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\(a,\Rightarrow12x-91=101\\ \Rightarrow12x=192\\ \Rightarrow x=16\\ b,\Rightarrow x:23+45=133\\ \Rightarrow x:23=88\\ \Rightarrow x=\dfrac{88}{23}\\ c,\Rightarrow\left(6x-39\right):7=3\\ \Rightarrow6x-39=21\\ \Rightarrow6x=60\\ \Rightarrow x=10\\ d,\Rightarrow3x-24=\dfrac{148}{73}\\ \Rightarrow3x=\dfrac{1900}{73}\\ \Rightarrow x=\dfrac{1900}{219}\\ e,\Rightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\\ f,\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ d,\left(9-x\right)^3=64=4^3\\ \Rightarrow9-x=4\\ \Rightarrow x=5\\ h,\Rightarrow x=27\\ i,\Rightarrow6x=312\cdot12=624\cdot6\\ \Rightarrow x=624\\ j,\Rightarrow\left(19x+104\right):14=25-42=-17\\ \Rightarrow19x+104=-238\\ \Rightarrow19x=-342\\ \Rightarrow x=-18\)
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1) 2x - 378 = 122
=> 2x = 122 + 378
=> 2x = 500
=> x = 500 : 2
=> x = 250
2) 8x - 4x = 1208
=> 4x = 1208
=> x = 1208 : 4
=> x = 302
3) x - 382 = 159 : 3
=> x - 382 = 53
=> x = 53 + 382
=> x = 435
4) 3x + 30 = 420
=> 3x = 420 - 30
=> 3x = 390
=> x = 390 : 3
=> x = 130
5) 12x - 13x - 500 = 1500
=> -x = 1500 + 500
=> -x = 2000
=> x = -2000
1,\(2x-378=122\)
\(2x=122+378\)
\(2x=500\)
\(x=500:2\)
\(x=250\)
2,\(8x-4x=1208\)
\(x.\left(8-4\right)=1208\)
\(x.4=1208\)
\(x=1208:4\)
\(x=302\)
3,\(x-382=159:3\)
\(x-382=53\)
\(x=52+382\)
\(x=434\)
4,\(3x+30=420\)
\(3x=420-30\)
\(3x=390\)
\(x=390:3\)
\(x=130\)
5,\(12x-13x-500=1500\)
như câu 2
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Bài 1:
a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)
hay \(x=\dfrac{50}{9}\)
Vậy: \(x=\dfrac{50}{9}\)
c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)
\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)
hay \(x=\dfrac{22}{15}\)
Vậy: \(x=\dfrac{22}{15}\)
d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)
\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)
hay \(x=\dfrac{15}{19}\)
Vậy:\(x=\dfrac{15}{19}\)
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a) 60 - 4(x + 5) = 12
4(x + 5) = 60 - 12
4(x + 5) = 48
x + 5 = 48 : 4
x + 5 = 12
x = 7
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+ x <0 => VP< 0 ; VT>0 loại
+ x>/ 0
=> x+1+x+3+x+5 =12x
=> 1+3+5 = 12x -3x
=>9x =9
=> x =1 (TM)
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* \(6x+3-2\left(x-5\right)=30\)
\(\Rightarrow6x+3-2x+10=30\)
\(\Rightarrow4x=17\Leftrightarrow x=\frac{17}{4}\)
* \(8\left(x-\frac{1}{8}\right)-5\left(x-\frac{1}{5}\right)=10\)
\(\Rightarrow3\left(x-\frac{1}{5}\right)=10\)
\(\Rightarrow x-\frac{1}{5}=\frac{10}{3}\)
\(\Rightarrow x=\frac{10}{3}+\frac{1}{5}=\frac{53}{15}\)
* \(13x-5-2\left(x+3\right)=1\)
\(\Rightarrow13x-5-2x-6=1\)
\(\Rightarrow11x=12\Leftrightarrow x=\frac{12}{11}\)
a)
\(\frac{x\left(x+1\right)}{2}=36\)(quy tắc tính tổng)
=>\(x\left(x+1\right)=72\)
=>\(x\left(x+1\right)=8.9\)
=>x=8
b)\(12x+13x=2000\Rightarrow\left(12+13\right)x=2000\Rightarrow25x=2000\Rightarrow x=80\)
c)
(x-1)(x-5)=0
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}}\)
cái ngoặc vuông là "hoặc " nhé
a ) x= 11
c)x = 1 hoặc x = 5