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( 3x - 24 ) . 73 = 2 . 74
( 3x - 24 ) = 2 . 74 : 73
( 3x - 24 ) = 2 . 7
3x - 24 = 14
3x = 14 + 24
3x = 14 + 16
3x = 30
x = 10
- Hok T -
\(16^x< 128^4\)
=> \(\left[2^4\right]^x< \left[2^7\right]^4\)
=> \(2^{4x}< 2^{28}\)
=> 4x < 28
=> x < 7
Đến đây tìm x được rồi
\(\left[3x^2-5\right]+3^4+6^0=5^3\)
=> \(\left[3x^2-5\right]=5^3-6^0-3^4=43\)
=> \(3x^2-5=43\)
=> \(3x^2=48\)
=> \(x^2=16\)
=> \(x=\pm4\)
\(3x+2x\left[2^3\cdot5-3^2\cdot4\right]+5^2=4^4\)
=> \(3x+2x\left[8\cdot5-9\cdot4\right]+25=256\)
=> \(3x+2x\cdot4+25=256\)
=> \(3x+2x\cdot4=231\)
Đến đây tìm x
\(a,3x-6=5x+2\)
\(3x-5x=2+6\)
\(-2x=8\)
\(x=-4\)
\(b,2\times\left(x-3\right)-3\times\left(x+7\right)=14\)
\(2x-6-3x-21=14\)
\(-x-27=14\)
\(x=-27-14\)
\(x=-41\)
a) 3x - 6 = 5x + 2
3x - 6 - 2 = 5x
3x - 8 = 5x
3x - 5x = 8
-2x = 8
x = -4
b) 2(x - 3) - 3(x + 7) = 14
2x - 6 - 3x - 21 = 14
(-x) - 27 = 14
(-x) = 41
x = -41
c) 3x2 - x - 2 = 0
x.(3x - 1) = 2
x.(3x - 1) = 2 = 1.2 = 2.1 = (-1).(-2) = (-2).(-1)
Xét 4 trường hợp ,ta có :
\(\left(1\right)\hept{\begin{cases}x=1\\3x-1=2\end{cases}\Rightarrow\hept{\begin{cases}x=1\\x=1\end{cases}}}\)(nhận)
\(\left(2\right)\hept{\begin{cases}x=2\\3x-1=1\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=\frac{2}{3}\end{cases}}}\)(loại)
\(\left(3\right)\hept{\begin{cases}x=-1\\3x-1=-2\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\x=-\frac{1}{3}\end{cases}}}\)(loại)
\(\left(4\right)\hept{\begin{cases}x=-2\\3x-1=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-2\\x=0\end{cases}}}\)(loại)
Ta có : ( 217 + 153 ) . ( 345 - 65 ) . ( 24 - 42 )
= ( 217 + 153 ) . ( 345 - 65 ) . (16 - 16)
= ( 217 + 153 ) . ( 345 - 65 ) . 0 = 0
\(S=3^0+3^2+3^4+3^6+.....+3^{2020}\)
\(3^2S=3^2+3^4+3^6+.....+3^{2020}+3^{2020}\)
\(9S-S=8S=3^{2020}-1\)
\(S=\frac{3^{2020}-1}{8}\)
\(\frac{4^6\cdot3^4\cdot9^5}{6^{12}}\)
\(=\frac{2^{12}\cdot3^4\cdot3^{10}}{2^{12}\cdot3^{12}}\)
\(=\frac{2^{12}\cdot3^{14}}{2^{12}\cdot3^{12}}\)
\(=3^2\)
\(=9\)
S = 1 + 32 + 34 + 36 + ... + 392 + 394 + 396 + 398
= (1 + 32) + (34 + 36) + ... + (392 + 394)+ (396 + 398)
= (1 + 32) + 34(1 + 32) + .... + 392(1 + 32) + 396(1 + 32)
= (1 + 9) + 34(1 + 9) + ..... + 392.( 1 + 9) + 396(1 + 9)
= 10 + 34.10 + ...... + 392.10 + 396.10
= 10(1 + 34 + ..... + 392 + 396) Chia hết cho 10
=> S Chia hết cho 10 (ĐPCM)
S=1+3^2+,,,,,,,+3^97+3^98
S=(1+3^2)+.............+(3^97+3^98)
S=(1+3^2)+............+3^97.(1+3^2)
S=(1+9)+........+3^97.(1+9)
S=10+......+3^97.10 \(⋮\)10
Vì (1+9=10\(⋮\)10)
=>S\(⋮10\)
\(\left(6^{2007}-6^{2006}\right):6^{2006}\)
\(=6^{2007}:6^{2006}-6^{2006}:6^{2006}\)
\(=6^{2007-2006}-1\)
\(=6^1-1\)
\(=6-1\)
\(=5\)
\(\left(7^3+7^5\right).\left(5^4+5^6\right).\left(3^3.3-9^2\right)\)
\(=\left(7^3+7^5\right).\left(5^4+5^6\right).\left(3^{3+1}-9^2\right)\)
\(=\left(7^3+7^5\right).\left(5^4+5^6\right).\left(3^4.9^2\right)\)
\(=\left(7^3+7^5\right).\left(5^4+5^6\right).\left[3^4-\left(3^2\right)^2\right]\)
\(=\left(7^3+7^5\right).\left(5^4+5^6\right).\left(3^4-3^4\right)\)
\(=\left(7^3+7^5\right).\left(5^4+5^6\right).0\)
\(=0\)
(3x-6).3=34
(3x-6)=34:3
(3x-6)=33
(3x-6)=27
3x=27+6
3x=33
=>x=11
t tôi nha bn