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\(a.2.x-138=2^3.3^2\)
\(2.x-138=8.9\)
\(2.x-138=72\)
\(2.x\) \(=72+138\)
\(2.x\) \(=210\)
\(x\) \(=210:2\)
\(x\) \(=105\)
\(b.231-\left(x-6\right)=1339:13\)
\(231-\left(x-6\right)=103\)
\(x-6=231-103\)
\(x-6=128\)
\(x\) \(=128+6\)
\(x\) \(=134\)
\(c.\left[\left(6.x-72\right):2-84\right].28=5628\)
\(\left[\left(6.x-72\right):2-84\right]\) \(=5628:28\)
\(\left(6.x-72\right):2-84\) \(=201\)
\(\left(6.x-72\right):2\) \(=201-84\)
\(\left(6.x-72\right):2\) \(=117\)
\(6.x-72\) \(=117.2\)
\(6.x-72\) \(=234\)
\(6.x\) \(=234+72\)
\(6.x\) \(=306\)
\(x\) \(=306:6\)
\(x\) \(=51\)
mik nè
\(\frac{121212}{161616}-\left(\frac{151515}{323232}-x\right)=2\)
=> \(\frac{3}{4}-\left(\frac{15}{32}-x\right)=2\)
=> \(\frac{15}{32}-x=\frac{3}{4}-2\)
=> \(\frac{15}{32}-x=-\frac{5}{4}\)
=> \(x=\frac{15}{32}-\frac{-5}{4}=\frac{15}{32}+\frac{5}{4}=\frac{55}{32}\)
b) \(\frac{x}{2}+\frac{x}{6}+\frac{x}{12}+\frac{x}{20}+\frac{x}{30}+\frac{x}{42}+\frac{x}{56}+\frac{x}{72}+\frac{x}{90}=\frac{9}{5}\)
=> \(\frac{x}{1\cdot2}+\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+\frac{x}{4\cdot5}+\frac{x}{5\cdot6}+\frac{x}{6\cdot7}+\frac{x}{7\cdot8}+\frac{x}{8\cdot9}+\frac{x}{9\cdot10}=\frac{9}{5}\)
=> \(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{9}-\frac{x}{10}=\frac{9}{5}\)
=> \(\frac{x}{1}-\frac{x}{10}=\frac{9}{5}\)
=> \(\frac{10x-x}{10}=\frac{9}{5}\)
=> \(\frac{9x}{10}=\frac{9}{5}\)
=> \(\frac{9x}{10}=\frac{9\cdot2}{5\cdot2}=\frac{18}{10}\)
=> x = 2
\(a)(2367-x)-1017=205\)
\(\Leftrightarrow(2367-x)=205+1017\)
\(\Leftrightarrow(2367-x)=1222\)
\(\Leftrightarrow x=2367-1222\)
\(\Leftrightarrow x=1145\)
\(b)(6x-39):3\cdot28=5628\)
\(\Leftrightarrow(6x-39):3=\frac{5628}{28}\)
\(\Leftrightarrow(6x-39):3=201\)
\(\Leftrightarrow(6x-39)=201\cdot3\)
\(\Leftrightarrow(6x-39)=603\)
\(\Leftrightarrow6x=603+39\)
\(\Leftrightarrow6x=642\)
\(c)5x-x+2x=42\)
\(\Rightarrow5x-1x+2x=42\)
\(\Rightarrow6x=42\)
\(\Rightarrow x=7\)
d\()\)Tương tự câu a và b
e\()(x-2)(x-3)=0\)
Xét có hai trường hợp :
TH1 : x - 2 = 0 => x = 2
TH2 : x - 3 = 0 => x = 3
Vậy : \(\hept{\begin{cases}x=2\\x=3\end{cases}}\)
\(\Leftrightarrow x=\frac{642}{6}=107\)
a)0,12.X-2=4
=>0,12.X=4+2
=>0,12.X=6
=>X=6:0,12
=>X=50
b)231-(X-6)=1339:13
=>231-(X-6)=103
=>X-6=231-103
=>X-6=128
=>X=128+6
=>X=134
c)7,99.X+X.2+X:100=7,9
=>X.(7,99+2+1/100)=7,9
=>X.10=7,9
=>X=7,9:10
=>X=0,79
d)(14-X:6).7=84
=>14-X:6=84:7
=>14-X:6=12
=>X:6=14-12
=>X:6=2
=>X=2.6
=>X=12
\(231-\left(x-6\right)=1339\div13\)
\(231-\left(x-6\right)=103\)
\(x-6=231-103\)
\(x-6=128\)
\(x=128+6\)
\(x=134\)
231-(x-6)=1339:13
=> 231-(x-6)=103
=> x-6=231-103
=> x-6=128
=> x=128+6
=> x=134
a)231-(x-6)=103
x-6=231-103=128
x=128+6=134
b)x+10-25=0
x+10=0+25=25
x=25-10=15
a/ \(2x^3=8x\)
\(2.8=2x^3\)
\(16=2x^3\)
\(x^3=16:2\)
\(x^3=8\)
\(x=2\)
phần b mk chưa nghiên cứu dc
a) \(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}=4\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-3}{97}-1+\frac{x-3}{96}-1=4-4\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
\(\Rightarrow x-1=0\) ( vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\) )
Vậy x = 1
b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=3\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=3-3\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)
\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)
=> x + 100 = 0
=> x = -100
c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)
=> x - 100 = 0
=> x = 100
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