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B = .................
Xét thừa số 63.1,2 - 21.3,6 = 0 nên B = 0
\(C=\left|\dfrac{4}{9}-\left(\dfrac{\sqrt{2}}{2}\right)^2\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{6}{7}}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{2\left(\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}\right)}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{1}{2}\right|=\dfrac{1}{18}+\dfrac{9}{10}=\dfrac{43}{45}\)
Mình làm câu 1,2 trước, câu 3 sau
Câu 1:
\(\sqrt{x^2}=0\)
=> \(\left(\sqrt{x^2}\right)^2=0^2\)
\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)
Câu 2:
\(A=\left(0,75-0,6+\dfrac{3}{7}+\dfrac{3}{12}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+2,75-2,2\right)\)
\(A=\left(\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)\cdot11\left(\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=33\cdot\dfrac{491}{1820}\cdot\dfrac{221}{420}=\dfrac{3580863}{764400}\)
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
câu 1 \(A=\dfrac{3^2}{5^2}.5^2-\dfrac{9^3}{4^3}:\dfrac{3^3}{4^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{\left(3^2\right)^3}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{3^6}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}=3^2-3^3+\dfrac{1}{2}=-18+\dfrac{1}{2}=-\dfrac{35}{2}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{22}.2\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{4^4}{8^2}\right)^{2009}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{11}\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{\left(2^2\right)^4}{\left(2^3\right)^2}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{1}{2^2}.\dfrac{2^8}{2^6}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{2^8}{2^8}\right)^{2009}\)
\(B=1^{2010}-1^{2009}=1-1=0\)
câu 2
a) \(2x-\dfrac{5}{4}=\dfrac{20}{15}\)
\(\Leftrightarrow2x=\dfrac{4}{3}+\dfrac{5}{4}\)
\(\Leftrightarrow2x=\dfrac{31}{12}\)
\(\Leftrightarrow x=\dfrac{31}{24}\)
b) \(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
a, \(\dfrac{3}{7}+\dfrac{4}{7}x=\dfrac{1}{3}\)
\(\Rightarrow\) \(\dfrac{4}{7}x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\Rightarrow\) \(\dfrac{4}{7}x=\dfrac{-2}{21}\)
\(\Rightarrow x=\dfrac{-2}{21}:\dfrac{4}{7}\)
\(\Rightarrow x=\dfrac{-1}{6}\)
b, \(25-\left(5-x\right)=-7\)
\(\Rightarrow\) \(5-x=25-\left(-7\right)\)
\(\Rightarrow5-x=32\)
\(\Rightarrow x=5-32\)
\(\Rightarrow x=-27\)
c, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}\)
\(\Rightarrow x=\dfrac{-5}{7}\)
d, \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\) \(\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0:2\\x=0+\dfrac{1}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
e, \(\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|-7=-3\)
\(\Rightarrow\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|=-3+7\)
\(\Rightarrow\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|=4\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{4}=4\\\dfrac{1}{2}x-\dfrac{3}{4}=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=4+\dfrac{3}{4}\\\dfrac{1}{2}x=-4+\dfrac{3}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{19}{4}\\\dfrac{1}{2}x=\dfrac{-13}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{4}:\dfrac{1}{2}\\x=\dfrac{-13}{4}:\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{2}\\x=\dfrac{-13}{2}\end{matrix}\right.\)
a)\(\dfrac{3}{7}+\dfrac{4}{7}x=\dfrac{1}{3}\)
\(\dfrac{4}{7}x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\dfrac{4}{7}x=\dfrac{-2}{21}\)
\(x=\dfrac{-2}{21}:\dfrac{4}{7}\)
\(x=\dfrac{-1}{6}\)
b)\(25-\left(5-x\right)=-7\)
\(5-x=25-\left(-7\right)\)
\(5-x=32\)
