2004}=0\)

\(\Leftrightarrow\left(x+2005\right).\left(\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}+\frac{1}{2004}\right)=0\)

Vì  \(\left(\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}+\frac{1}{2004}\right)\ne0\)

\(\Rightarrow x+2004=0\)

\(\Rightarrow x=0-2004=-2004\)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)

\(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)

Ta có : \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{75}\), \(\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\)nên :

\(A>\frac{1}{75}.25+\frac{1}{100}.25=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

\(A< \frac{1}{51}.25+\frac{1}{76}.25< \frac{1}{50}.25+\frac{1}{75}.25=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

Vậy \(\frac{7}{12}< A< \frac{5}{6}\)

+A=1/(1.2)+1/(3.4)+...+1/(99.100) 
=1/1-1/2+1/3-1/4+....+1/99-1/100 
=1/2+1/3-1/4+1/5-1/6+1/7+...-1/98+1/99... 
=(1/2+1/3)+(1/5-1/4)+(1/7-1/6)+..+(1/9... 
=5/6-(1/4.5+1/6.7+..1/98.99+1/100)<5/6 
do -(1/4.5+1/6.7+..1/98.99+1/100)<0 
+A=1/(1.2)+1/(3.4)+...+1/(99.100) 
=1/2+1/12+1/(5.6)+...+1/(99.100) 
=7/12+[1/(5.6)+...1/(99.100)] 
>7/12 do [1/(5.6)+...1/(99.100)]>0

\(\left(\frac{3}{5}\right)^{18}:\left(\frac{9}{25}\right)^8\)

= \(\left(\frac{3}{5}\right)^{18}:\left(\frac{3}{5}\right)^{16}\)

= \(\left(\frac{3}{5}\right)^2\)

(3/5)¹⁸:(9/25)⁸

Ta có

\(\left(\frac{3}{5}\right)^{18}:\left(\frac{9}{25}\right)^8=\frac{\frac{3}{5}.\frac{3}{5}.\frac{3}{5}.....\frac{3}{5}}{\frac{9}{25}.\frac{9}{25}........\frac{9}{25}}\)

\(=\frac{\frac{15}{25}.\frac{15}{25}.........\frac{15}{25}}{\frac{9}{25}.\frac{9}{25}..........\frac{9}{25}}\)

\(=\frac{15.\left(\frac{1}{25}.\frac{1}{25}.\frac{1}{25}......\frac{1}{25}\right)}{9.\left(\frac{1}{25}.\frac{1}{25}.........\frac{1}{25}\right)}\)

\(=\frac{15.\left(\frac{1}{25}\right)^{10}}{9}=\frac{5.\left(\frac{1}{25}\right)^{10}}{3}\)