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A = 2 - 3|2x - 1|
có |2x - 1| ≥ 0 => -3|2x - 1| ≤ 0
=> 2 - 3|2x -1| ≤ 2
dấu = xảy ra <=> 2x - 1 = 0<=> x = 1/2
vậy max A = 2 khi x = 1/2
a/
Đặt $\frac{a-1}{2}=\frac{b-2}{3}=\frac{c-3}{4}=k$
$\Rightarrow a=2k+1; b=3k+2; c=4k+3$
Khi đó:
$3a+3b-c=50$
$\Rightarrow 3(2k+1)+3(3k+2)-(4k+3)=50$
$\Rightarrow 11k+6=50$
$\Rightarrow 11k=44\Rightarrow k=4$
Ta có:
$a=2k+1=2.4+1=9$
$b=3k+2=3.4+2=14$
$c=4k+3=4.4+3=19$
b/
$2a=3b; 5b=7c\Rightarrow \frac{a}{3}=\frac{b}{2}; \frac{b}{7}=\frac{c}{5}$
$\Rightarrow \frac{a}{21}=\frac{b}{14}=\frac{c}{10}$
Áp dụng TCDTSBN:
$\frac{a}{21}=\frac{b}{14}=\frac{c}{10}=\frac{3a}{63}=\frac{7b}{98}=\frac{5c}{50}=\frac{3a-7b+5c}{63-98+50}=\frac{45}{15}=3$
$\Rightarrow a=21.3=63; b=14.3=42; c=10.3=30$
\(\frac{4^{20}.20^{10}}{80^{10}.5^7}\)\(=\frac{4^{10}.4^{10}.20^{10}}{4^{10}.20^{10}.5^7}\)\(=\frac{4^{10}}{5^7}\)
\(\frac{9^{10}.6^3}{36^7.3^2}\)\(=\frac{3^5.3^2.3^{10}.6^3}{6^3.6^4.6^7.3^2}\)\(=\frac{3^{15}}{6^{11}}\)\(=\frac{3^{11}.3^4}{3^{11}.2^{11}}\)\(=\frac{3^4}{2^{11}}\)
\(2^2+4^2+...+20^2\)
\(=\left(1.2\right)^2+\left(2.2\right)^2+...+\left(2.10\right)^2\)
\(=1^2.2^2+2^2.2^2+...+2^2+10^2\)
\(=2^2.\left(1^2+2^2+3^2+...+10^2\right)\)
\(=4.385\)
\(=1540\)
1/ \(\Rightarrow x\left(\frac{1}{4}x-\frac{1}{3}\right)=0\Rightarrow x=0\) hoặc \(\frac{1}{4}x-\frac{1}{3}=0\Rightarrow\frac{1}{4}x=\frac{1}{3}\Rightarrow x=\frac{4}{3}\)
Vậy x = 0 ; x = 4/3
b/ \(\Rightarrow2^x.2^2-2^x=96\Rightarrow2^x\left(2^2-1\right)\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)
mấy câu còn lại tương tự
\(\frac{28}{14}=2\)
\(\frac{5}{2}:2=\frac{5}{4}\)
\(\frac{8}{4}=2\)
\(\frac{1}{2}:\frac{2}{3}=\frac{3}{4}\)
\(\frac{3}{10}\)
\(\frac{21}{10}:7=\frac{3}{10}\)
\(3:\frac{3}{10}=\frac{1}{10}\)
từ đó ta có các tỉ lệ thức bằng nhau là:
28:14=8:4
3:10=2,1:7
\(G\)\(=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(G=\frac{1}{4}\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
Đặt S = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\)
Ta thấy : \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};......;\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}\)
=> S < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right).n}\)
=> S <\(1-\frac{1}{n}\)
Thay S vào G ta có :
G < \(\frac{1}{4}\left(1-\frac{1}{n}\right)\)
G< \(\frac{1}{4}-\frac{1}{4n}< \frac{1}{4}\)( đpcm )
Học tốt
#Dương