x= -27
c)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)
\(x=\dfrac{-5}{7}\)
d)\(2x\left(x-\dfrac{1}{7}\right)=0\)
⇒\(\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
e)\(|\dfrac{1}{2}x-\dfrac{3}{7}|-7=-3\)
\(\left|\dfrac{1}{2}x-\dfrac{3}{7}\right|=-3+7\)
\(\left|\dfrac{1}{2}x-\dfrac{3}{7}\right|=4\)
⇒\(\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{4}=4\\\dfrac{1}{2}x-\dfrac{3}{4}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=4\dfrac{3}{4}\Rightarrow x=9\dfrac{1}{2}=\dfrac{19}{2}\\\dfrac{1}{2}x=-3\dfrac{1}{4}\Rightarrow x=\dfrac{-13}{2}\end{matrix}\right.\)
a,
\(\dfrac{1}{4}x-1+\dfrac{1}{3}\left(\dfrac{5}{2}x-7\right)-\left(\dfrac{5}{8}x-2\right)=\dfrac{7}{2}\)
\(\Rightarrow\dfrac{1}{4}x-1+\dfrac{5}{6}x-\dfrac{7}{3}-\dfrac{5}{8}x+2=\dfrac{7}{2}\)
\(\Rightarrow\dfrac{1}{4}x+\dfrac{5}{6}x-\dfrac{5}{8}x=\dfrac{7}{2}+1+\dfrac{7}{3}-2\)
\(\Rightarrow\dfrac{11}{24}x=\dfrac{29}{6}\)
\(\Rightarrow x=\dfrac{116}{11}\)
b,
\(\left|2-\dfrac{3}{2}x\right|-4=x+2\)
\(\Rightarrow\left|2-\dfrac{3}{2}x\right|=x-2\)
\(\Rightarrow\left[{}\begin{matrix}2-\dfrac{3}{2}x=x+2\\2-\dfrac{3}{2}x=-\left(x+2\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2-\dfrac{3}{2}x=x+2\\2-\dfrac{3}{2}x=-x-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2-2=x+\dfrac{3}{2}x\\2+2=-x+\dfrac{3}{2}x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{5}{2}x=0\\\dfrac{1}{2}x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
c,
\(-3\left(\dfrac{2}{5}x-\dfrac{1}{5}\right)-x\left(x-\dfrac{1}{2}\right)=\dfrac{1}{6}-x^2\)
\(\Rightarrow-\dfrac{6}{5}x+\dfrac{3}{5}-x^2+\dfrac{1}{2}x=\dfrac{1}{6}-x^2\)
\(\Rightarrow-\dfrac{7}{10}x=\dfrac{1}{6}-\dfrac{3}{5}-x^2+x^2\)
\(\Rightarrow-\dfrac{7}{10}x=-\dfrac{13}{30}\Leftrightarrow x=\dfrac{13}{21}\)
Bài 2:
\(\left(\dfrac{2}{5}\right)^x>\left(\dfrac{5}{2}\right)^{-3}.\left(\dfrac{-2}{5}\right)^2\)
\(\Rightarrow\left(\dfrac{2}{5}\right)^x>\left(\dfrac{2}{5}\right)^3.\left(\dfrac{2}{5}\right)^2\)
\(\Rightarrow\left(\dfrac{2}{5}\right)^x>\left(\dfrac{2}{5}\right)^5\)
Vì \(\dfrac{2}{5}\ne\pm1;\dfrac{2}{5}\ne0\) nên \(x>5\)
Vậy \(x>5\) thoả mãn yêu cầu đề bài.
Chúc bạn học tốt!!!
Bài 1:
\(C=\left(\dfrac{1}{2^2-1}\right)\left(\dfrac{1}{3^2-1}\right).....\left(\dfrac{1}{100^2-1}\right)\)
\(C=\left(\dfrac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(\dfrac{1}{\left(3-1\right)\left(3+1\right)}\right).....\left(\dfrac{1}{\left(100-1\right)\left(100+1\right)}\right)\)
\(C=\dfrac{1}{1.3}\dfrac{1}{2.4}.....\dfrac{1}{99.101}=\dfrac{1}{101!}\)
Chúc bạn học tốt!!!
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
Bài 1:
\(\dfrac{\left(\dfrac{2}{5}\right)^7\cdot5^7+\left(\dfrac{9}{4}\right)^3:\left(\dfrac{3}{16}\right)^3}{2^7\cdot5^2+512}\)
\(=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}:\dfrac{3}{16}\right)^3}{2^7\cdot5^2+512}\)
\(=\dfrac{2^7+12^3}{2^7\cdot5^2+512}\)
\(=\dfrac{1856}{3712}\)
\(=0,5\)
Bài 2:
\(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Rightarrow5x+1=\dfrac{6}{7}\)
\(\Rightarrow5x=\dfrac{-1}{7}\)
\(\Rightarrow x=\dfrac{-1}{35}\)
a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)
=>x+1=0
hay x=-1
b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
\(\Leftrightarrow\left(\dfrac{13}{4}-x\right)\cdot\dfrac{101}{25}-\dfrac{1213}{100}=2\cdot\left[\left(x-\dfrac{10}{7}\right)\cdot\dfrac{49}{50}+\dfrac{2}{5}\right]\)
\(\Leftrightarrow\left(\dfrac{13}{4}-x\right)\cdot\dfrac{101}{25}=\dfrac{49}{25}\left(x-\dfrac{10}{7}\right)+\dfrac{4}{5}+\dfrac{1213}{100}\)
\(\Leftrightarrow\dfrac{1313}{100}-\dfrac{101}{25}x=\dfrac{49}{25}x-\dfrac{490}{175}+\dfrac{1293}{100}\)
=>-6x=13/5
hay x=-13/30