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Bài làm
\(A=\frac{2^2.10+2^3.6}{2^2.15-2^4}\)
\(A=\frac{2^2.10+2.2^2.6}{2^2.15-2^2.2^2.1}\)
\(A=\frac{2^2.\left(10+6\right).2}{2^2.\left(15-1\right).2^2}\)
\(A=\frac{2^2.16.2}{2^2.14.2^2}\)
\(A=\frac{16}{14.2}\)
\(A=\frac{8}{7.2}\)
\(A=\frac{8}{14}\)
\(A=\frac{4}{7}\)
Vậy \(A=\frac{4}{7}\)
\(B=\frac{2^9.15^{17}.75^3}{18^8.5^{24}.9^2}\)
\(B=\frac{2^9.\left(3.5\right)^{17}.\left(3.5^2\right)^3}{\left(2.3^2\right)^8.5^{24}.\left(3^2\right)^2}\)
\(B=\frac{2^9.3^{17}.5^{17}.3^3.5^6}{2.3^{19}.5^{24}.3^4}\)
\(B=\frac{2^8.1.1.1.5}{1.3^2.1.3}\)
\(B=\frac{2^8.5}{3^3}\)
\(B=\frac{1280}{27}\)
Cách tui lẹ hơn cách bạn Nguyễn Duy Khánh nè!
Ta có: \(C=\frac{7^{28}+7^{24}+....+7^4+7^0}{\left(7^{30}+7^{26}+...+7^6+7^2\right)+\left(7^{28}+7^{24}+...+7^4+7^0\right)}\)
\(=\frac{7^{28}+7^{24}+...+7^4+7^0}{7^2\left(7^{28}+7^{24}+...+7^4+7^0\right)+\left(7^{28}+7^{24}+...+7^4+7^0\right)}\)
\(=\frac{7^{28}+7^{24}+...+7^4+7^0}{\left(7^{28}+7^{24}+...+7^4+7^0\right)\left(7^2+1\right)}=\frac{1}{7^2+1}=\frac{1}{50}\)
P/s: Easy đúng không?
\(C=\frac{7^{28}+7^{2\text{4}}+...+7^{\text{4}}+7^0}{7^{30}+7^{28}+...+7^2+7^0}\)
Đặt A là tử số ,B là mẫu số.Ta có:
\(7^{\text{4}}A=7^{32}+7^{28}+...+7^8+7^{\text{4}}+7^0\)
\(20\text{4}1A-A=\left(7^{32}+7^{28}+7^{2\text{4}}+...+7^8+7^{\text{4}}\right)-\left(7^{28}+7^{2\text{4}}+...+7^{\text{4}}+7^0\right)\)
\(2\text{4}00A=7^{32}-7^0=7^{32}-1\)
\(\Rightarrow A=\left(7^{32}-1\right):2\text{4}00\)
\(7^2B=\left(7^{32}+7^{30}+7^{28}+...+7^{\text{4}}+7^2\right)\)
49B-B= ....tự..điền......như A nhé.....
48B=732-1 =>B=[7232-1]:48
=>\(C=\frac{A}{B}=\frac{\left(7^{32}-1\right):2\text{4}00}{\left(7^{32}-1\right):\text{4}8}\)
Tui nghĩ vậy đc r á
p/s:ko chắc
.
Bài 2 :
1) \(x-70=-45\) 2) \(\frac{4}{7}:x=\frac{12}{28}\)
\(\Rightarrow\) \(x=-45+70\) \(\Rightarrow x=\frac{4}{7}:\frac{12}{28}\)
\(\Rightarrow\) \(x=25\) \(\Rightarrow x=\frac{4}{3}\)
Vậy \(x=25\) Vậy \(x=\frac{4}{3}\)
3) Giống câu c) ở bài 1
4) \(x-50=-35\) 5) \(\frac{4}{7}.x=\frac{11}{18}\)
\(\Rightarrow x=-35+50\) \(\Rightarrow x=\frac{11}{28}:\frac{4}{7}\)
\(\Rightarrow x=15\) \(\Rightarrow x=\frac{77}{72}\)
Vậy \(x=15\) Vậy \(x=\frac{77}{72}\)
6) \(\left(\frac{2}{3}x+2,5\right):2\frac{2}{6}=6\)
\(\Rightarrow\)\(\left(\frac{2}{3}x+2,5\right):\frac{14}{6}=6\)
\(\Rightarrow\) \(\frac{2}{3}x+2,5=6.\frac{14}{6}\)
\(\Rightarrow\frac{2}{3}x+2,5=14\)
\(\Rightarrow\frac{2}{3}x=\frac{23}{2}\)
\(\Rightarrow x=\frac{23}{2}:\frac{2}{3}\)
\(\Rightarrow x=\frac{69}{4}\)
Vậy \(x=\frac{69}{4}\)
Bài 1:
1) \(\frac{7}{5}+\frac{-8}{5}=-\frac{1}{5}\)
2) \(-\frac{6}{5}.\frac{15}{24}=-\frac{3}{4}\)
3) \(\left(\frac{2}{3}+1,5\right)-3,5:7\frac{1}{2}=\)\(\frac{13}{6}-\frac{7}{15}=\frac{17}{10}\)
4) \(\frac{5}{8}-\frac{-7}{9}=\frac{5}{8}+\frac{7}{9}=\frac{101}{72}\)
5)\(\frac{-7}{3}.\frac{12}{28}=-1\)
bài 1:a,
\(3^9.3:3^{10}+\left|2010^0\right|\)
=> \(3^9.3:3^{10}+\left|1\right|\)
=> \(3^9.3:3^{10}+1\)
=> \(3^{10}:3^{10}+1\)
=> 1+1
=> 2
b, \([\left(4^9:4^7\right):8-735^0]^{2011}\)
=> \([4^2:8-735^0]^{2011}\)
=> \([2^4:2^3-735^0]^{2011}\)
=> \([2-1]^{2011}\)
=> 1
