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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: =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5
=>x=3/5:2/3=3/5x3/2=9/10
b: \(\Leftrightarrow x\cdot2.8-50=34\)
=>2,8x=84
=>x=30
c: \(\Leftrightarrow\dfrac{1}{6}x=\dfrac{5}{12}\)
hay x=5/2
d: \(\Leftrightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}+\dfrac{7}{4}=\dfrac{41}{4}\)
=>2x-3/4=41/4 hoặc 2x-3/4=-41/4
=>2x=44/4=11 hoặc 2x=-19/2
=>x=11/2 hoặc x=-19/4
Tìm x :
a) \(x-5=49:7\)
\(\Leftrightarrow x-5=7\)
\(\Leftrightarrow x=7+5\)
\(\Leftrightarrow x=12\)
Vậy : \(x=12\)
b) \(2x+6=24\)
\(\Leftrightarrow2x=24-6=18\)
\(\Leftrightarrow x=18:2\)
\(\Leftrightarrow x=9\)
Vậy : \(x=9\)
c) \(\frac{1}{3}:x+\frac{1}{2}=5\)
\(\Leftrightarrow\frac{1}{3}:x=5-\frac{1}{2}=\frac{9}{5}\)
\(\Leftrightarrow x=\frac{1}{3}:\frac{9}{5}\)
\(\Leftrightarrow x=\frac{5}{27}\)
Vậy : \(x=\frac{5}{27}\)
d) \(\frac{1}{6}.x-\frac{1}{3}=2\)
\(\Leftrightarrow\frac{1}{6}.x=2-\frac{1}{3}=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{5}{3}:\frac{1}{6}\)
\(\Leftrightarrow x=10\)
Vậy : \(x=10\)
e) \(\frac{x}{27}=\frac{3}{x}\)
\(\Leftrightarrow x.x=27.3\)
\(\Leftrightarrow x^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=9\end{matrix}\right.\) mà \(x\in N\)
\(\Rightarrow x=9\)
Vậy : \(x=9\)
g) \(1200:24-\left(17-x\right)=36\)
\(\Leftrightarrow50-17+x=36\)
\(\Leftrightarrow33+x=36\)
\(\Leftrightarrow x=36-33\)
\(\Leftrightarrow x=3\)
Vậy : \(x=3\)
h) \(674-\left(12+x\right)=427\)
\(\Leftrightarrow12+x=674-427=247\)
\(\Leftrightarrow x=247-12\)
\(\Leftrightarrow x=230\)
Vậy : \(x=230\)
k) \(36.\left(x-9\right)=900\)
\(\Leftrightarrow x-9=900:36\)
\(\Leftrightarrow x-9=25\)
\(\Leftrightarrow x=25+9\)
\(\Leftrightarrow x=34\)
m) \(1,2:x+3,8:x=2,5\)
\(\Leftrightarrow\left(1,2-3,8\right):x=2,5\)
\(\Leftrightarrow-2,6:x=2,5\)
\(\Leftrightarrow x=\frac{-2,6}{2,5}=-\frac{26}{25}\)
Vậy : \(x=-\frac{26}{25}\)
n) \(\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right).x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right).x=\frac{1}{3}\)
\(\Leftrightarrow\left[\frac{1}{2}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\right].x=\frac{1}{3}\)
\(\Leftrightarrow\left[\frac{1}{2}.\left(1-\frac{1}{10}\right)\right].x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}.\frac{9}{10}.x=\frac{1}{3}\)
\(\Leftrightarrow\frac{9}{20}.x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}:\frac{9}{20}\)
\(\Leftrightarrow x=\frac{20}{27}\)
Vậy : \(x=\frac{20}{27}\)
b) \(-2\dfrac{2}{3}:x=1\dfrac{7}{9}:0.8\\ -\dfrac{8}{3}:x=\dfrac{16}{9}:\dfrac{4}{5}\\ -\dfrac{8}{3}:x=\dfrac{16}{9}.\dfrac{5}{4}\\ -\dfrac{8}{3}:x=\dfrac{20}{9}\\ x=-\dfrac{8}{3}:\dfrac{20}{9}\\ x=-\dfrac{8}{3}.\dfrac{9}{20}\\ x=-\dfrac{6}{5}\)
\(2\dfrac{2}{3}x+3\dfrac{3}{4}x=\dfrac{385}{24}\)
\(\dfrac{8}{3}x+\dfrac{15}{4}x=\dfrac{385}{24}\)
\(x\left(\dfrac{8}{3}+\dfrac{15}{4}\right)=\dfrac{385}{24}\)
\(x\left(\dfrac{32}{12}+\dfrac{45}{12}\right)=\dfrac{385}{24}\)
\(x\left(\dfrac{77}{12}\right)=\dfrac{385}{24}\)
\(x=\dfrac{385}{24}:\dfrac{77}{12}\)
\(x=\dfrac{385}{24}.\dfrac{12}{77}\)
<Bạn tự viết phép tính nha>
\(x=\dfrac{5}{2}\)
Vậy \(x=\dfrac{5}{2}\)
Chúc bn hk tốt!!
