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5.(3x+8)-7.(2x+3)=16
=> 15x+40-14x-21=16
=>(15x-14x)+(40-21)
=> x+19=16
=>x=-3. Vậy x=-3
k cho mik nha
Chúc bạn hok tốt
\(5\left(3x+8\right)-7\left(2x+3\right)=16\)
\(15x+40-14x-21=16\)
\(x+19=16\)
\(x+19-16=0\)
\(x+3=0\)
\(x=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(3\left(2x-5\right)+125=134\)
\(\Leftrightarrow3\left(2x-5\right)=9\)
\(\Leftrightarrow2x-5=3\)
\(\Leftrightarrow2x=8\Leftrightarrow x=4\)
b) \(\left(2x+5\right)+\left(2x+3\right)+\left(2x+1\right)=27\)
\(\Leftrightarrow6x+9=27\)
\(\Leftrightarrow6x=18\Leftrightarrow x=3\)
d) \(27\left(x-27\right)-27=0\)
\(\Leftrightarrow27\left(x-27\right)=27\)
\(\Leftrightarrow x-27=1\Leftrightarrow x=28\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bằng 0 chứ nhỉ em ?
(x-5) . (2x-4)= 0
\(\left[{}\begin{matrix}x-5=0\\2x-4=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=5\\2x=4\end{matrix}\right.< =>\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
(x-5) . (2x-4)= 5
<=> 2x^2 - 4x - 10x + 20 = 5
<=> 2x^2 - 14x + 15 = 0
Giải pt bâc hai ra đc : \(x=\dfrac{7+\sqrt{19}}{2}\)và \(x2=\dfrac{7-\sqrt{19}}{2}\)
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a, \(\dfrac{5}{9}.\dfrac{10}{11}+\dfrac{5}{9}.\dfrac{14}{11}-\dfrac{5}{9}.\dfrac{15}{11}=\dfrac{5}{9}.\left(\dfrac{10}{11}+\dfrac{14}{11}-\dfrac{15}{11}\right)=\dfrac{5}{9}.\dfrac{9}{11}=\dfrac{5}{11}\)
b, \(\dfrac{6}{7}.\dfrac{8}{13}+\dfrac{6}{13}.\dfrac{9}{7}-\dfrac{3}{13}.\dfrac{6}{7}\)\(=\dfrac{6}{7}.\left(\dfrac{8}{13}-\dfrac{3}{13}\right)+\dfrac{6}{13}.\dfrac{9}{7}=\dfrac{6}{7}.\dfrac{5}{13}+\dfrac{54}{91}=\dfrac{30}{91}+\dfrac{54}{91}=\dfrac{84}{91}=\dfrac{12}{13}\)
Bạn ơi, gõ latex cho dễ nhìn nhé!
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\(\left(x+2\right)-2=0\)
\(\Rightarrow x+2-2=0\)
\(\Rightarrow x=0\)
\(\left(x+3\right)+1=7\)
\(\Rightarrow x+3+1=7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)
\(\Rightarrow3x=12\)
\(\Rightarrow x=4\)
\(\left(5x+4\right)-1=13\)
\(\Rightarrow5x+4-1=13\)
\(\Rightarrow5x+3=13\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
\(\left(4x-8\right)-3=5\)
\(\Rightarrow4x-8-3=5\)
\(\Rightarrow4x-11=5\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
\(8-\left(2x+4\right)=2\)
\(\Rightarrow8-2x-4=2\)
\(\Rightarrow4-2x=2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
\(7+\left(5x+2\right)=14\)
\(\Rightarrow7+5x+2=14\)
\(\Rightarrow9+5x=14\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=1\)
\(5-\left(3x-11\right)=1\)
\(\Rightarrow5-3x+11=1\)
\(\Rightarrow16-3x=1\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
24 chia hết cho 2x-1
=> 2x-1 \(\in\) Ư(24)
=> 2x-1 \(\in\) {1,2,3,4,6,8,12,24,-1,-2,-3,-4,-6,-8,-12,-24}
=> 2x \(\in\) { 2; 3;4;5;7;9;13;25;0;-1;-2;-3;-5;-7;-11;-23}
=> x \(\in\) {1;\(\dfrac{3}{2}\); 2; \(\dfrac{5}{2}\); \(\dfrac{7}{2}\);\(\dfrac{9}{2}\); \(\dfrac{13}{2}\); \(\dfrac{25}{2}\); 0;\(\dfrac{-1}{2}\); -1; \(\dfrac{-3}{2}\); \(\dfrac{-5}{2}\); \(\dfrac{-7}{2}\); \(\dfrac{-11}{2}\); \(\dfrac{-23}{2}\)}
Vậy...
\(2x-24=8^5:8^3\)
\(2x-24=8^2\)
\(2x-24=64\)
\(2x=88\)
\(x=44\)
\(\left(2x-24\right).8^3=8^5\)
\(\left(2x-24\right)=8^5:8^3\)
\(\left(2x-24\right)=8^2\)
\(\left(2x-24\right)=64\)
\(2x=64+24\)
\(2x=88\)
\(\Rightarrow x=44\)