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1. Ta có :
275. 498 = 315 . 716
2115 = 715.315
Do 315 . 716 > 715.315
Nên 275. 498 > 2115
Câu 2 đang làm
21^15 và 27^5 . 49^8
27^5 . 49^8 = (3^3)^5 . (7^2)^8 = 3^15 . 7^16 = 3^15 . 7^15 . 7 = 21^15 . 7
Vì 21^15 < 21^15 . 7 nên 21^15 < 27^5 . 49^8
a, 5x - 19 = 106
5x = 106 + 19
5x = 125
Mà 125 = 53
=> x = 3
Vậy x = 3
\(a,TH1:x-2021=0=>x=2021\)
\(Th2:x-2022=0=>x=2022\)
Vậy \(x\in\left\{2021;2022\right\}\)
\(b,x\left(8-5\right)=1080\)
\(x.3=1080\)
\(x=360\)
\(c,x^3=216< =>6^3=216=>x=3\)
\(d,5^5=3125\)
a) ( x- 2021) * ( x- 2022) = 0
=> \(\orbr{\begin{cases}x-2021=0\\x-2022=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2021\\x=2022\end{cases}}}\)
b) b. 8x - 5x = 2022
=> 3x = 2022
=> x = 674
c) \(5\cdot x^3=1080\)
=> \(x^3=216\)
=> \(x^3=6^3\)
=> x = 6
d) \(5^x=3125\)
=> \(5^x=5^5\)
=> x = 5
a. 2x . 7 = 56
2x = 56 : 7
2x = 8
\(\Rightarrow\)23 = 8
Vậy x = 8
a)-6x=6
x=-1
b)8x=35->x=35/8
c)8|x|=35->|x|=35/2->x=35/2;x=-35/2
mấy ý kia tương tự bạn ạ!
\(a)\)\(\left(50-6.x\right).18=2^3.3^2.5\)
\(\Leftrightarrow\)\(\left(50-6.x\right).18=8.9.5\)
\(\Leftrightarrow\)\(\left(50-6.x\right).18=360\)
\(\Leftrightarrow\)\(\left(50-6.x\right)=360\div18\)
\(\Leftrightarrow\)\(50-6.x=20\)
\(\Leftrightarrow\)\(6.x=50-20\)
\(\Leftrightarrow\)\(6.x=30\)
\(\Leftrightarrow\)\(x=5\)
\(b)\)\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=7450\)
\(\Leftrightarrow\)\(100x+\left(1+2+3+...+100\right)=7450\)
\(\Leftrightarrow\)\(100x+5050=7450\)
\(\Leftrightarrow\)\(100x=7450-5050\)
\(\Leftrightarrow\)\(100x=2400\)
\(\Leftrightarrow\)\(x=24\)
b.
(x+1)+(x+2)+...+(x+100)=7450
=> 100x + (1+2+3+...+100)=7450
=>100x + (100+1).50=7450
=>100x=2400
=>x=24
a) -12.(x - 5) + 7(3 - x) = 5
=> -12x + 60 + 21 - 7x = 5
=> -19x + 81 = 5
=> -19x = 5 - 81
=> -19x = -76
=> x = -76 : (-19)
=> x = 4
b) (x + 1) + (x + 2) + (x + 3) + ... + (x + 20) = 250
=> (x + x + x + ... + x) + (1 + 2 + 3 + ... + 20) = 250
=> 20x + 210 = 250
=> 20x = 250 - 210
=> 20x = 40
= > x = 40 : 20
=> x = 2
\(-12\left(x-5\right)+7\left(3-x\right)=5\)
\(\Leftrightarrow-12x+60+21-7x=5\)
\(\Leftrightarrow-19x+81=5\)
\(\Leftrightarrow81-5=19x\)
\(\Leftrightarrow19x=76\)
\(\Leftrightarrow x=4\)
Bài 1 :
\(A=3^0+3^1+3^2+3^3+...+3^{98}\)
\(A=\left(1+3+3^2\right)+.....+\left(3^{97}+3^{98}+3^{99}\right)\) ( Nhóm 3 số 1 nhé )
\(A=13+.....+3^{97}.13⋮13\left(\text{đ}pcm\right)\)
Bài 2 :
Theo ý a ta có :
\(A=13+.....+3^{97}.13+3^{99}+3^{100}\)
\(A=13+.....+3^{97}.13+3^{99}.4⋮̸13\)
Bài 3 :
Để D chia hết cho 2 thì x chia hết cho 2
1. \(A=3^0+3^1+3^2+...+3^{98}\)
\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\)
\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+...+3^{96}\right)\)chia hết cho \(13\).
2. \(B=3^0+3^1+3^2+3^3+...+3^{100}\)
\(=1+3+\left(3^2+3^3+3^4\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)
\(=4+3^2\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)
\(=4+13\left(3^2+3^5+...+3^{98}\right)\)không chia hết cho \(13\).
3. \(D=\left(12.3+26.b+2022.c+x\right)\)chia hết cho \(2\)
\(\Leftrightarrow x⋮2\)(vì \(12.3⋮2,26b⋮2,2022c⋮2\))
a) x2.x3:7=224
=>x5 :7=224
=>x5 =32
=>x5 =25 => x=2
b)x3 :xx +7=8
=>x3-x =1
=>x3-x =13-x
=> x=1
c) xn =1
=> xn=1n
=> x=1
k cho minh nhee:3