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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
\(\Rightarrow\left|7x+1\right|=\left|5x+6\right|\)
\(\Rightarrow\orbr{\begin{cases}7x+1=5x+6\\7x+1=-\left(5x+6\right)\end{cases}\Rightarrow\orbr{\begin{cases}7x-5x=6-1\\7x+1=-5x-6\end{cases}\Rightarrow}\orbr{\begin{cases}2x=5\\7x+5x=-6-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{7}{12}\end{cases}}}\)
1. Giải:
Do \(5x+13B\in\left(2x+1\right)\Rightarrow5x+13⋮2x+1.\)
\(\Rightarrow2\left(5x+13\right)⋮2x+1\Rightarrow10x+26⋮2x+1.\)
\(\Rightarrow5\left(2x+1\right)+21⋮2x+1.\)
Do 5(2x+1)⋮2x+1⇒ Ta cần 21⋮2x+1.
⇒ 2x+1 ϵ B(21)=\(\left\{1;3;7;21\right\}.\)
Ta có bảng:
2x+1 | 1 | 3 | 7 | 21 |
x | 0 | 1 | 3 | 10 |
TM | TM | TM | TM |
Vậy xϵ\(\left\{0;1;3;10\right\}.\)
2. Giải:
Do (2x-18).(3x+12)=0.
⇒ 2x-18=0 hoặc 3x+12=0.
⇒ 2x =18 3x =-12.
⇒ x =9 x =-4.
Vậy xϵ\(\left\{-4;9\right\}.\)
3. S= 1-2-3+4+5-6-7+8+...+2021-2022-2023+2024+2025.
S= (1-2-3+4)+(5-6-7+8)+...+(2021-2022-2023+2024)+2025 Có 506 cặp.
S= 0 + 0 + ... + 0 + 2025.
⇒S= 2025.
a) 3x2 + 12x =0
<=> 3x( x+ 4)=0
<=> \(\orbr{\begin{cases}3x=0\\x+4=0\end{cases}}\) <=>\(\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
d) \(4x^3=4x\)
<=> \(4x^3-4x=0\)
<=> 4x( x2 -1) =0
<=> 4x ( x - 1) ( x+ 1) =0
<=> 4x=0 hoac x-1=0 hoac x+1=0
<=> x=0 hoac x=1 hoac x=-1
1) \(3\left(5x-2\right)-9=6\)
\(15x-6=6+9\)
\(15x=15+6\)
\(15x=21\)
\(x=\frac{7}{5}\)
2) \(4\left(3x-2\right)+7\left(3x-2\right)=22\)
\(\left(3x-2\right)\left(4+7\right)=22\)
\(\left(3x-2\right)\cdot11=22\)
\(\left(3x-2\right)=2\)
\(3x=4\)
\(x=\frac{4}{3}\)
3) \(2x^2-6x=0\)
\(2x\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Vậy,............
\(1)3\left(5x-1\right)-9=6\)
\(\Rightarrow3\left(5x-2\right)=15\)
\(\Rightarrow5x-2=5\Rightarrow5x=7\Rightarrow x=\frac{7}{5}\)
\(2)4\left(3x-2\right)+7\left(3x-2\right)=22\)
\(\Rightarrow11\left(3x-2\right)=22\)
\(\Rightarrow33x-22=22\Rightarrow33x=44\Rightarrow x=\frac{4}{3}\)
\(3)2x^2-6x=0\)
\(\Rightarrow2x\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
a) <=> 2x - 1 = 0 hoặc 5x + 2 = 0
<=> 2x = 1 hoặc 5x = -2
<=> x = \(\frac{1}{2}\) hoặc x = \(-\frac{2}{5}\)
b) <=> 3/7 . (1 + 1/x) = 1/4
=> 1 + 1/x = 7/12 <=> 1/x = -5/12
<=> -5/-5x = -5/12 <=> -5x = 12
<=> x = \(-\frac{12}{5}\)
c) Dễ thấy 3x + 5 > 2x - 3
Để (3x + 5)( 2x - 3) < 0 thì 3x + 5 > 0 và 2x - 3 < 0
<=> x > -5/3 và x < 3/2
Vậy \(-\frac{5}{3}< x< \frac{3}{2}\)
a) (2x-1).(5x+2) = 0
\(\Leftrightarrow\) 2x-1 = 0 hoặc 5x+2 = 0
\(\Leftrightarrow\) 2x = 1 hoặc 5x = -2
\(\Leftrightarrow\) x = \(\frac{1}{2}\) hoặc x = \(\frac{-2}{5}\)
b) \(\frac{3}{7}+\frac{3}{7}:x=\frac{-1}{2}-\left(\frac{-3}{4}\right)\)
\(\frac{3}{7}+\frac{3}{7}:x=\frac{-1}{2}+\frac{3}{4}\)
\(\frac{3}{7}+\frac{3}{7}:x=\frac{1}{4}\)
\(\frac{3}{7}:x=\frac{1}{4}-\frac{3}{7}\)
\(\frac{3}{7}:x=\frac{-5}{28}\)
\(x=\frac{3}{7}:\frac{-5}{28}\)
\(x=\frac{-12}{5}\)
\((x-6)(3x-9)>0\)
TH1:
\(\orbr{\begin{cases}x-6< 0\\3x-9< 0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x< 6\\x< 3\end{cases}}\)\(\Rightarrow x< 3\)
TH2:
\(\orbr{\begin{cases}x-6>0\\3x-9>0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>6\\x>3\end{cases}}\)\(\Rightarrow x>6\)
Vậy \(x< 3\) hoặc \(x>6\)thì \((x-6)(3x-9)>0\)
Học tốt!
20.
\((2x-1)(6-x)>0\)
TH1:
\(\orbr{\begin{cases}2x-1>0\\6-x>0\end{cases}\Rightarrow\orbr{\begin{cases}x< \frac{1}{2}\\x< 6\end{cases}}\Rightarrow x< 6}\)
TH2
\(\orbr{\begin{cases}2x-1< 0\\6-x< 0\end{cases}\Rightarrow\orbr{\begin{cases}x>\frac{1}{2}\\x>6\end{cases}}\Rightarrow x>\frac{1}{2}}\)
Vậy \(x< 6\)hoặc \(x>\frac{1}{2}\)thì \((2x-1)(6-x)>0\)
2x - 3 = 0 hoặc 6 - 1/5x = 0
=> x = 3/2 => x = 30
Vậy x = 3/2 hoặc 30
=> 2x-3=0 hoặc 6-1/5x=0
TH1: 2x-3=0 TH2: 6-1/5x=0
2x=3 1/5x=6
x=3/2 x=30
=>x=3/2 hoặc x=30