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(20-3.x):5=55-51
(20-3.x):5=4
20-3.x=4.5
20-3.x=20
3.x=20-20
3.x=0
x=0:3
x=0
không đc đừng ném mik nha
\(\left|x+5\right|=5\)
<=> \(\hept{\begin{cases}x+5=5\\x+5=-5\end{cases}}\)
<=> \(\hept{\begin{cases}x=0\\x=-10\end{cases}}\)
\(\left|x+1\right|+7=10\)
<=> \(\left|x+1\right|=3\)
<=> \(\hept{\begin{cases}x+1=3\\x+1=-3\end{cases}}\)
<=> \(\hept{\begin{cases}x=2\\x=-4\end{cases}}\)
\(\left|x-3\right|-6=5\)
<=> \(\left|x-3\right|=11\)
<=> \(\hept{\begin{cases}x-3=11\\x-3=-11\end{cases}}\)
<=> \(\hept{\begin{cases}x=14\\x=-8\end{cases}}\)
\(\left|x+2\right|-6\left(x-4\right)=20-6x\)
<=> \(\left|x+2\right|-6x+24=20-6x\)
<=> \(\left|x+2\right|=-4\)
<=> \(\hept{\begin{cases}x+2=-4\\x+2=4\end{cases}}\)
<=> \(\hept{\begin{cases}x=-2\\x=2\end{cases}}\)
a) \(|x+5|=5\)
\(\Rightarrow\orbr{\begin{cases}x+5=5\\x+5=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-10\end{cases}}\)
Vậy x = 0 hoặc x = -10
b) \(|x+1|+7=10\)
\(\Rightarrow|x+1|=10-7\)
\(\Rightarrow|x+1|=3\)
\(\Rightarrow\orbr{\begin{cases}x+1=3\\x+1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)
Vậy x = 2 hoặc x = -4
c) \(|x-3|-6=5\)
\(\Rightarrow|x-3|=5+6\)
\(\Rightarrow|x-3|=11\)
\(\Rightarrow\orbr{\begin{cases}x-3=11\\x-3=-11\end{cases}}\Rightarrow\orbr{\begin{cases}x=14\\x=-8\end{cases}}\)
Vậy x = 14 hoặc x = -8
d) \(|x+2|-6\left(x-4\right)=20-6x\)
\(\Rightarrow|x+2|-6x+24=20-6x\)
\(\Rightarrow|x+2|=20-6x-24+6x\)
\(\Rightarrow|x+2|=\left(20-24\right)+\left(-6x+6x\right)\)
\(\Rightarrow|x+2|=-4\)
Vì \(|x|\ge0\)mà \(|x+2|=-4\)
\(\Rightarrow\)Không có giá trị x thỏa mãn
_Chúc bạn học tốt_
a) Ta có: \(71+\frac{26-3x}{5}=75\)
\(\Leftrightarrow\frac{26-3x}{5}=4\)
\(\Leftrightarrow26-3x=20\)
\(\Leftrightarrow3x=6\)
hay x=2
Vậy: x=2
b) Ta có: \(\left|x-12\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-12=5\\x-12=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-5+12=7\end{matrix}\right.\)
Vậy: \(x\in\left\{17;7\right\}\)
c) Ta có: \(\frac{7}{8}+x=\frac{3}{5}\)
\(\Leftrightarrow x=\frac{3}{5}-\frac{7}{8}\)
\(\Leftrightarrow x=\frac{24}{40}-\frac{35}{40}\)
hay \(x=-\frac{11}{40}\)
Vậy: \(x=-\frac{11}{40}\)
d) Ta có: \(\frac{1}{2}x-\frac{2}{5}=\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{2}\cdot x=\frac{1}{5}+\frac{2}{5}=\frac{3}{5}\)
\(\Leftrightarrow x=\frac{3}{5}:\frac{1}{2}=\frac{3}{5}\cdot2\)
hay \(x=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