a: \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}\cdot x=\dfrac{16}{5}\)
=>2/5x=8/5
=>x=4
b: \(\Leftrightarrow\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{39}-\dfrac{1}{40}\right)\cdot120+\dfrac{1}{3}x=-4\)
\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-4\)
=>1/3x=-6
=>x=-18
c: =>2|x-1/3|=0,24-4/5=-0,56<0
Các bạn ơi giúp mk với 
các bạn ơi mk sắp phải đi học rồi giúp mk với
a) \(\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{36}\) \(\Leftrightarrow\left(\dfrac{-1}{6}\right)^{x-1}=\dfrac{1}{36}\)
\(\Leftrightarrow\left(\dfrac{-1}{6}\right)^{x-1}=\left(\dfrac{1}{6}\right)^2\)
\(\Leftrightarrow x-1=2\Rightarrow x=3\)
b) \(\dfrac{25}{5^x}=\dfrac{1}{125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{3125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{5^5}\Rightarrow x=5\)
a) \(\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{36}\Leftrightarrow\left(-\dfrac{1}{6}\right)^{x-1}=\left(-\dfrac{1}{6}\right)^2\)
\(\Leftrightarrow x-1=2\Rightarrow x=2+1=3\)
b) \(\dfrac{25}{5^x}=\dfrac{1}{125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{3125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{5^5}\Rightarrow x=5\)
Giờ mới đúng thật nè
3) \(\left(x+\dfrac{1}{5}\right)^2\) + \(\dfrac{17}{25}\) = \(\dfrac{26}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{26}{25}\) - \(\dfrac{17}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{9}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{3}{5}.\dfrac{3}{5}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\left(\dfrac{3}{5}\right)^2\)
=> \(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\)
=> \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\)
=> \(x\) = \(\dfrac{2}{5}\)
4) -1\(\dfrac{5}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-24}{27}\)
=> \(\dfrac{-32}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-32}{27}\) - \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{27}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\left(\dfrac{-2}{3}\right)^3\)
=> \(3x-\dfrac{7}{9}=\dfrac{-2}{3}\)
=> \(3x=\dfrac{-2}{3}+\dfrac{7}{9}\)
=> \(3x=\dfrac{1}{9}\)
=> \(x=\dfrac{1}{9}:3\)
=> \(x=\dfrac{1}{27}\)
\(\left(\frac{x}{10}-\frac{2}{3}\right)^2-\frac{1}{25}=0\)
\(\Leftrightarrow\left(\frac{x}{10}-\frac{2}{3}\right)^2=\frac{1}{25}\)
\(\Leftrightarrow\left(\frac{x}{10}-\frac{2}{3}\right)^2=\left(\pm\frac{1}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{x}{10}-\frac{2}{3}=\frac{1}{5}\\\frac{x}{10}-\frac{2}{3}=-\frac{1}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{26}{3}\\x=\frac{14}{3}\end{cases}}\)
a) \(1200:24-\left(17-x\right)=36\)
\(50-\left(17-4\right)=36\)
\(50-\left(17-x\right)=36\)
\(17-x=50-36\)
\(17-x=14\)
\(x=17-14\)
\(x=3\)
Vậy \(x=3\) là giá trị cần tìm
b) \(x+\dfrac{1}{2}.x-25\%.x=10\)
\(x+\dfrac{1}{2}x-\dfrac{25}{100}x=10\)
\(x+\dfrac{1}{2}x-\dfrac{1}{4}x=10\)
\(x\left(1+\dfrac{1}{2}-\dfrac{1}{4}\right)=10\)
\(x\left(1+\dfrac{1}{2}+\dfrac{-1}{4}\right)=10\)
\(x.\dfrac{5}{4}=10\)
\(x=10:\dfrac{5}{4}\)
\(x=8\)
Vậy \(x=8\) là giá trị cần tìm