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=>x:x2-5x<0
=>x:x.x-5x<0
=>1.x-5x<0
=>x(1-5)<0
=>x.(-4)<0
=>x là một số nguyên bất kì
Dấu “=” xảy ra khi x=0 (Không chắc sai thì kệ :P)
\(x^2-5x\le0\)
\(\Leftrightarrow x\left(x-5\right)\le0\)
\(TH:x\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
\(TH:x\left(x-5\right)< 0\)
\(\Leftrightarrow\hept{\begin{cases}x\\x-5\end{cases}}\) trái dấu
Mà x > x - 5 nên \(\hept{\begin{cases}x>0\\x-5< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>0\\x< 5\end{cases}}\Leftrightarrow x\in\left\{1;2;3;4\right\}\)
Ta có : \(\left|5x-4\right|=\left|x+2\right|\)
\(\Leftrightarrow\orbr{\begin{cases}5x-4=x+2\\5x-4=-x-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x-x=2+4\\5x+x=-2+4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=6\\6x=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)
b) \(\left|2x-3\right|-\left|3x+2\right|=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}\Rightarrow\orbr{\begin{cases}2x-3x=2+3\\2x+3x=-2+3\end{cases}\Rightarrow}\orbr{\begin{cases}-x=5\\5x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}}\)
c)/2+3x/=/4x-3/
\(\Rightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=-\left(4x-3\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x-4x=-3-2\\3x+4x=3-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-x=-5\\7x=1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}}\)
d)/7x+1/-/5x+6|=0
\(\Rightarrow\left|7x+1\right|=\left|5x+6\right|\)
\(\Rightarrow\orbr{\begin{cases}7x+1=5x+6\\7x+1=-\left(5x+6\right)\end{cases}\Rightarrow\orbr{\begin{cases}7x-5x=6-1\\7x+1=-5x-6\end{cases}\Rightarrow}\orbr{\begin{cases}2x=5\\7x+5x=-6-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{7}{12}\end{cases}}}\)
\(23\left(x-1\right)+19=65\)
\(23\left(x-1\right)=65-19\)
\(23\left(x-1\right)=46\)
\(x-1=46:23\)
\(x-1=2\)
\(x=2+1\)
\(x=3\)
\(5x+3x=88\)
\(x\left(5+3\right)=88\)
\(x.8=88\)
\(x=88:8\)
\(x=11\)
\(x^3=64\)
\(x^3=4^3\)
\(\Rightarrow x=4\)
\(\left(5x-4\right):7-2=6\)
\(\left(5x-4\right):7=6+2\)
\(\left(5x-4\right):7=8\)
\(5x-4=8.7\)
\(5x-4=56\)
\(5x=56+4\)
\(5x=60\)
\(x=60:5\)
\(x=12\)
\(x^{50}=x\)
\(\Rightarrow x=1\)
\(4.2^x-3=125\)
\(4.2^x=125+3\)
\(4.2^x=128\)
\(2^x=128:4\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
k mk nha
`@` `\text {Ans}`
`\downarrow`
`1)`
\(2x+\dfrac{1}{2}=\dfrac{5}{3}\)
`\Rightarrow`\(2x=\dfrac{5}{3}-\dfrac{1}{2}\)
`\Rightarrow`\(2x=\dfrac{7}{6}\)
`\Rightarrow`\(x=\dfrac{7}{6}\div2\)
`\Rightarrow`\(x=\dfrac{7}{12}\)
Vậy, `x = 7/12`
`2)`
\(\dfrac{1}{7}+\dfrac{4}{5}x=\dfrac{5}{3}\)
`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{5}{3}-\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{32}{21}\)
`\Rightarrow`\(x=\dfrac{32}{21}\div\dfrac{4}{5}\)
