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moa moa
Áp dụng t/c dãy tỉ số bằng nhau:
a.
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x}{6}=\dfrac{4y}{20}=\dfrac{2x+4y}{6+20}=\dfrac{28}{26}=\dfrac{14}{13}\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.\dfrac{14}{13}=\dfrac{52}{13}\\y=5.\dfrac{14}{13}=\dfrac{70}{13}\end{matrix}\right.\)
(Em có nhầm đề 26 thành 28 ko nhỉ, số xấu quá)
b.
\(4x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{3x}{15}=\dfrac{-2y}{-8}=\dfrac{3x-2y}{15-8}=\dfrac{35}{7}=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=5.5=25\\y=4.2=20\end{matrix}\right.\)
c.
\(\dfrac{x}{-3}=\dfrac{y}{-7}=\dfrac{2x}{-6}=\dfrac{4y}{-28}=\dfrac{2x+4y}{-6-28}=\dfrac{68}{-34}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}x=-3.\left(-2\right)=6\\y=-7.\left(-2\right)=14\end{matrix}\right.\)
d.
\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{-3y}{9}=\dfrac{-2z}{-8}=\dfrac{4x-3y-2z}{8+9-8}=\dfrac{16}{9}\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.\dfrac{16}{9}=\dfrac{32}{9}\\y=-3.\dfrac{16}{9}=-\dfrac{48}{9}\\z=4.\dfrac{16}{9}=\dfrac{64}{9}\end{matrix}\right.\)
Ta có: 290=(29)10=(23.3)10=(23)3.10=83.10
360=(36)10=(33.2)10=(32)3.10=93.10
Vì 83.10 < 93.10
Nên 290< 360
Tham khảo thôi nhé mk cx ko chắc
Ta có:290=(23)30=830
360=(32)30=930
Vì 830<930
nên 290<360
Vậy 290<360
\(D=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{100}-1\right)\)
\(=\left(\frac{1}{2}-\frac{2}{2}\right)\left(\frac{1}{3}-\frac{3}{3}\right)...\left(\frac{1}{100}-\frac{100}{100}\right)\)
\(=\frac{\left(-1\right)}{2}.\frac{\left(-2\right)}{3}...\frac{\left(-99\right)}{100}\)
\(=\left[\left(-1\right).\left(-1\right)...\left(-1\right)\text{ 99 thưa số -1 }\right].\left(\frac{1}{2}.\frac{2}{3}...\frac{99}{100}\right)\)
\(=-\frac{1}{100}\)\(>-\frac{1}{99}\)
\(\text{Vậy }D>-\frac{1}{99}\)
\(\frac{\left(-1\right)}{2}.\frac{\left(-2\right)}{3}...\frac{\left(-99\right)}{100}=\left(-1\right).\frac{1}{2}.\left(-1\right).\frac{2}{3}....\left(-1\right).\frac{99}{100}\)
dùng tính chất kếp hợp nhóm 99 thừa số -1 lại
\(=\left[\left(-1\right).\left(-1\right)...\left(-1\right)\text{ 99 thừa số -1}\right].\left(\frac{1}{2}.\frac{2}{3}...\frac{99}{100}\right)\)
òi còn j nữa ko
Ta có a+2/b+2 = a/b+2 + 2/b+2 = a(b+2)/b+2 + 2(b+2)/b+2 = a+2
Do a+2 > a/b => a+2/b+2 >a/b
mik làm đại ko bik đúng hay sai đâu nha
Xét tích : \(a\left(b+2\right)=ab+2a\)
\(b\left(a+2\right)=ab+2b\)
Nếu \(a>b\)thì \(ab+2a>ab+2b\)
hay \(a\left(b+2\right)>b\left(a+2\right)\)
\(\Rightarrow\frac{a}{b}>\frac{a+2}{b+2}\)
Nếu \(a< b\)thì \(ab+2a< ab+2b\)
hay \(a\left(b+2\right)< b\left(a+2\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+2}{b+2}\)
Nếu \(a=b\)thì \(ab+2a=ab+2b\)
hay \(a\left(b+2\right)=b\left(a+2\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a+2}{b+2}\)
M=\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{17}{8^2.9^2}+\dfrac{19}{9^2.10^2}\)
=\(\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{17}{64.81}+\dfrac{19}{81.100}\)
=\(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{64}-\dfrac{1}{81}+\dfrac{1}{81}-\dfrac{1}{100}\)
=1-\(\dfrac{1}{100}\)=\(\dfrac{99}{100}\)<\(\dfrac{100}{100}=1\)
Ta có : \(\left\{{}\begin{matrix}2^{24}=\left(2^3\right)^8=8^8\\3^{16}=\left(3^2\right)^8=9^8\end{matrix}\right.\)
Thấy : \(9>8\)
\(\Leftrightarrow9^8>8^8\)
\(\Leftrightarrow3^{16}>2^{24}\)
Vậy ...
ta có
\(2^{24}=\left(2^3\right)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
mà 8 < 9 nên 8^8<9^8 hay \(2^{24}\)< \(3^{16}\)
vậy ...