30+{96-3.[54-2.(3 mũ 2+6)]}.x=79
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30+{96-3.[54-2.(32+6)]}.x =79
30+{96-3.[54-2.15]}.x=79
30+{96-3.[54-30]}.x=79
30+{96-3.24}.x=79
30+{96-72}.x=79
30+24.x=79
24.x=79-30
24x=49
x=49/24
\(30+\left\{96-3.\left[54-2\left(3^2+6\right)\right]\right\}.x=78\)
\(30+\left\{96-3.\left[56-2.15\right]\right\}.x=78\)
\(30+\left\{96-3.\left[56-30\right]\right\}.x=78\)
\(30+\left\{96-3.26\right\}.x=78\)
\(30+\left\{96-78\right\}.x=78\)
\(30+18.x=78\)
\(\Rightarrow18.x=48\)
\(\Rightarrow x=\frac{8}{3}\)
học tốt
a, (-0,2)2 \(\times\) 5 - \(\dfrac{2^{13}\times27^3}{4^6\times9^5}\)
= 0,04 \(\times\) 5 - \(\dfrac{2^{13}\times3^9}{2^{12}\times3^{10}}\)
= 0,2 - \(\dfrac{2}{3}\)
= \(\dfrac{2}{10}\) - \(\dfrac{2}{3}\)
= - \(\dfrac{7}{15}\)
b, \(\dfrac{5^6+2^2.25^3+2^3.125^2}{26.5^6}\)
= \(\dfrac{5^6+4.5^6+8.5^6}{26.5^6}\)
= \(\dfrac{5^6.\left(1+4+8\right)}{26.5^6}\)
= \(\dfrac{1}{2}\)
\(10^{30}vs\)\(2^{100}\)
\(10^{30}=\left(10^3\right)^{10}=1000^{10}\)
\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)
Vì \(1000^{10}< 1024^{10}=>10^{30}< 2^{100}\)
\(3^{54}vs2^{81}\)
\(3^{54}=\left(3^6\right)^9=729^9\)
\(2^{81}=\left(2^9\right)^9=512^9\)
Vì \(729^9>512^9=>3^{54}>2^{81}\)
\(3^{200}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
Vì \(8^{100}< 9^{100}=>2^{300}< 3^{200}\)
12x - 10x + 54 = 86
\(\Rightarrow\)12x - 10x = 86 - 54
\(\Rightarrow\)2x = 32
\(\Rightarrow\)x = 32 : 2
\(\Rightarrow\) x= 16
12x-10x+54=86
=(12-10)x+54=86
(12-10)x =32
2x =32
x =32chia2
x =16
Ta có:
3/2 - 1/3 + 1/6 = 4/3
1/5 + 1/6 + 79/30 = 3
⇒ 4/3 < 2 < 3
Và 2 là số nguyên tố
Vậy có 1 số nguyên tố là x = 2 thỏa mãn đề bài
ta có:
3/2-1/3+1/6<x<1/5+1/6+79/30
=45/30-10/30+5/30<x<6/30+5/30+79/30
=40/30<x<90/30
=>4/3<x<9/3
=>x có 4 số nguyên tố thỏa mãn
Không chắc lắm nha :((
a) \(x^2-5x+6\)
\(=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
b) \(x^2-9x+18=x^2-3x-6x+18\)
\(=x\left(x-3\right)-6\left(x-3\right)=\left(x-3\right)\left(x-6\right)\)
c) \(x^2-6x+5=x^2-x-5x+5\)
\(=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)
d) \(3x^2+5x-30=3\left(x^2+\dfrac{5x}{3}-10\right)=3\left(x^2+2.x.\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{5347}{500}\right)\)
Câu này bạn xem lại đề nha
e) \(3x^2-5x-2=3x^2-6x+x-2\)
\(3x\left(x-2\right)+x-2=\left(x-2\right)\left(3x+1\right)\)
\(\Leftrightarrow x:3+367\cdot\left(-2\right)=-60\)
=>x:3=674
hay x=2022
\(30+\left\{96-3.\left[54-2.\left(3^{2+6}\right)\right]\right\}.x=79\)
\(30+\left\{96-3.\left[54-2.6561\right]\right\}.x=79\)
\(30+\left\{96-3.\left[54-13122\right]\right\}.x=79\)
\(30+\left\{96-3.-13068\right\}.x=79\)
\(30+\left\{96--39204\right\}.x=79\)
\(30+39300.x=79\)
\(39300.x=79-30\)
\(39300.x=49\)
\(\Rightarrow x=\frac{39300}{49}\)