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Giản Nguyên mình cũng ko biết bạn nhé, đề bài ghi như vậy nên mình mới ko hiểu
1/h=1/2(1/a+1/b)=1/2a+1/2b=(a+b)/2ab
=>(a+b/)2ab-1/h=0
quy dong len ta co
(a+b)h/2abh-2ab/2abh=0=> (ah+bh-2ab)/2abh=0 =>ah+bh-2ab=0
=>ah+bh-ab-ab=0
=>a(h-b)-b(a-h)=0
=>a(h-b)=b(a-h)
=>a/b=(a-h)(h-b)
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{56}+\frac{1}{72}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{8}{9}-\left(1-\frac{1}{9}\right)\)
\(=\frac{8}{9}-\frac{8}{9}=0\)
Ta có
3344=(3.11)44=344.1144=(34)11.1144=8111.1144
4433=(4.11)33=433.1133=(43)11.1133=6411.1133
=> 3344>4433
KL:
b) 52222=(52)1111=251111
25555=(25)1111=321111
=> 52222<25555
KL
Ta có \(\widehat{S}+\widehat{SGQ}+\widehat{Q}=180^0\Rightarrow\widehat{S}+\widehat{Q}=180^0-\widehat{SGQ}\)
Mà \(\widehat{S}-\widehat{Q}=12^0\Rightarrow\left\{{}\begin{matrix}\widehat{S}=\dfrac{180^0-\widehat{SGQ}+12^0}{2}=96^0-\dfrac{\widehat{SGQ}}{2}\\\widehat{Q}=\dfrac{180^0-\widehat{SGQ}-12^0}{2}=84^0-\dfrac{\widehat{SGQ}}{2}\end{matrix}\right.\)
Mà GP là p/g nên \(\widehat{QGP}=\widehat{PGS}=\dfrac{\widehat{SGQ}}{2}\)
\(\Rightarrow\widehat{Q}=84^0-\widehat{QGP}\)
Ta có \(\widehat{GPS}=\widehat{Q}+\widehat{QGP}=84^0-\widehat{QGP}+\widehat{QGP}=84^0\) (tc góc ngoài)
\(P=\frac{\left(\frac{10203}{125}:\frac{24}{5}-\frac{901}{200}\right)^2+125\cdot\frac{3}{4}}{\left\{\left[\left(\frac{11}{25}\right)^2+\frac{353}{100}\right]^2-\left(\frac{11}{4}\right)^2\right\}:\frac{13}{25}}\)
\(P=\frac{\left(\frac{10203}{125}\cdot\frac{5}{24}-\frac{901}{100}\right)^2+\frac{375}{4}}{\left[\left(\frac{121}{625}+\frac{353}{100}\right)^2-\frac{121}{16}\right]\cdot\frac{25}{13}}\)
\(P=\frac{\left(\frac{3401}{200}-\frac{1802}{200}\right)^2+\frac{18750}{200}}{\left[\left(\frac{484}{2500}+\frac{8825}{2500}\right)^2-\frac{121}{16}\right]\cdot\frac{25}{13}}\)
\(P=\frac{\frac{1599}{200}^2+\frac{18750}{200}}{\left(\frac{9309}{2500}^2-\frac{121}{16}\right)\cdot\frac{25}{13}}\)
\(P=\frac{\frac{2556801}{40000}+\frac{3750000}{40000}}{\left(\frac{86657481}{6250000}-\frac{47265625}{6250000}\right)\cdot\frac{25}{13}}\)
\(P=\frac{\frac{6306801}{40000}}{\frac{39391856}{6250000}\cdot\frac{25}{13}}\)
\(P=\frac{\frac{6306801}{40000}}{\frac{2461991}{203125}}\)