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theo mk nghĩ là bài này áp dụng dãy tỉ số = nhau
a: \(\Leftrightarrow x^2=\dfrac{-5}{2}\cdot\dfrac{-10}{9}=\dfrac{50}{18}=\dfrac{25}{9}\)
=>x=5/3hoặc x=-5/3
c: \(\Leftrightarrow4\left(x-\dfrac{5}{8}\right)=\dfrac{1}{4}+\dfrac{3}{4}=1\)
=>x-5/8=1/4
hay x=2/8+5/8=7/8
d: \(\Leftrightarrow\left|x-3\right|=\dfrac{2}{5}+\dfrac{3}{5}=1\)
=>x-3=1 hoặc x-3=-1
=>x=4 hoặc x=2
e: =>1-1/2x=-3
=>1/2x=4
hay x=8
Tacó :
B = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{9^2}\) \(\Rightarrow\)Đặt D=\(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)<B
\(\Rightarrow\)D= \(\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{3}-.....+\dfrac{1}{9}-\dfrac{1}{10}\) \(\Rightarrow D=\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow D=\dfrac{2}{5}\)
Vì D =\(\dfrac{2}{5}\) =\(\dfrac{2}{5}\)
mà D<B
\(\Rightarrow\)B>\(\dfrac{2}{5}\)(dpcm)
tuyệt đói ko chép mạng thề 100%
Câu 1:
\(A\in Z\Rightarrow6n-1⋮3n+2\)
\(\Rightarrow6n+4-5⋮3n+2\)
\(\Rightarrow2\left(3n+2\right)-5⋮3n+2\)
\(\Rightarrow5⋮3n+2\)
đến đây tự lm nốt nhé
1. Để A có giá trị nguyên thì \(6n-1⋮3n+2\)
Ta có: \(\left\{{}\begin{matrix}6n-1⋮3n+2\\3n+2⋮3n+2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}6n-1⋮3n+2\\2\left(3n+2\right)⋮3n+2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}6n-1⋮3n+2\\6n+4⋮3n+2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}6n-1⋮3n+2\\6n-1+5⋮3n+2\end{matrix}\right.\)
\(\Rightarrow\left(6n-1+5\right)-\left(6n-1\right)⋮3n+2\)
\(\Rightarrow5⋮3n+2\)
\(\Rightarrow3n+2\inƯ\left(5\right)\)
\(\Rightarrow3n+2\in\left\{\pm1;\pm5\right\}\)
\(\Rightarrow3n\in\left\{-7;\pm3;-1;\right\}\)
\(\Rightarrow n\in\left\{\pm1\right\}\)
Vậy để \(A\in Z\) thì n nhận các giá trị là: \(\pm1\)
Bài 1:
a) \(\dfrac{2}{5}\cdot x-\dfrac{1}{4}=\dfrac{1}{10}\)
\(\dfrac{2}{5}\cdot x=\dfrac{1}{10}+\dfrac{1}{4}\)
\(\dfrac{2}{5}\cdot x=\dfrac{7}{20}\)
\(x=\dfrac{7}{20}:\dfrac{2}{5}\)
\(x=\dfrac{7}{8}\)
Vậy \(x=\dfrac{7}{8}\).
b) \(\dfrac{3}{5}=\dfrac{24}{x}\)
\(x=\dfrac{5\cdot24}{3}\)
\(x=40\)
Vậy \(x=40\).
c) \(\left(2x-3\right)^2=16\)
\(\left(2x-3\right)^2=4^2\)
\(\circledast\)TH1: \(2x-3=4\\ 2x=4+3\\ 2x=7\\ x=\dfrac{7}{2}\)
\(\circledast\)TH2: \(2x-3=-4\\ 2x=-4+3\\ 2x=-1\\ x=\dfrac{-1}{2}\)
Vậy \(x\in\left\{\dfrac{7}{2};\dfrac{-1}{2}\right\}\).
