
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


dài :vv
a) \(\left|2x-5\right|=4\Leftrightarrow\hept{\begin{cases}2x-5=4\\2x-5=-4\end{cases}\Leftrightarrow\hept{\begin{cases}2x=9\\2x=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{9}{2}\\x=\frac{1}{2}\end{cases}}}\)
b) \(\frac{1}{3}-\left|\frac{5}{4}-2x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left|\frac{5}{4}-2x\right|=\frac{1}{12}\Leftrightarrow\hept{\begin{cases}\frac{5}{4}-2x=\frac{1}{12}\\\frac{5}{4}-2x=-\frac{1}{12}\end{cases}\Leftrightarrow\hept{\begin{cases}2x=\frac{7}{6}\\2x=\frac{4}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{7}{12}\\x=\frac{2}{3}\end{cases}}}\)
Bài 1 :
a) \(|2x-5|=4\)
\(\Rightarrow\orbr{\begin{cases}2x-5=4\\2x-5=-4\end{cases}\Rightarrow}\orbr{\begin{cases}2x=9\\2x=1\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x=\frac{1}{2}\end{cases}}}\)
b) \(\frac{1}{3}-\left|\frac{5}{4}-2x\right|=\frac{1}{4}\)
\(\Rightarrow\left|\frac{5}{4}-2x\right|=\frac{1}{12}\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{12}\\\frac{5}{4}-2x=-\frac{1}{12}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=\frac{7}{6}\\2x=\frac{4}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=\frac{2}{3}\end{cases}}}\)
c) \(\left|\frac{-2}{3}\right|+\left|x-\frac{1}{3}\right|=\left|-1\right|-\left|\frac{-1}{3}\right|\)
\(\Rightarrow\frac{2}{3}+\left|x-\frac{1}{3}\right|=1-\frac{1}{3}\)
\(\Rightarrow\frac{2}{3}+\left|x-\frac{1}{3}\right|=\frac{2}{3}\)
\(\Rightarrow\left|x-\frac{1}{3}\right|=0\Rightarrow x-\frac{1}{3}=0\Rightarrow x=\frac{1}{3}\)
d) \(\left|-\frac{1}{2}\right|-\left|x+\frac{1}{4}\right|=\left|-\frac{3}{4}\right|\)
\(\Rightarrow\frac{1}{2}-\left|x+\frac{1}{4}\right|=\frac{3}{4}\)
\(\Rightarrow\left|x+\frac{1}{4}\right|=-\frac{1}{4}\)
Vì \(\left|x\right|\ge0\Rightarrow\)ko có gtri nào của x thỏa mãn đề bài
Bài 2 :
a) \(\left|x-1\right|=3x+2\)
\(\Rightarrow\orbr{\begin{cases}x-1=3x+2\\x-1=-3x-2\end{cases}\Rightarrow\orbr{\begin{cases}x-3x=2+1\\x+3x=-2+1\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}-2x=3\\4x=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{-1}{4}\end{cases}}\)
b|) \(\left|9+x\right|=2x\Rightarrow\orbr{\begin{cases}9+x=2x\\9+x=-2x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-2x=-9\\x+2x=-9\end{cases}\Rightarrow\orbr{\begin{cases}-x=-9\\3x=-9\end{cases}\Rightarrow}\orbr{\begin{cases}x=9\\x=-3\end{cases}}}\)
c) \(\left|x+6\right|-9=2x\Rightarrow\left|x+6\right|=2x+9\)
\(\Rightarrow\orbr{\begin{cases}x+6=2x+9\\x+6=-2x-9\end{cases}\Rightarrow}\orbr{\begin{cases}x-2x=9-6\\x+2x=-9-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-x=3\\3x=-15\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}}\)
Cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt ^^

Bài giải:
a) Ta có: \(-8⋮x\) và \(12⋮x\Rightarrow x\inƯC\left(-8;12\right)\)
\(Ư\left(-8\right)=\left\{-1;-2;-4;-8;1;2;4;8\right\}\)
\(Ư\left(12\right)=\left\{-1;-2;-3;-4;-6;-12;1;2;3;4;6;12\right\}\)
\(\RightarrowƯC\left(-8;12\right)=\left\{-1;-2;-4;1;2;4\right\}\)
Vậy: \(x\in\left\{-1;-2;-4;1;2;4\right\}\)
b)

