Chứng minh các đẳng thức sau :
a) \(^{12^8}\) .\(^{9^{12}}\)=\(^{18^{16}}\)
b)\(75^{20}\)=\(^{45^{10}}\).\(5^{30}\)
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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~^^
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
\(\widehat{bac}\) = \(\widehat{\text{dac }}\)= 90o
=> tam giác abc = tam giác adc
\(\frac{x+4}{2001}+\frac{x+3}{2002}=\frac{x+2}{2003}+\frac{x+1}{2004}\)
\(\Leftrightarrow\left(\frac{x+4}{2001}+1\right)+\left(\frac{x+3}{2002}+1\right)=\left(\frac{x+2}{2003}+1\right)+\left(\frac{x+1}{2004}+1\right)\)
\(\Leftrightarrow\frac{x+2005}{2001}+\frac{x+2005}{2002}=\frac{x+2005}{2003}+\frac{x+2005}{2004}\)
\(\Leftrightarrow\frac{x+2005}{2001}+\frac{x+2005}{2002}-\frac{x+2005}{2003}-\frac{x+2005}{2004}=0\)
\(\Leftrightarrow\left(x+2005\right).\left(\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}+\frac{1}{2004}\right)=0\)
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
\(\left(\frac{3}{5}\right)^{18}:\left(\frac{9}{25}\right)^8=\frac{\frac{3}{5}.\frac{3}{5}.\frac{3}{5}.....\frac{3}{5}}{\frac{9}{25}.\frac{9}{25}........\frac{9}{25}}\)
\(=\frac{\frac{15}{25}.\frac{15}{25}.........\frac{15}{25}}{\frac{9}{25}.\frac{9}{25}..........\frac{9}{25}}\)
\(=\frac{15.\left(\frac{1}{25}.\frac{1}{25}.\frac{1}{25}......\frac{1}{25}\right)}{9.\left(\frac{1}{25}.\frac{1}{25}.........\frac{1}{25}\right)}\)
\(=\frac{15.\left(\frac{1}{25}\right)^{10}}{9}=\frac{5.\left(\frac{1}{25}\right)^{10}}{3}\)
Vì \(\Delta ABC=\Delta A'B'C'\Rightarrow\) AB = A'B' ; BC = B'C'
Ta co: BM=1/2BC ; B'M'=1/2B'C' mà BC = B'C' => BM =B'M'
a, \(\Delta AMB=\Delta A'M'B'\left(ccc\right)\)vì có AB = A'B' ; BM =B'M' ; AM = A'M'
b, => \(\widehat{AMB}=\widehat{A'M'B'}\)
Ta co: \(\widehat{AMB}+\widehat{AMC}=180^O\) ; \(\widehat{A'M'B'}+\widehat{A'M'C'}=180^o\)
mà \(\widehat{AMB}=\widehat{A'M'B'}\) => \(\widehat{AMC}=\widehat{A'M'C'}\)
a/ Ta có: \(\Delta ABC=\Delta A'B'C'\)
\(\Rightarrow AB=A'B'\left(1\right)\)
\(\Rightarrow BC=B'C'\)
\(\Rightarrow BM=B'M'\left(2\right)\)
Xét \(\Delta AMB\)và \(\Delta A'M'B'\) có
\(AB=A'B'\)(theo )
\(BM=B'M'\)(theo 2)
\(AM=A'M'\)(gt)
\(\Rightarrow\Delta AMB=\Delta A'M'B'\)
b/ Ta có: \(\Delta AMB=\Delta A'M'B'\)
\(\Rightarrow\widehat{AMB}=\widehat{A'M'B'}\)
Mà \(\hept{\begin{cases}\widehat{AMC}=180^o-\widehat{AMB}\\\widehat{A'M'C'}=180^o-\widehat{A'M'B'}\end{cases}}\)
\(\Rightarrow\widehat{AMC}=\widehat{A'M'C'}\)
a, 3/2 - 10x - 4/5 +3x = 0
<=> 7/10 -7x =0
<=> x = 1/10
b, 3/2 - 1 + 4x - 2/3 + 7x =0
<=> -1/6 + 11x = 0
<=> x = 1/66
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