`\Rightarrow`\(x=\dfrac{40}{21}\)
Vậy, `x = 40/21`
`3)`
\(\dfrac{3}{5}-\dfrac{3}{5}x=\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{3}{5}-\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{16}{35}\)
`\Rightarrow`\(x=\dfrac{16}{35}\div\dfrac{3}{5}\)
`\Rightarrow`\(x=\dfrac{16}{21}\)
Vậy, `x = 16/21`
`4)`
\(\dfrac{5}{6}-3x=\dfrac{3}{4}\)
`\Rightarrow`\(3x=\dfrac{5}{6}-\dfrac{3}{4}\)
`\Rightarrow`\(3x=\dfrac{1}{12}\)
`\Rightarrow`\(x=\dfrac{1}{12}\div3\)
`\Rightarrow`\(x=\dfrac{1}{36}\)
Vậy, `x = 1/36`
`5)`
\(\dfrac{5}{3}-\dfrac{1}{2}x=\dfrac{3}{7}\)
`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{5}{3}-\dfrac{3}{7}\)
`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{26}{21}\)
`\Rightarrow`\(x=\dfrac{26}{21}\div\dfrac{1}{2}\)
`\Rightarrow`\(x=\dfrac{52}{21}\)
Vậy, `x = 52/21`
`6)`
\(5x+\dfrac{1}{2}=\dfrac{2}{3}\)
`\Rightarrow`\(5x=\dfrac{2}{3}-\dfrac{1}{2}\)
`\Rightarrow`\(5x=\dfrac{1}{6}\)
`\Rightarrow`\(x=\dfrac{1}{6}\div5\)
`\Rightarrow`\(x=\dfrac{1}{30}\)
Vậy, `x = 1/30.`
\(5x+1\text{⋮}2x+3\)
\(\text{⇒}2\left(5x+1\right)\text{⋮}2x+3\)
\(\text{⇒}10x+2\text{⋮}2x+3\)
\(\text{⇒}10x+15-13\text{⋮}2x+3\)
\(\text{⇒}5\left(2x+3\right)-13\text{⋮}2x+3\)
\(\text{⇒}13\text{⋮}2x+3\)
\(\text{⇒}2x+3\text{∈}Ư\left(13\right)=\left\{13;-13;1;-1\right\}\)
\(\text{⇒}2x\text{∈ }\left\{10;-16;-2;-4\right\}\)
\(\text{⇒}x\text{∈}\left\{5;-8;-1;-2\right\}\)
Vậy \(x\text{∈}\left\{5;8;-1;-2\right\}\)
a, 100 - 7 ( x - 5 ) = 31 + 33
100 - 7 ( x - 5 ) = 31 + 27
100 - 7 ( x - 5 ) = 58
7 ( x - 5 ) = 100 - 58
7 ( x - 5 ) = 42
x - 5 = 42 : 7
x - 5 = 6
=> x = 6 +5
=> x = 11
Vậy x = 11
b, 12 ( x - 1 ) : 3 = 43 + 23
12 ( x - 1 ) : 3 = 64 + 8
12 ( x - 1 ) : 3 = 72
12 ( x - 1 ) = 72 . 3
12 ( x - 1 ) = 216
x - 1 = 216 : 12
x - 1 = 18
=> x = 18 + 1
=> x = 19
Vậy x = 19
c, 24 + 5x = 75 : 73
24 + 5x = 72
24 + 5x = 49
5x = 49 - 24
5x = 25
=> x = 25 : 5
=> x = 5
Vậy x = 5
d, 5x - 206 = 24 . 4
5x - 206 = 16 . 4
5x - 206 = 64
5x = 64 + 206
5x = 270
=> x = 270 : 5
=> x = 54
Vậy x = 54
e, 125 = x3
53 = x3
=> x = 5
Vậy x = 5
g, 64 = x2
82 = x2
=> x = 8
Vậy x = 8
2 - x - 49 = 5 . 32
2 - x - 49 = 5 . 9
2 - x - 49 = 45
2 - x = 45 + 49
2 - x = 94
x = 2 - 94
x = - 92
\(\frac{5x+4}{2006}+\frac{5x+3}{2007}=\frac{5x+2}{2008}+\frac{5x+1}{2009}\)
\(\Leftrightarrow\frac{5x+4}{2006}+1+\frac{5x+3}{2007}+1=\frac{5x+2}{2008}+1+\frac{5x+1}{2009}+1\)
\(\Leftrightarrow\frac{5x+2010}{2006}+\frac{5x+2010}{2007}=\frac{5x+2010}{2008}+\frac{5x+2010}{2009}\)
\(\Leftrightarrow\left(5x+2010\right)\left(\frac{1}{2006}+\frac{1}{2007}\right)=\left(5x+2010\right)\left(\frac{1}{2008}+\frac{1}{2009}\right)\)
\(\Leftrightarrow5x+2010=0\)
\(\Leftrightarrow5x=-2010\)
\(\Leftrightarrow x=-402\)
2 4 6 và 8