\(a,2\frac{x}{7}=\frac{75}{35}\)
\(\frac{75}{35}=2\frac{5}{35}=2\frac{1}{7}\)
\(\Rightarrow x=1\)
\(b,47:4=...\left(+3\right)\)
\(\Rightarrow\left(47-3\right):4=11\)
\(\Rightarrow x=11\)
\(c,\frac{112}{5}=\frac{336}{15}=22\left(+6\right)\)
\(\Rightarrow x=22\frac{x=6}{15}\)
\(d,7,5x=\frac{75x}{100}:\left(9-6\frac{13}{21}\right)=2\frac{13}{25}\)
\(\frac{75x}{100}:\left(9-\frac{139}{21}\right)=\frac{63}{25}\)
\(\frac{75x}{100}:\frac{50}{21}=\frac{63}{25}\)
\(\Rightarrow\frac{63^3}{25^{\left(1\right)}}\cdot\frac{50^{\left(2\right)}}{21^1}=6=\frac{6}{1}\)
\(\Rightarrow\frac{6}{1}=\frac{600}{100}\)
Con d sai đề
a, \(2.x^x=10.3^{12}+8.27^4\)
\(2.x^x=10.3^{12}+8.3^{12}\)
\(2.x^x=3^{12}.\left(10+8\right)\)
\(2.x^x=3^{12}.18\)
\(2.x^x=3^{12}.2.3^3\)
\(2.x^x=3^{15}.2\)
\(x^x=3^{15}\)( Hình như sai đề )
b,\(3^{2x+2}=9^{x+3}\)
\(3^{2x+2}=3^{2x+3}\)
a,|x|+3=5
\(\Leftrightarrow\left|x\right|=5-3=2\)
\(\Rightarrow x=\left\{{}\begin{matrix}2\\-2\end{matrix}\right.\)
b,|x+3|=5
\(\Rightarrow\left\{{}\begin{matrix}x+3=5\\x+3=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2\\-8\end{matrix}\right.\)
c,|x-7|+13=25
<=>|x-7|=25-13=12
\(\Rightarrow\left\{{}\begin{matrix}x-7=12\\x-7=-12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=19\\x=-5\end{matrix}\right.\)
d,\(26-\left|x+9\right|=-13\)
\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)=39\)
\(\Rightarrow\left\{{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=30\\x=-48\end{matrix}\right.\)
e,8-|x|=15
<=>|x|=8-15=-7
\(\Rightarrow\left\{{}\begin{matrix}x=-7\\x=7\end{matrix}\right.\)
f,6-|-3+x|=-15
\(\Leftrightarrow\left|-3+x\right|=6-\left(-15\right)=21\)
\(\Rightarrow\left\{{}\begin{matrix}-3+x=21\\-3+x=-21\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=24\\x=-18\end{matrix}\right.\)
a, |x| + 3 = 5
|x| = 5 - 3
|x| = 2
|x| = 2 hoặc |x| = -2
Vậy |x| thuộc {2; -2}
b,|x + 3| = 5
|x + 3| = 5 hoặc |x + 3| = -5
x = 5 - 3 x = (-5) - 3
x = 2 x = -8
Vậy x thuộc {2; -8}
c,|x - 7| + 13 = 25
|x - 7| = 25 - 13
|x - 7| = 12
|x - 7| = 12 hoặc |x - 7| = -12
x = 12 + 7 x = (-12) + 7
x = 19 x = -5
Vậy x thuộc {19 ; -5}
giải
a)51+(20-3.x):5=55
(20-3.x):5=55-51
(20-3x):5=4
20-3x=5.4
20-3x=20
3x=20-20
3x=0
x=0:3
x=0
b)71+(26-3.x):5=75
(26-3.x):5=75-71
(26-3x):5=4
26-3x=4.5
26-3x=20
3x=26-20
3x=6
x=6:3
x=2
mik chỉ giải đc 2 câu đầu thôi
mong bạn thông cảm
mk giải tiếp câu c
(x-6)^2=5^2
x-6=5
x=5+6
x=11