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1.Tim x:
a)| x + 1 | = 5 -> Th1: x+1=5-> x= 5-1=4
Th2: x+1=-5-> x= (-5) -1=-6(Loại. vì x lớn hơn hoặc bằng 0)
Vậy x= 4
b)| x - 3 | = 7 -> TH1: x-3=7-> x=7+3=10(Loại. Vì x<3)
TH2: x-3=-7-> x=-7+3=-4
Vậy x= -4
c) x + | 2 - x | = 6
-> | 2 - x | =6 -x
-> TH1: 2-x = 6-x
-> -x+ x= 2-6
-> 0x =-4(LOẠI)
TH2: 2-x= -6+x
->(-x)-x= 2+6
-> -2.x=8
-> x=8: -2=-4
Vậy x=-4
Tick cho mik nha!!!
2. Tìm x
a) | x | = 7-> x=-7 hoặc x=7
b) | x | < 7.Vì| x | lớn hơn hoặc bằng 0
-> | x | =(0;1;2;3;4;5;6)
-> x= (-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6)
c) | x | > 7
-> | x | =(8;9;10;11;12;13.............)
-> x= (...............;-9;-8;8;9;10;.............)
\(M=2+2^3+2^5+2^7+....+2^{51}\)
\(=\left(2+2^3\right)+\left(2^5+2^7\right)+....+\left(2^{49}+2^{51}\right)\)
\(=10+2^4\left(2+2^3\right)+....+2^{48}\left(2+2^3\right)\)
\(=10+2^4.10+...+2^{48}.10\)
\(=10\left(1+2^4+...+2^{48}\right)\Rightarrow M⋮10\)
\(=2.5.\left(1+2^4+...+2^{48}\right)\Rightarrow M⋮5\)
\(M=2+2^3+2^5+2^7+....+2^{51}.\)
\(M+2^{ }=2+2+2^3+2^5+2^7+.....+2^{51}\)
\(=\left(2+2+2^3\right)+\left(2^5+2^7+2^9\right)+....+\left(2^{47}+2^{49}+2^{51}\right)\)
\(=12+2^4\left(2+2^3+2^5\right)+......+2^{46}\left(2+2^3+2^5\right)\)
\(=12+2^4.42+....+2^{46}.42\)
\(=12+7.3.2\left(2^4+...+2^{46}\right)\)
\(\Rightarrow M=\left[12+7.3.2\left(2^4+.....+2^{46}\right)\right]-2\)
\(=10+7.3.2\left(2^4+....+2^{46}\right)\)
Ta có: \(7.3.2\left(2^4+...+2^{46}\right)⋮7\)mà 10 không chia hết cho 7
Suy M không chia hết cho 7
|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|
\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)
\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)
\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)
Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}
|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0
|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|
\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)
\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)
\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)
Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}
a)x+3/6=7/2
x =7/2-3/6
x =3
b)2,8+(3x-5/6)=2/2/5
14/5+(3x-5/6)=12/5
(3x-5/6)=12/5-14/5
3x+5/6=-2/5
3x =-2/5-5/6
3x =-37/30
x =-37/36:3
x =-37/90
sai đề phải ko?