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a)
\(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=255\\ x+x-1+x-2+...+x-50=255\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+50\right)\\ 51x-1275=255\\ 51x=1530\\ x=30\)
e)
\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\\ x+x+1+x+2+...+x+30=1240\\ \left(x+x+x+...+x\right)+\left(1+2+3+...+30\right)=1240\\ 31x+465=1240\\ 31x=775\\ x=25\)
f)
\(\left(x-1\right)+\left(x-2\right)+...+\left(x-19\right)+\left(x-20\right)=-610\\ x-1+x-2+...+x-19+x-20=-610\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+20\right)=-610\\ 20x-210=-610\\ 20x=-400\\ x=-20\)
b) Ta có : \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Rightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}\left(x-\frac{1}{3}\right)^2=\left(\frac{1}{2}\right)^2\\\left(x-\frac{1}{3}\right)^2=\left(-\frac{1}{2}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{2}\\x-\frac{1}{3}=-\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=-\frac{1}{6}\end{cases}}\)
b) \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{4}\\x-\frac{1}{3}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=\frac{1}{12}\end{cases}}\)
d) \(\frac{x+5}{2}=\frac{8}{x+5}\)
\(\Rightarrow\left(x+5\right)^2=16\)
\(\Rightarrow\orbr{\begin{cases}x+5=16\\x+5=-16\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=-21\end{cases}}}\)
a) \(\left(\dfrac{1}{2}x-3\right)\left(-\dfrac{1}{3}+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=0+3\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{1}{2}\\x=0-\left(-\dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)
d) \(9x^2=1\)
\(\Leftrightarrow x^2=1:9\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
\(\Leftrightarrow x^2=\left(\dfrac{1}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
Bài 2:
a) \(\left(x-3\right)^3+27=0\)
\(\Leftrightarrow\left(x-3\right)^3=0-27\)
\(\Leftrightarrow\left(x-3\right)^3=-27\)
\(\Leftrightarrow\left(x-3\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x-3=-3\)
\(\Leftrightarrow x=\left(-3\right)+3\)
\(\Leftrightarrow x=0\)
b) \(-125-\left(x+1\right)^3=0\)
\(\Leftrightarrow\left(x+1\right)^3=-125-0\)
\(\Leftrightarrow\left(x+1\right)^3=-125\)
\(\Leftrightarrow\left(x+1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow x+1=-5\)
\(\Leftrightarrow x=\left(-5\right)-1\)
\(\Leftrightarrow x=-6\)
c) \(\left(2x-\dfrac{1}{4}\right)^2-\dfrac{1}{16}=0\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=0+\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Leftrightarrow2x-\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{4}+\dfrac{1}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{2}:2\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
d) \(2^x+2^{x+1}=24\)
\(\Leftrightarrow2^x+2^x.2=24\)
\(\Leftrightarrow2^x\left(1+2\right)=24\)
\(\Leftrightarrow2^x.3=24\)
\(\Leftrightarrow2^x=24:3\)
\(\Leftrightarrow2^x=8\)
\(\Leftrightarrow2^x=2^3\)
\(\Rightarrow x=3\)
e) \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=1\)
\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=1+\dfrac{1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=-\dfrac{3}{2}\\x+\dfrac{1}{5}=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)
g) \(\left|x-3\right|+2x=10\)
\(\Leftrightarrow\left|x-3\right|=10-2x\)
\(\Leftrightarrow\left|x-3\right|=2.5-2x\)
\(\Leftrightarrow\left|x-3\right|=2\left(5-x\right)\)
(không chắc có nên làm tiếp câu g không, thấy đề cứ là lạ, có j sai sai...)
