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\(a.\)\(10x+2^2.5=10\)
\(\Leftrightarrow10x+4.5=10\)
\(\Leftrightarrow10x+20=10\)
\(\Leftrightarrow10x=10-20\)
\(\Leftrightarrow10x=-10\)
\(\Leftrightarrow x=-10:10\)
\(\Leftrightarrow x=-1\)
\(b.\)\(125-5\left(4+x\right)=15\)
\(\Leftrightarrow5\left(4+x\right)=125-15\)
\(\Leftrightarrow5\left(4+x\right)=110\)
\(\Leftrightarrow4+x=110:5\)
\(\Leftrightarrow4+x=22\)
\(\Leftrightarrow x=22-4\)
\(\Leftrightarrow x=18\)
\(c.\)\(2^6+\left(218-x\right)=73\)
\(\Leftrightarrow64+\left(218-x\right)=73\)
\(\Leftrightarrow218-x=73-64\)
\(\Leftrightarrow218-x=9\)
\(\Leftrightarrow x=218-9\)
\(\Leftrightarrow x=209\)
\(76-\left\{2\cdot\left[2\cdot5^2-\left(31-2\cdot3\right)\right]\right\}\)
\(=76-\left\{2\cdot\left[2\cdot25-\left(31-6\right)\right]\right\}\)
\(=75-\left[2\cdot\left(50-25\right)\right]\)
\(=76-\left(2\cdot25\right)\)
\(=76-50\)
\(=26\)
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\(6^2\cdot10:\left\{780:\left[10^3-\left(2\cdot5^3+35\cdot14\right)\right]\right\}\)
\(=36\cdot10:\left\{780:\left[1000-\left(2\cdot125+490\right)\right]\right\}\)
\(=360:\left\{780:\left[1000-\left(250+490\right)\right]\right\}\)
\(=360:\left[780:\left(1000-740\right)\right]\)
\(=360:\left(780:260\right)\)
\(=360:3\)
\(=120\)
a: =76-{2*[2*25-31+6]}
=76-{2*[50-31+6]}
=76-2*25
=76-50=26
b: \(=360:\left\{\dfrac{780}{1000-2\cdot125-490}\right\}\)
\(=360:\dfrac{780}{260}\)
=360/3=120
\(142-\left[50-\left(2^3.10-2.5\right)\right]\)
= \(142-\left[50-\left(80-10\right)\right]\)
= \(142-\left(50-70\right)\)
= \(142-\left(-20\right)\)
= \(162\)
\(142-\left[50-\left(2^3\cdot10-2\cdot5\right)\right]\)
\(=142-50+80+10\)
\(=72+90\)
=162
Chữa lại câu 1
\(2.5^{10}\): \(5^{10}\)có quy luật chữ số tận cùng là : \(5;5;5;5;...;5\)( 10 số 5 )
\(\Rightarrow\)Tận cùng của \(5^{10}\)là 5
\(\Rightarrow\)Chữ số tận cùng của \(2.5^{10}\)là : \(2.5=10\)
\(\Rightarrow\)Vậy chữ số tận cùng của \(2.5^{10}\)là 0
1) \(\left(2.5\right)^{10}=10^{10}=100...00\) Suy ra tận cùng \(\left(2.5\right)^{10}\)là \(0\)
2) \(S=1+2+2^2+2^3+...+2^{10}\)
\(\Rightarrow2S=2+2^2+2^3+2^4+...+2^{10}+2^{11}\)
\(\Rightarrow2S-S=S=\left(2+2^2+2^3+...+2^{10}+2^{11}\right)-\left(1+2+2^2+...+2^{10}\right)\)
\(\Rightarrow S=2^{11}-1\)
3) \(1.2.3.4.5.6.7.8.9.10=\left(2.2^2.2.2^3.2\right).1.3.5.3.7.9.5\)
\(=2^8.1.3.5.3.7.9.5\Rightarrow⋮2^8\Rightarrow1.2.3.4.5.6.7.8.9.10⋮2^8\Rightarrow dpcm\)
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
b) \(9x-2:3^2=3^4\)
\(9x-2:9=81\)
\(2:9=9x-81\)
\(\dfrac{2}{9}=9x-81\)
\(9x=81+\dfrac{2}{9}\)
\(9x=\dfrac{731}{9}\)
\(x=\dfrac{731}{9}:9\)
\(x=\dfrac{731}{81}\)
\(a.5x-5^2=10\) \(b.9x-2:3^2=3^4\)
\(5x=10+5^2\) \(9x-2=3^4.3^2\)
\(5x=35\) \(9x-2=729\)
\(x=35:5=7\) \(9x=729+2=731\)
\(x=731:9\)
\(x=\dfrac{731}{81}\)
\(c=10x+\left(2^2\right).5=10^2\)
\(10x+20=100\)
\(10x=100-20\)
\(10x=80\)
\(x=80:10=8\)
\(10x+2^2.5=10^2\)
\(10x+4.5=100\)
\(10x+20=100\)
\(10x=100-20\)
\(10x=80\)
\(x=80:10\)
\(x=8\)
x = 8