tìm x:
a,3x-1+3x+3x+1=1053
b, (x+3)5= (x+3)7
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a) 7(x - 5) + 2 = 51
\(\Leftrightarrow\) 7(x - 5) = 51 - 2
\(\Leftrightarrow\) 7(x - 5) = 49
\(\Leftrightarrow\) x - 5 = 49 : 7
\(\Leftrightarrow\) x - 5 = 7
\(\Leftrightarrow\) x = 7 + 5
\(\Leftrightarrow\) x = 12.
Vậy x = 12.
k) 2412 : (3x + 147) = |-38| + (-26)
\(\Leftrightarrow\) 2412 : (3x + 147)= 38 + (-26)
\(\Leftrightarrow\) 2412 : (3x + 147)= 12
\(\Leftrightarrow\) 3x + 147 = 2412 : 12
\(\Leftrightarrow\) 3x + 147 = 201
\(\Leftrightarrow\) 3x = 201 - 147
\(\Leftrightarrow\) 3x = 54
\(\Leftrightarrow\) x = 54 : 3
\(\Leftrightarrow\) x = 18.
Vậy x = 18.
I) 4824 : (4x + 137) = |-59| + (-35)
\(\Leftrightarrow\) 4824 :(4x + 137) = 59 + (-35)
\(\Leftrightarrow\) 4824 :(4x + 137) = 24
\(\Leftrightarrow\) 4x + 137 = 4824 : 24
\(\Leftrightarrow\) 4x + 137 = 201
\(\Leftrightarrow\) 4x = 201 - 137
\(\Leftrightarrow\) 4x = 64
\(\Leftrightarrow\) x = 64 : 4
\(\Leftrightarrow\) x = 16.
Vậy x = 16.
c) |-123| - 5(x - 3) = (-28) + 66
\(\Leftrightarrow\) 123 - 5(x - 3) = 38
\(\Leftrightarrow\) 5(x - 3) = 123 - 38
\(\Leftrightarrow\) 5(x - 3) = 85
\(\Leftrightarrow\) x - 3 = 85 : 5
\(\Leftrightarrow\) x - 3 = 17
\(\Leftrightarrow\) x = 17 + 3
\(\Leftrightarrow\) x = 20.
Vậy x = 20.
m) 7x - 4 . 6 = 2058
\(\Leftrightarrow\) 7x - 24 = 2058
\(\Leftrightarrow\) 7x = 2058 + 24
\(\Leftrightarrow\) 7x = 2082
\(\Leftrightarrow\) x = 2082 : 7
\(\Leftrightarrow\) x = \(\frac{2058}{7}\).
Vậy x = \(\frac{2058}{7}\).
Phần b) ra phân số nên mình để mai làm và cả những bài còn lại nữa.
Chúc bạn học tốt !!!
---------------NHANH NHA MK ĐANG CẦN GẤP--------------
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\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)
\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)
2:
a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8
=>x^2-x-12-x^2+4x+5=8
=>3x-7=8
=>3x=15
=>x=5
b: =>3x^2+3x-2x-2-3x^2-21x=13
=>-20x=15
=>x=-3/4
c: =>x^2-25-x^2-2x=9
=>-2x=25+9=34
=>x=-17
d: =>x^3-1-x^3+3x=1
=>3x-1=1
=>3x=2
=>x=2/3
a)<=>|3x+1|-|2x-5|+|x-12|=2x+3
=>x=15/2
b)<=>|x-1|+2|3x+2|-|5x-3|=-(|5x-3|-2|3x+2|-|x-1|)
=>-(|5x-3|-2|3x+2|-|x-1|)=|3x+7|
=>x=4/3 hoặc x=2/3
a, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)
\(\Leftrightarrow\left(x+1\right)\left(3x-5-3x+1\right)=x-4\Leftrightarrow-4\left(x+1\right)=x-4\)
\(\Leftrightarrow-4x-4=x-4\Leftrightarrow-4x-x=0\Leftrightarrow x=0\)
b, \(\left(x-2\right)\left(x+3\right)-\left(x+4\right)\left(x-7\right)=5-x\)
\(\Leftrightarrow x^2+x-6-x^2-3x+28=5-x\Leftrightarrow-2x+22=5-x\Leftrightarrow x=17\)
c, thiếu đề
d, \(3\left(x-7\right)\left(x+7\right)-\left(x-1\right)\left(3x+2\right)=13\)
\(\Leftrightarrow3x^2-147-3x^2+x+2=13\Leftrightarrow x=11+147=158\)
a.\(3x^2-2x-5-\left(3x^2+2x-1\right)=x-4\)
\(\Leftrightarrow-5x=0\Leftrightarrow x=0\)
b.\(x^2+x-6-\left(x^2-3x-28\right)=5-x\)
\(\Leftrightarrow5x=-17\Leftrightarrow x=-\frac{17}{5}\)
c.\(5\left(x^2-10x+21\right)-\left(5x^2-9x-2\right)=0\)
\(\Leftrightarrow-41x+107=0\Leftrightarrow x=\frac{107}{41}\)
d.\(3\left(x^2-49\right)-\left(3x^2-x-2\right)=13\Leftrightarrow x=158\)
a,6x-3-5x+15+18x-24=24
19x-12=24
19x=36
x=36/19
c,10x-6x2+6x2-10x+21=3
0x=-18
không có x
d,3x2+3x-2x2-4x=-1-x
x2-x=-1-x
x2-x+x=-1
x2=-1
không có x thỏa mãn
\(a,3^{x-1}+3^x+3^{x+1}=1053\\ \Leftrightarrow3^x.\dfrac{1}{3}+3^x+3^x.3=1053\\ \Leftrightarrow3^x\left(\dfrac{1}{3}+1+3\right)=1053\\ \Leftrightarrow\dfrac{3^x.13}{3}=1053\\ \Leftrightarrow3^x=243\\ \Leftrightarrow3^x=3^5\\ \Leftrightarrow x=5\)
\(b,\left(x+3\right)^5=\left(x+3\right)^7\\ \Leftrightarrow\left(x+3\right)^7-\left(x+3\right)^5=0\\ \Leftrightarrow\left(x+3\right)^5\left(\left(x+3\right)^2-1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x+3\right)^5=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3-0\\\left(x+3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-3\\x+3=\pm1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x\in\left\{-4;-2\right\}\end{matrix}\right.\)