c, \(8^{2x}:8=512\)
=> \(8^{2x}:8=8^3\)
=> \(8^{2x}=8^4\)
=> 2x=4
=> x=2
bài 2:
Theo đề ta có:
\(\left(7^0+7^1+7^2+7^3+......+7^{2010}+7^{2011}\right)\)
=> \((7^0+7^1)+(7^2+7^3)+......+(7^{2010}+7^{2011})\)
=> \(7^0.\left(1+7\right)+7^2\left(1+7\right)+..+7^{2010}\left(1+7\right)\)
=> \(7^0.8+7^2.8+..+7^{2010}.8\)
Mà \(7^0.8+7^2.8+..+7^{2010}.8\) \(⋮\) 8 ( vì có thừa số 8 nên chia hết cho 8)
nên \(\left(7^0+7^1+7^2+7^3+......+7^{2010}+7^{2011}\right)\)\(⋮\) 8
\(a)\dfrac{3}{4}-\dfrac{-5}{2}-\dfrac{7}{-24}\)
\(=\dfrac{13}{4}-\dfrac{7}{-24}\)
\(=\dfrac{85}{24}\)
\(b)\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)
\(=\dfrac{-3}{56}-\dfrac{3}{28}\)
\(=\dfrac{-9}{56}\)
\(c)\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}\)
\(=\dfrac{13}{12}\)\(+\dfrac{-2}{3}\)
\(=\dfrac{5}{12}\)
\(d)\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)
\(=\dfrac{-1}{14}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)
\(=\dfrac{-23}{126}+\dfrac{-7}{18}+\dfrac{4}{7}\)
\(=\dfrac{-4}{7}+\dfrac{4}{7}\)
\(=0\)
\(e)\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)
\(=\dfrac{-5}{56}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)
\(=\dfrac{83}{56}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)
\(=\dfrac{305}{168}+\dfrac{1}{7}+\dfrac{5}{-8}\)
\(=\dfrac{47}{24}+\dfrac{5}{-8}\)
\(=\dfrac{4}{3}\)
Bài 2 : Tính
a) \(\dfrac{3}{4}-\dfrac{-5}{2}-\dfrac{7}{-24}\)
\(=\dfrac{18}{24}-\dfrac{-60}{24}-\dfrac{-4}{24}\)
\(=\dfrac{18-\left(-60\right)-\left(-7\right)}{24}\)
\(=\dfrac{85}{24}\)
b) \(\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)
\(=\dfrac{32}{56}+\dfrac{-35}{56}-\dfrac{6}{56}\)
\(=\dfrac{32+\left(-35\right)-6}{56}\)
\(=\dfrac{-9}{56}\)
c) \(\dfrac{7}{36}-\dfrac{8}{9}+\dfrac{-2}{3}\)
\(=\dfrac{7}{36}-\dfrac{32}{36}+\dfrac{-24}{36}\)
\(=\dfrac{7-32+\left(-24\right)}{36}\)
\(=\dfrac{-49}{36}\)
d) \(\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)
\(=\dfrac{-9}{18}+\dfrac{3}{7}-\dfrac{2}{18}+\dfrac{-7}{18}+\dfrac{4}{7}\)
\(=\left(\dfrac{-9}{18}+\dfrac{-7}{18}-\dfrac{2}{18}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)
\(=\left(-1\right)+1\)
\(=0\)
e) \(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)
\(=\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{11}{7}\right)+\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\dfrac{1}{3}\)
\(=2+\left(-1\right)+\dfrac{1}{3}\)
\(=1+\dfrac{1}{3}\)
\(=\dfrac{4}{3}\)
a) -30
b)\(\frac{3}{2}\)
c)\(\frac{153.\left(24-11\right)}{-153}\)=\(\frac{153.13}{-153}\)=\(\frac{13}{-1}\)= -13
a) \(\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}=\frac{\frac{37}{42}}{\frac{-37}{28}}=\frac{37}{42}.\frac{28}{-37}=\frac{-2}{3}\)
b) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0-\frac{1}{2}=\frac{-1}{2}\\x=\left(\frac{2}{3}-0\right):2=\frac{1}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
a,\(\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}=\frac{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}{3.\left(-\frac{1}{3}-\frac{3}{7}+\frac{1}{28}\right)}=\frac{-2}{3}\)
cách này k cần dùng máy tính (hok tốt)
ở chỗ \(7^0\)
phần tử minhf viết thừa số 0 cuối nhé
save me giúp mình nhé