Bài 1:
a) \(2^7+2^9⋮10\)
Ta có: \(2^7+2^9=2^{4.1}.2^3+2^{4.2}.2\)
\(\Leftrightarrow\overline{A6}.2^3+\overline{B6}.2\)
\(\Leftrightarrow\overline{A6}.8+\overline{B6}.2\)
\(\Leftrightarrow\overline{C8}+\overline{D2}\)
\(\Leftrightarrow\overline{E0}\)
Mà \(\overline{E0}⋮10\) \(\Rightarrow2^7+2^9⋮10\)
b) \(8^{24}.25^{10}⋮2^{36}.5^{20}\)
Ta có: \(8^{24}.25^{10}=\left(2^3\right)^{24}.\left(5^2\right)^{10}\)
\(\Leftrightarrow2^{72}.5^{20}\)
Do \(2^{72}⋮2^{36}\) và \(5^{20}⋮5^{20}\) \(\Rightarrow8^{24}.25^{10}⋮2^{36}.5^{20}\)
c) \(3^{10}+3^{12}⋮30\)
Ta có: \(3^{10}+3^{12}=3^{4.2}.3^2+3^{4.3}\)
\(\Leftrightarrow\overline{A1}.3^2+\overline{B1}\)
\(\Leftrightarrow\overline{A1}.9+\overline{B1}\)
\(\Leftrightarrow\overline{C9}+\overline{B1}\)
\(\Leftrightarrow\overline{D0}⋮10\)
(Chứng minh chia hết cho 10 rồi chứng minh chia hết cho 3, mình chưa tìm được cách làm, chờ chút)
a: \(\Leftrightarrow\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{-3}{2}x+\dfrac{3}{4}\)
=>13/6x=3/4+5/6
=>13/6x=9/12+10/12=19/12
hay x=19/26
b: \(\left(5x-3\right)\left(2x+5\right)=0\)
=>5x-3=0 hoặc 2x+5=0
=>x=3/5 hoặc x=-5/2
c: \(\left(\dfrac{5}{6}:x-\dfrac{5}{4}\right)^4=\dfrac{81}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{6}:x-\dfrac{5}{4}=\dfrac{3}{2}\\\dfrac{5}{6}:x-\dfrac{5}{4}=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{6}:x=\dfrac{11}{4}\\\dfrac{5}{6}:x=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{33}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
d: \(\left|\dfrac{2}{5}x-\dfrac{1}{5}\right|\cdot\dfrac{5}{4}-2=\dfrac{3}{2}\)
\(\Leftrightarrow\left|\dfrac{2}{5}x-\dfrac{1}{5}\right|\cdot\dfrac{5}{4}=\dfrac{3}{2}+2=\dfrac{7}{2}\)
\(\Leftrightarrow\left|\dfrac{2}{5}x-\dfrac{1}{5}\right|=\dfrac{7}{2}:\dfrac{5}{4}=\dfrac{7}{2}\cdot\dfrac{4}{5}=\dfrac{28}{10}=\dfrac{14}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}x-\dfrac{1}{5}=\dfrac{14}{5}\\\dfrac{2}{5}x-\dfrac{1}{5}=-\dfrac{14}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{2}{5}=\dfrac{15}{2}\\x=-\dfrac{13}{5}:\dfrac{2}{5}=\dfrac{-13}{2}\end{matrix}\right.\)
\(a,\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\Leftrightarrow\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}\Leftrightarrow\dfrac{2}{3}x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{2}{3}:\dfrac{2}{3}=1\)\(b,5\dfrac{4}{7}:x=13\Leftrightarrow\dfrac{39}{7}:x=13\Leftrightarrow x=\dfrac{39}{7}:13=\dfrac{3}{7}\)\(c,\left(2\dfrac{4}{5}x-50\right):\dfrac{2}{3}=51\Leftrightarrow\left(\dfrac{14}{5}x-50\right).\dfrac{3}{2}=51\Leftrightarrow\dfrac{21}{5}x-75=51\Leftrightarrow\dfrac{21}{5}x=51+75=126\Leftrightarrow x=126:\dfrac{21}{5}=30\)
d,\(\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{-1}{2}\\2x=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Bài 1:
a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\)
=>x+4/15=8/5 hoặc x+4/15=-8/5
=>x=4/3 hoặc x=-28/15
b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c: \(\Leftrightarrow\left|x-1\right|-1=1\)
=>|x-1|=2
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
Bài 2:
b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)
Bài 3:
a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\)
Dấu '=' xảy ra khi x=-15/19
b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\)
Dấu '=' xảy ra khi x=4/7
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
a)
Ta có: 5 = 2+3; 9 = 4+5 ; 13 = 6+7 ; 17 = 8+9;.....
Do vậy x = a + ( a + 1) ( a thuộc N )
Nên 1 + 5 + 9 + 13 + 16 + ....+ x = 1+2+3+4+5+6+7+.....+a+ ( a + 1 ) = 501501
Hay (a + 1)( a + 2) = 1003002 = 1001 . 1002
Suy ra : a = 1000
Do đó : x = 1000 + ( 1000+ 1) = 2001