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\(a)\)
Ta có :
\(1-\frac{2}{3}=\frac{1}{3};1-\frac{4}{5}=\frac{1}{5};1-\frac{7}{8}=\frac{1}{8};1-\frac{3}{4}=\frac{1}{4}\)
\(1-\frac{9}{10}=\frac{1}{10};1-\frac{8}{9}=\frac{1}{9};1-\frac{5}{6}=\frac{1}{6};1-\frac{6}{7}=\frac{1}{7}\)
Do \(\frac{1}{3}>\frac{1}{4}>\frac{1}{5}>\frac{1}{6}>\frac{1}{7}>\frac{1}{8}>\frac{1}{9}>\frac{1}{10}\)
\(\Rightarrow1-\frac{1}{3}< 1-\frac{1}{4}< 1-\frac{1}{5}< 1-\frac{1}{6}< 1-\frac{1}{7}< 1-\frac{1}{8}< 1-\frac{1}{9}< 1-\frac{1}{10}\)
\(\Rightarrow\frac{2}{3}< \frac{3}{4}< \frac{4}{5}< \frac{5}{6}< \frac{6}{7}< \frac{7}{8}< \frac{8}{9}< \frac{9}{10}\)
Nếu \(\frac{a}{b}\)là 1 số thuộc dãy trên thì số tiếp theo là :
\(\frac{a+1}{b+1}\)
\(b)\)
Ta có :
\(a\left(a+2\right)=a^2+2a\)
\(b\left(a+1\right)=ab+b\)
Sorry , đến bước này mik chịu
~ Ủng hộ nhé
Phần b) Ý bạn là so sánh \(\frac{a}{b}\)và \(\frac{a+1}{b+2}\)
1) Chỉ tìm được Max thôi nhé
a) \(C=\frac{4}{5}+\frac{20}{\left|3x+5\right|+\left|4y+5\right|+8}\le\frac{4}{5}+\frac{20}{8}=\frac{33}{10}\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|3x+5\right|=0\\\left|4y+5\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{5}{3}\\y=-\frac{5}{4}\end{cases}}\)
b) \(E=\frac{2}{3}+\frac{21}{\left(x+3y\right)^2+5\left|x+5\right|+14}\le\frac{2}{3}+\frac{21}{14}=\frac{13}{6}\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+3y\right)^2=0\\5\left|x+5\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-5\\y=\frac{5}{3}\end{cases}}\)
2) Thì chỉ tìm được GTNN thôi nhé
a) \(A=5+\frac{-8}{4\left|5x+7\right|+24}\ge5-\frac{8}{24}=\frac{14}{3}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(4\left|5x+7\right|=0\Rightarrow x=-\frac{7}{5}\)
Vậy Min(A) = 14/3 khi x = -7/5
b) \(B=\frac{6}{5}-\frac{14}{5\left|6y-8\right|+35}\ge\frac{6}{5}-\frac{14}{35}=\frac{4}{5}\left(\forall y\right)\)
Dấu "=" xảy ra khi: \(5\left|6y-8\right|=0\Rightarrow x=\frac{4}{3}\)
Vậy Min(B) = 4/5 khi x = 4/3
3. S= -1/6 + -1/20 + 1/10 + 1/6
=0
4. A= -1 -1 -1 -1 -.... -1 [ có (50-2): 2 +1 = 25 số -1)
=-25
a) \(x:\left(\frac{3}{4}\right)^3=\left(\frac{3}{4}\right)^2\)
\(x=\left(\frac{3}{4}\right)^2.\left(\frac{3}{4}\right)^3\)
\(x=\left(\frac{3}{4}\right)^5\)
\(x=\frac{243}{1024}\)
vay \(x=\frac{243}{1024}\)
b) \(\left(\frac{2}{5}\right)^5.x=\left(\frac{2}{5}\right)^8\)
\(x=\left(\frac{2}{5}\right)^8:\left(\frac{2}{5}\right)^5\)
\(x=\left(\frac{2}{5}\right)^3\)
\(x=\frac{8}{125}\)
vay \(x=\frac{8}{125}\)
4) \(\left(0,36\right)^8=\left(0,6^2\right)^8=\left(0,6\right)^{16}\)
\(\left(0,216\right)^4=\left(0,6^3\right)^4=\left(0,6\right)^{12}\)
5) a) \(\left(3,5\right)^3=42,875\)
b) \(\left(-\frac{4}{11}\right)^2=\frac{16}{121}\)
c) \(\left(0,5\right)^4.6^4=3^4=81\)
d) \(\left(-\frac{1}{3}\right)^5:\left(\frac{1}{6}\right)^5=\left(-2\right)^5=-32\)
3,
a) (−23+37):45+(−13+47):45
= \(-\frac{5}{21}:\frac{4}{5}+\frac{5}{21}:\frac{4}{5}\)
= \(\left(-\frac{5}{21}+\frac{5}{21}\right):\frac{4}{5}\)
= \(0:\frac{4}{5}=0\)
2,
a) \(\frac{-3}{4}\).\(\frac{12}{-5}\).(\(\frac{-25}{6}\))
= \(\frac{-3.4.3.\left(-5\right).5}{4.\left(-5\right).3.3}\)
= \(-5\)
b) (−2).\(\frac{-38}{21}\).\(\frac{-7}{4}\).(\(\frac{-3}{8}\))
= \(\frac{-2.\left(-38\right)\left(-7\right)\left(-3\right)}{\left(-7\right)\left(-3\right)\left(-2\right)\left(-2\right).8}\)
= \(\frac{19}{8}\)
c) (\(\frac{11}{12}:\frac{33}{16}\)).\(\frac{3}{5}\)
= \(\left(\frac{11}{12}.\frac{16}{33}\right).\frac{3}{5}\)
= \(\frac{4}{9}.\frac{3}{5}\)
= \(\frac{4}{15}\)
d) \(\frac{7}{23}\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left(\frac{-41}{10}\right)\)
= \(\frac{-287}{203}\)
3. Tính:
a) (\(\frac{-2}{3}+\frac{3}{7}\)):\(\frac{4}{5}\)+(\(\frac{-1}{3}+\frac{4}{7}\)):\(\frac{4}{5}\)
= (\(\frac{-2}{3}+\frac{3}{7}\)\(+\)\(\frac{-1}{3}+\frac{4}{7}\)) : \(\frac{4}{5}\)
= 0 : \(\frac{4}{5}\)
= 0
b) \(\frac{5}{9}\):(\(\frac{1}{11}-\frac{5}{22}\))+\(\frac{5}{9}\):(\(\frac{1}{15}-\frac{2}{3}\))
= \(\frac{5}{9}\): \(\frac{-3}{22}\)+ \(\frac{5}{9}\): \(\frac{-3}{5}\)
= \(\frac{5}{9}\): \(\frac{-81}{110}\)
= \(\frac{-550}{729}\)
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
Số hạng thứ 5 của dãy là: \(\frac{-100}{a^6}\)
-120/a^6