a) \(\left(6\frac{2}{7}x+\frac{3}{7}\right)\cdot\frac{11}{5}-\frac{3}{7}=-2\)
=> \(\left(\frac{44}{7}x+\frac{3}{7}\right)\cdot\frac{11}{5}=-\frac{11}{7}\)
=> \(\frac{44}{7}x+\frac{3}{7}=-\frac{5}{7}\)
=> \(\frac{44}{7}x=-\frac{8}{7}\)
=> \(\frac{44x}{7}=-\frac{8}{7}\)
=> 44x = -8 => 11x = -2 => \(x=-\frac{2}{11}\)
b) \(3\frac{1}{4}x+\left(-\frac{7}{6}\right)-1\frac{2}{3}=\frac{5}{12}\)
=> \(\frac{13}{4}x-\frac{7}{6}-1\frac{2}{3}=\frac{5}{12}\)
=> \(\frac{13}{4}x-\frac{7}{6}=\frac{25}{12}\)
=> \(\frac{13}{4}x=\frac{13}{4}\)
=> x = 1
c) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
=> \(\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{1}{3}\end{cases}}\)
d) \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
=> \(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)
=> \(\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)
e) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=\frac{-24}{27}\)
=> \(\left(3x-\frac{7}{9}\right)^3=-1\frac{5}{27}-\left(-\frac{24}{27}\right)=-\frac{32}{27}+\frac{24}{27}=-\frac{8}{27}\)
=> \(\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
=> \(3x-\frac{7}{9}=-\frac{2}{3}\)
=> \(x=\frac{-\frac{2}{3}+\frac{7}{9}}{3}=\frac{1}{27}\)
g) \(\frac{x}{1\cdot2}+\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+...+\frac{x}{99\cdot100}=\frac{99}{100}\)
=> \(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{99}-\frac{x}{100}=\frac{99}{100}\)
=> \(\frac{x}{1}-\frac{x}{100}=\frac{99}{100}\)
=> \(\frac{100x-x}{100}=\frac{99}{100}\)
=> \(\frac{99x}{100}=\frac{99}{100}\)
=> x = 1
h) \(\frac{x}{3}+\frac{x}{6}+\frac{x}{10}+\frac{x}{15}=3x-1\)
=> \(\frac{2x}{6}+\frac{2x}{12}+\frac{2x}{20}+\frac{2x}{30}=3x-1\)
=> \(\frac{2x}{2\cdot3}+\frac{2x}{3\cdot4}+\frac{2x}{4\cdot5}+\frac{2x}{5\cdot6}=3x-1\)
=> \(2\left(\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+\frac{x}{4\cdot5}+\frac{x}{5\cdot6}\right)=3x-1\)
=> \(2\left(\frac{x}{2}-\frac{x}{6}\right)=3x-1\)
=> \(2\left(\frac{3x}{6}-\frac{x}{6}\right)=3x-1\)
=> \(2\cdot\frac{2x}{6}=3x-1\)
=> \(\frac{x}{3}=\frac{3x-1}{2}\)
=> 2x = 3(3x - 1)
=> 2x - 9x + 3 = 0
=> -7x = -3
=> x = 3/7

a) Ta có: \(x⋮4;x⋮7;x⋮8\Rightarrow x\in BCNN\left(4;7;8\right)=56\)
b) Tương tự câu a
c)Ta có: \(x\in BC\left(9;8\right)\) và x nhỏ nhất
\(\Rightarrow x\in BCNN\left(9;8\right)=72\)
d) Ta có: \(B\left(6\right)=\left\{0;6;12;18;24;30;36;42;48;54;....\right\}\)
\(B\left(4\right)=\left\{0;4;8;12;16;20;24;28;32;36;40;44;48;52;...\right\}\)
Mà \(16\le x< 50\Rightarrow x=\left\{24;36;48\right\}\)
e;f;g;h Tương tự

a) 45 ⋮ x
Vì 45 ⋮ x nên x E Ư( 45 )
= { 1;3;5;9;15;45 }
mà x E Ư(45)
=> x E { 1;3;5;9;15;45 }
b) 24 ⋮ x ; 36 ⋮ x ; 160 ⋮ x và x lớn nhất
Vì 24 ⋮ x ; 36 ⋮ x ; 160 ⋮ x nên x E ƯC ( 24;36;160)
mà x lớn nhất
=> x E ƯCLN ( 24;36;160 )
Ta có
24 = 23 . 3
36 = 22.32
160 = 25 . 5
=> ƯCLN ( 24;36;160 ) = 22 = 4

a) 3x = 81 b) 2x . 16 = 128 c) 3x : 9 = 27 d) x4 = x
3x = 34 2x = 128 : 16 3x = 27 : 9 => x = 1
=> x = 4 2x = 8 3x = 3 Vậy x= 1
Vậy x = 4 x = 4 => x = 1
Vậy x = 4 Vậy x = 1
e) ( 2x + 1 )3 = 27
( 2x + 1 )3 = 33
=> 2x + 1 = 3
2x = 3 - 1
2x = 2
x = 1
Vậy x = 1
a, 3x=81 b, 2x*16=128 c, 3x:9=27 d, x4=x
=> 3x=34 => 2x=128:16 => 3x=27.9 => x=0 hoặc x=1
=> x=4 => 2x=8 => 3x=243
=> x=4 => 3x=35
=> x=5
e, (2x+1)3=27 f, (x-2)2=(x-2)4
=> (2x+1)3=33 +, TH1: x-2=1 => (x-2)2=(x-2)4=1 => x-2=x-2=1 => x=3
=> 2x+1=3 +, TH2: x-2=0 => (x-2)2=(x-2)4=0 => x-2=x-2=0 => x=2
=> 2x=2
=> x=2:2
=> x=1
g, 25 <= 5x < 625
=> 52 <= 5x < 54
=> x={2;3}
\(\frac27-\frac89x=\frac23\)
\(\frac89x=\frac27-\frac23=-\frac{8}{21}\)
\(x=-\frac{8}{21}:\frac89\)
\(x=-\frac{8}{21}\cdot\frac98=-\frac37\)
8/9.x=2/7-2/3
8/9.x=14/21-6/21
8/9.x=8/21
x=8/9;8/21
x=8/9.21/8
x=21/9
x=7/3