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a: =>(x-2)^3*[(x-2)^2-1]=0
=>(x-2)(x-3)(x-1)=0
=>\(x\in\left\{1;2;3\right\}\)
b: =>(x-3)^2*(x-3-1)=0
=>(x-3)(x-4)=0
=>x=3 hoặc x=4
c: =>\(11\cdot\dfrac{6^x}{6}+2\cdot6^x\cdot6=6^{11}\left(11+2\cdot6^2\right)\)
=>6^x(11/6+12)=6^12(11/6+12)
=>x=12
a: \(\Leftrightarrow\left[\left(3x+14\right):4-3\right]:2=1\)
=>(3x+14):4-3=2
=>(3x+14):4=5
=>3x+14=20
=>3x=6
hay x=2
b: \(\Leftrightarrow\left[\left(x:4+17\right):10+3\cdot16\right]:10=5\)
\(\Leftrightarrow\left(x:4+17\right):10=50-48=2\)
=>x:4+17=20
=>x:4=3
hay x=12
c: \(\Leftrightarrow2\cdot15^2+\left[2\cdot125-\left(2x+4\right)\cdot5\right]:19=453\)
\(\Leftrightarrow250-\left(2x+4\right)\cdot5=\left(453-450\right)\cdot19=57\)
=>5(2x+4)=197
=>2x+4=197/5
=>2x=177/5
hay x=177/10
d: \(\Leftrightarrow\left(19x+50\right):14=5^2-4^2=9\)
=>19x+50=126
=>19x=76
hay x=4
e: \(\Leftrightarrow2\cdot3^x=10\cdot3^{12}+8\cdot3^{12}=18\cdot3^{12}\)
\(\Leftrightarrow3^x=3^2\cdot3^{12}=3^{14}\)
hay x=14
f: \(\Leftrightarrow3\left(x+2\right):7=30\)
=>3(x+2)=210
=>x+2=70
hay x=68
g: \(2480-1570+200-x+5=1010\)
=>1115-x=1010
hay x=105
dễ mk bn cho mình hỏi nhé câu 4 là \(\frac{1}{2\cdot3}\)hay là\(\frac{1}{2}\cdot3\)
Câu 1 :
a, 8.( -5 ).( -4 ).2
= [ 8.2 ].[( -5 ).(-4 ]
= 16.20
= 320
b, \(1\frac{3}{7}+\frac{-1}{3}+2\frac{4}{7}\)
\(=\frac{10}{7}+\frac{-1}{3}+\frac{18}{7}\)
\(=\frac{11}{3}\)
c, \(\frac{8}{5}.\frac{2}{3}+\frac{-5.5}{3.5}\)
\(=\frac{8}{3}+\frac{-5}{3}\)
\(=\frac{3}{3}=1\)
d, \(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}.\left(-2\right)^2\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{16}.4\)
\(=\frac{55}{56}-\frac{3}{4}\)
\(=\frac{13}{56}\)
Câu 2 :
a, 2x + 10 = 16
2x = 16 + 10
2x = 26
x = 26 : 2
x = 13
b, \(x-\frac{1}{3}=\frac{5}{4}\)
\(x=\frac{5}{4}+\frac{1}{3}\)
\(x=\frac{19}{12}\)
c, \(2x+3\frac{1}{3}=7\frac{1}{3}\)
\(2x+\frac{10}{3}=\frac{22}{3}\)
\(2x=\frac{22}{3}-\frac{10}{3}\)
\(2x=4\)
\(x=4:2\)
\(x=2\)
d, \(\left(\frac{2}{11}+\frac{1}{3}\right)x=\left(\frac{1}{7}-\frac{1}{8}\right).56\)
\(\frac{17}{33}x=1\)
\(x=1-\frac{17}{33}\)
\(x=\frac{16}{33}\)
c) \(3^2+2^4-\left(6^8:6^6-6^2\right)< 5^x< 125\)
\(=9+16-\left(6^{8-6}-36\right)< 5^x< 5^3\)
\(=25-\left(6^2-36\right)< 5^x< 5^3\)
\(=25-\left(36-36\right)< 5^x< 5^3\)
\(=25-0< 5^x< 5^3\)
\(=25< 5^x< 5^3\)
\(=5^2< 5^x< 5^3\)
Vì \(5^2=25\) và \(5^3=125\) nên \(x\) không thể thỏa mãn đề bài
⇒ \(x\) không thỏa mãn đề bài