a,

\(\dfrac{\left(3^3\right)^{15}.5^3.\left(2^3\right)^4}{\left(5^2\right)^2.\left(3^4\right)^{11}.2^{11}}=\dfrac{3^{45}.5^3.2^{12}}{5^4.3^{44}.2^{11}}=\dfrac{6}{5}\)

b, \(\left(-\dfrac{14}{25}\right)^2.\dfrac{125}{49}+\left(-3\dfrac{11}{36}\right).2\dfrac{2}{17}=\dfrac{4}{5}.\left(-7\right)=-\dfrac{28}{5}\)

c, \(\dfrac{1}{3}-2.1=-\dfrac{5}{3}\)

Áp dụng BĐT cosi:

\(\left(a+b+b+c+c+a\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\\ \ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\cdot3\sqrt[3]{\dfrac{1}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=9\\ \Leftrightarrow2\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\\ \Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{9}{2}\left(đpcm\right)\)

Dấu \("="\Leftrightarrow a=b=c\)

a) \(=\dfrac{\left(-1\right)^4}{3^4}=\dfrac{1}{81}\)

b) \(=\dfrac{\left(-9\right)^3}{4^3}=\dfrac{-729}{64}\)

c) \(=\left(-\dfrac{2}{10}\right)^2=\left(-\dfrac{1}{5}\right)^2=\dfrac{1}{25}\)

d) \(=1\)

\(a,=\dfrac{1}{81}\\ b,=\dfrac{729}{64}\\ c,=0,04\\ d,=1\)

a) Áp dụng BĐT AM - GM:

\(\dfrac{a}{b}+\dfrac{b}{a}\) >= 2\(\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}\) =2

Dấu '=' xảy ra <=> a=b=1

b) Cũng áp dụng BĐT AM- GM nhưng cho 3 số

c) Áp dụng BĐT AM- GM a+b>= 2\(\sqrt{ab}\)

\(\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\) >= 8\(\sqrt{ab.bc.ca}\) = 8abc

Dấu '=' xảy ra <=> a=b=c

Bài 1

;;;

a) \(5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^{6-5}+1=5+1=6\)

b) \(\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6\)

\(=\left(\dfrac{3}{7}\right)^{21-6}=\left(\dfrac{3}{7}\right)^{15}\)

c) \(\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)

\(=\dfrac{8}{27}-1+\dfrac{4}{9}\)

\(=\dfrac{8-27+12}{27}=-\dfrac{7}{27}\)

\(a)5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^1+1=6\)

\(b,\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{49-40}{49}\right)^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3=\left(\dfrac{3}{7}\right)^{21}:[\left(\dfrac{3}{7}\right)^2]^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6=\left(\dfrac{3}{7}\right)^{21-6}\)

\(=\left(\dfrac{3}{7}\right)^{15}\)

\(c,3.\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)

\(=3.\dfrac{8}{27}-1+\dfrac{4}{9}\)

\(=\dfrac{8}{9}-1+\dfrac{4}{9}\)

\(=\dfrac{8-9+4}{9}=\dfrac{1}{3}\)

a) 56:55+(49)0=56−5+1=5+1=656:55+(94​)0=56−5+1=5+1=6

b) (37)21:(1−4049)3(73​)21:(1−4940​)3

=(37)21:(949)3=(73​)21:(499​)3

=(37)21:[(37)2]3=(73​)21:[(73​)2]3

=(37)21:(37)6=(73​)21:(73​)6

=(37)21−6=(37)15=(73​)21−6=(73​)15

c) (23)3−(−523)0+49(32​)3−(3−52​)0+94​

=827−1+49=278​−1+94​

=8−27+1227=−727=278−27+12​=−277​
 

a) Ta có: \(2\dfrac{3}{3}\cdot4\cdot\left(-0.4\right)+1\dfrac{3}{5}\cdot1.75+\left(-7.2\right):\dfrac{9}{11}\)

\(=-4.8+\dfrac{8}{5}\cdot\dfrac{7}{4}-\dfrac{36}{5}\cdot\dfrac{11}{9}\)

\(=\dfrac{-24}{5}+\dfrac{14}{5}-\dfrac{44}{5}\)

\(=\dfrac{-54}{5}\)

b) Ta có: \(\left(\dfrac{1}{24}-\dfrac{5}{16}\right):\dfrac{-3}{8}+1^{10}\cdot\left(-5\right)^0\)

\(=\left(\dfrac{2}{48}-\dfrac{15}{48}\right)\cdot\dfrac{8}{-3}+1\cdot1\)

\(=\dfrac{-13}{48}\cdot\dfrac{-8}{3}+1\)

\(=\dfrac{13}{18}+\dfrac{18}{18}=\dfrac{31}{18}\)

Kết quả học tập kì I của lớp 6A xếp thành 3 loại:giỏi,khá,trung bình.Số học sinh giỏi chiếm \(\dfrac{1}{3}\) số học sinh của lớp,số học sinh khá chiếm 40% số học sinh cả lớp,số học sinh trung bình là 12 em.Tính số học sinh lớp 6A và tỉ số phần trăm của học sinh giỏi so với học sinh cả lớp.

a) \(\left|x-\dfrac{4}{11}\right|+\left|5+y\right|=0\)

<=>\(\left[{}\begin{matrix}x-\dfrac{4}{11}=0\\5+y=0\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}x=\dfrac{4}{11}\\y=-5\end{matrix}\right.\)

phần b, c tương tự