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\(1.\left(x-1\right)^2=4=\left(-2\right)^2=2^2\)
\(TH1:x-1=2\Rightarrow x=3\)
\(TH2:x-1=-2\Rightarrow x=-1\)
Vậy:...
\(2.\left(1+x\right)^2=9=\left(-3\right)^2=3^2\)
\(TH1:1+x=3\Rightarrow x=2\)
\(TH2:1+x=-3\Rightarrow x=-4\)
Vậy:....
\(3,\left(x+2019\right)^4=1\Rightarrow\left(x+2019\right)^4=1^4\)
\(\Rightarrow\orbr{\begin{cases}x+2019=1\\x+2019=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=-2018\\x=-2020\end{cases}}}\)
\(4,\left(x+10\right)^3=1\Rightarrow\left(x+10\right)^3=1^3\)
\(\Rightarrow x+10=1\)
\(\Rightarrow x=-9\)
\(A = 1 + 4 + 4^2 + ... + 4\)\(20\)
\(4A = 4 + 4^2 + 4^3 + ...+ 4\)\(21\)
\(4A - A = ( 4+ 4^2 + 4^3 + ... + 4\)\(21\)\()\)\(- ( 1 + 4 + 4^2 + ... + 4\)\(20\) \()\)
\(3A = 2\)\(21\) \(- 1\)
\(\Leftrightarrow\)\(3A + 1 = 2\)\(21\)\(= ( 2^3)^7\)\(= 8^7\)
\(Ta có : 8^7 < 63^7 \)
\(Nên 3A + 1 < 63^7\)
Vì A= 4^0 + 4^1 + 4^2+ 4^3+....+4^20
Suy ra: 4A= 4^1+4^2+4^3+4^4+......+ 4^21
Suy ra:4A-A= 4^21 - 4^0
Suy ra: 3A = 4^21-1
Suy ra: A= (4^21-1) : 3
Suy ra: 3A+1= 3. [ ( 4^21-1) : 3] +1
Suy ra: 3A+1 = ( 4^21-1)+1
Suy ra: 3A + 1 = 4^21= (4^3)^7=64^7
Vì 64 > 63; 7=7
Suy ra: 64^7 > 63^7 hay 3A+1 > 63^7
a) \(\frac{4^2.25^2+32,125}{2^3.5^2}=\frac{\left(4.25\right)^2+32,125}{8.25}=\frac{100^2+32,125}{200}=\frac{10000+32,125}{200}=\frac{10032,125}{200}=50,160625\)
b) \(\frac{4^5.9^4.2.6^9}{2^{10}.3^8+20}=\frac{2^{10}.3^8.2.2^9.3^9}{2^{10}.3^8+2^2.5}=\frac{2^{10}.3^9}{2^2.5}=\frac{20155392}{20}=1007769,6\)
Mình không chắc là đúng hay sai đâu nhé ! Nếu sai mong bạn thứ lỗi!
a) \(\frac{18^4.3^2.8^3}{27^3.16^2}=\frac{\left(2.3^2\right)^4.3^2.\left(2^3\right)^3}{\left(3^3\right)^3.\left(2^4\right)^2}=\frac{2^4.2^9.3^8.3^2}{3^9.2^8}=\frac{2^{13}.3^{10}}{3^9.2^8}=3.2^5=96\)
b) \(\frac{35^5.9^3.8^5}{81^4.32^5}=\frac{35^5.\left(3^2\right)^3.\left(2^3\right)^5}{\left(3^4\right)^4.\left(2^5\right)^5}=\frac{35^5.3^6.2^{15}}{3^{16}.2^{25}}=\frac{35^5}{3^{10}.2^{10}}=\frac{35^5}{6^{10}}\)
c) \(\frac{48^5.18^2}{81^2.34^4}=\frac{\left(2^4.3\right)^5.\left(2.3^2\right)^2}{\left(3^4\right)^2.\left(2.17\right)^4}=\frac{2^{20}.3^5.2^2.3^4}{3^8.2^4.17^4}=\frac{2^{22}.3^9}{3^8.2^4.17^4}=\frac{2^{18}.3}{17^4}\)
d) \(\frac{54^7.27^3.16^2}{243^2.64^3}=\frac{\left(2.3^3\right)^7.\left(3^3\right)^3.\left(2^4\right)^2}{\left(3^5\right)^2.\left(2^6\right)^3}=\frac{2^7.3^{21}.3^9.2^8}{3^{10}.2^{18}}=\frac{2^{15}.3^{30}}{3^{10}.2^{18}}=\frac{3^{20}}{2^3}\)
\(30^{20}:\left(3^{15}\cdot2^3+3^{15}\cdot2020^0\right)=30^{20}:\left(3^{15}\cdot2^3+3^{15}\cdot1\right)\)
\(=30^{20}:\left[3^{15}\cdot\left(2^3+1\right)\right]\)
\(=30^{20}:\left(3^{15}\cdot3^2\right)\)
\(=\left(3\cdot10\right)^{20}:3^{17}\)
\(=3^{20}:3^{17}\cdot10^{20}\)
\(=3^3\cdot10^{20}\)
\(=27\cdot100000000000000000000\)
\(=2700000000000000000000\)
1) 6x+2=216=63
=>x+2=3
=>x=1
2)72-(15+x)=5.22
49-15-x=5.4
34-x=20
x=14
3)[(6x-72):2-84].28=5628
(3x-36-84).28=5628
3x-36-84=201
3x-120=201
3x=321
x=107
4)3x-2.4=324
3x-2=81=34
=>x-2=4
x=6
\(6^{x+2}=216\Leftrightarrow6^x=216:6^2=6;x=1\)\(7^2-\left(15+x\right)=5.2^2\Leftrightarrow49-\left(15+x\right)=20\)
\(15+x=49-20=29;x=14\)
2
a) 5 X - 5 mu 3=5
5 X - 125=5
5 X=5+125
5 X=130
X=130:5
X=26
mk chi biet moi bai do thoi sorry ban
- Bài 1:
a)1117-1116:{1240-(2^4-5)^2+[39-9(3^2-7)]:7}
=1117-1116:{1240-121+[39-9.2]:7}
=1117-1116:{1240-121+21:7}
=1117-1116:1122
=\(\frac{208693}{187}\)
b)7+10+13+16+...+2014+2017
Số số hạng của tổng là: (2017-7):3+1=671
Tổng: (2017+7).671=1358104
- Bài 2:
a)5x- 5^3=5 b)3(x-7)-128=157 c)611-11(5x+37)=39 d)3x.3x+1.3x+2=31.32.33.34.35
5x=5+5^3 3(x-7)=157+128 11(5x+37)=611-39 33x+3=315
5x=130 3(x-7)=285 11(5x+37)=572 => 3x+3=15
x=130:5 x-7=285:3 5x+37=572:11 3x=15-3
x=26 x-7=95 5x+37=52 3x=12
x=95+7 5x=52-37=15 x=12:3
x=102 x=15:3=5 x=4
Thấy đúng thì k cho mình nha ^^
360 : [ (3\(x\) + 48): \(x\)] = 24
(3\(x\) + 48):\(x\) = 360 : 24
(3\(x\) + 48) : \(x\) = 15
3\(x\) + 48 = 15\(x\)
15\(x\) - 3\(x\) = 48
12\(x\) = 48
\(x\) = 48 : 12
\(x\) = 4
ai biết ko cứu tớ với