Bài 2:
a) \(25\%-4\dfrac{2}{5}+0.3:\dfrac{6}{5}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}:\dfrac{6}{5}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}\cdot\dfrac{5}{6}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{1}{4}\)
\(=\dfrac{5}{20}-\dfrac{88}{20}+\dfrac{5}{20}\)
\(=\dfrac{5-88+5}{20}\)
\(=\dfrac{78}{20}=\dfrac{39}{10}\)
b) \(\left(\dfrac{1}{6}-\dfrac{1}{5^2}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{1}{6}-\dfrac{1}{25}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{1}{6}-\dfrac{1}{5}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{5}{30}-\dfrac{6}{30}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{5-6+1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=0\cdot\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=0\)
Bài 3:
a) \(\dfrac{4}{19}\cdot\dfrac{-3}{7}+\dfrac{-3}{7}\cdot\dfrac{15}{19}\)
\(=\dfrac{-3}{7}\left(\dfrac{4}{19}+\dfrac{15}{19}\right)\)
\(=\dfrac{-3}{7}\cdot1\)
\(=\dfrac{-3}{7}\)
b) \(7\dfrac{5}{9}-\left(2\dfrac{3}{4}+3\dfrac{5}{9}\right)\)
\(=\dfrac{68}{9}-\dfrac{11}{4}-\dfrac{32}{9}\)
\(=\dfrac{68}{9}-\dfrac{32}{9}-\dfrac{11}{4}\)
\(=4-\dfrac{11}{4}\)
\(=\dfrac{16}{4}-\dfrac{11}{4}\)
\(\dfrac{5}{4}\)
Bài 4:
\(\dfrac{4}{12\cdot14}+\dfrac{4}{14\cdot16}+\dfrac{4}{16\cdot18}+...+\dfrac{4}{58\cdot60}\)
\(=2\left(\dfrac{1}{12\cdot14}+\dfrac{1}{14\cdot16}+\dfrac{1}{16\cdot18}+...+\dfrac{1}{58\cdot60}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{18}+...+\dfrac{1}{58}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{5}{60}-\dfrac{1}{60}\right)\)
\(=2\cdot\dfrac{1}{15}\)
\(=\dfrac{2}{15}\)
a, =\(3^4+2^5=81+32=113\)
b, =\(3.\left(4^2-2.3\right)=3.\left(16-6\right)=3.10=30\)
c, =\(\dfrac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}=\dfrac{2^{12}.3^{14}}{2^{12}.3^{12}}=3^2=9\)
d, =\(\dfrac{3^2.7^2.2.7.5^3}{5^3.7^3.2.3}=3\)
e, =\(\dfrac{3^6.5^3.2^8.5^4.2^2.3^4}{2^{10}.3^{10}.5^5}=\dfrac{3^{10}.2^{10}.5^7}{2^{10}.3^{10}.5^5}=5^2=25\)
g, =\(\dfrac{2^5.\left(2^8+1\right)}{2^2.\left(2^8+1\right)}=\dfrac{2^5}{2^2}=2^3=8\)
a)\(\dfrac{3}{10}\)-x=\(\dfrac{25}{30}\)-\(\dfrac{4}{30}\)
\(\dfrac{3}{10}-x=\dfrac{7}{10}\)
x = \(\dfrac{3}{10}-\dfrac{7}{10}\)
x=\(\dfrac{-4}{10}\)
b)\(\dfrac{-5}{8}+x=\dfrac{4}{9}-\dfrac{63}{9}\)
\(\dfrac{-5}{9}+x=\dfrac{-59}{9}\)
\(x=\dfrac{-59}{9}-\dfrac{-5}{9}\)
\(x=\dfrac{-64}{9}\)
c)=>2.18=(x-3).(x-3)
=>36=(x-3)\(^2\)
=>6\(^2\)=(x-3)\(^2\)
6= x-3
x=6+3=9
\(\dfrac{2x+1}{5}-\dfrac{4x-2}{4}=10x\Leftrightarrow\dfrac{4\left(2x+1\right)}{20}-\dfrac{5\left(4x-2\right)}{20}=\dfrac{10x.20}{20}\)
\(\Leftrightarrow4\left(2x+1\right)-5\left(4x-2\right)=200x\)
\(\Leftrightarrow8x+4-20x+10=200x\)
\(\Leftrightarrow8x-20x-200x=-4-10\)
\(\Leftrightarrow-228x=-14\)\(\Leftrightarrow-\dfrac{14}{-228}=\dfrac{7}{114}\)
\(\dfrac{2x+1}{5}-\dfrac{4x-2}{4}=10x\)
\(\rightarrow\dfrac{8x+4}{20}-\dfrac{20x-10}{20}=10x\)
\(\rightarrow\dfrac{\left(8x+4\right)-\left(20x-10\right)}{20}=10x\)
\(\rightarrow\dfrac{8x+4-20x+10}{20}=10x\)
\(\rightarrow\dfrac{\left(8x-20x\right)+\left(4+10\right)}{20}=10x\)
\(\rightarrow\dfrac{-12x+14}{20}=10x\)
\(\rightarrow\left(-12x\right)+14=10x.20\)
\(\rightarrow\left(-12x\right)+14=200x\)
\(\rightarrow14=200x-\left(-12x\right)\)
\(\rightarrow14=200x+12x\)
\(\rightarrow14=\left(200+12\right)x\)
\(\rightarrow14=212x\)
\(\rightarrow14:212=x\)
\(\rightarrow\dfrac{14}{212}=x\)
\(\rightarrow\dfrac{7}{106}=x\)
Vậy ... ... ...