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\(2⋮x+1\\ \\ x+1\inƯ\left(2\right)=\left\{1;2;-1;-2\right\}\\ \\ \Rightarrow x\in\left\{0;1;-2;-3\right\}\)
Vì \(x\in N\)\(\Rightarrow x\in\left\{0;1\right\}\)
Vậy \(x\in\left\{0;1\right\}\)
a, A=8+12+x = 20 + x
Vì: 20:3 dư 2 => x:3 dư 1 thì A chia hết cho 3
b, A không chia hết cho 3 khi x chia hết cho 3 hoặc x:3 dư 2
`@` `\text {Ans}`
`\downarrow`
`676 . 17 + 17.160 + 164.17`
`= 17.(676 + 160 + 164)`
`= 17. (840 + 160)`
`= 17.1000`
`= 17000`
676 . 17 + 17 . 160 + 164 . 17
= (676 + 160 + 164) . 17
= 1000 . 17
= 17 000
`@` `\text {Ans}`
`\downarrow`
`9)`
\(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\)
`\Rightarrow`\(3+2^{x-1}=24-\left[16-\left(4-1\right)\right]\)
`\Rightarrow`\(3+2^{x-1}=24-\left(16-3\right)\)
`\Rightarrow`\(3+2^{x-1}=24-13\)
`\Rightarrow`\(3+2^{x-1}=11\)
`\Rightarrow`\(2^{x-1}=11-3\)
`\Rightarrow`\(2^{x-1}=8\)
`\Rightarrow`\(2^{x-1}=2^3\)
`\Rightarrow x - 1 = 3`
`\Rightarrow x = 3 + 1`
`\Rightarrow x = 4`
Vậy, `x = 4`
`10)`
\(7^{x+1} +7^x\cdot42=7^{27}\)
`\Rightarrow 7^x . 7 + 7^x . 7 . 6 =`\(7^{27}\)
`\Rightarrow 7^x . 7 , (1 + 6) =`\(7^{27}\)
`\Rightarrow`\(7^{x+1}\cdot7=7^{27}\)
`\Rightarrow`\(7^{x+1}=7^{27}\div7\)
`\Rightarrow`\(7^{x+1}=7^{26}\)
`\Rightarrow x + 1 = 26`
`\Rightarrow x = 26 - 1`
`\Rightarrow x = 25`
Vậy, `x = 25.`
Câu 9:
3 + 2\(^{x-1}\) = 24 - [ 42 - (22 - 1)]
3 + 2\(^{x-1}\) = 24 - [ 16 - (4 - 1)]
3 + 2\(^{x-1}\) = 24 - [ 16 - 3]
3 + 2\(^{x-1}\) = 24 - 13
3 + 2\(^{x-1}\) = 11
2\(^{x-1}\) = 11 - 3
\(2^{x-1}\) = 8
2\(^{x-1}\) = 23
\(x-1\) = 3
\(x\) = 3 + 1
\(x\) = 4
10; 7\(^{x+1}\) + 7\(^x\) . 42 = 727
7\(^x\).( 7 + 42) = 727
7\(^x\) . 49 = 727
7\(^x\) = 727 : 49
7\(^x\) = 725
\(x\) = 25
\(A=5^3+5^4+5^5+...+5^{100}\)
\(A=5^3\left(1+5^1+5^2+...+5^{97}\right)\)
\(A=5^3.\dfrac{5^{97+1}-1}{5-1}=\dfrac{5^3}{4}.\left(5^{98}-1\right)\)
Đẻ \(\dfrac{\sqrt[]{x}-2}{\sqrt[]{x}+3}\) là số nguyên khi
\(\left(\sqrt[]{x}-2\right)⋮\left(\sqrt[]{x}+3\right)\)
\(\Rightarrow\sqrt[]{x}-2-\left(\sqrt[]{x}+3\right)⋮\sqrt[]{x}+3\)
\(\Rightarrow\sqrt[]{x}-2-\sqrt[]{x}-3⋮\sqrt[]{x}+3\)
\(\Rightarrow-5⋮\sqrt[]{x}+3\)
\(\Rightarrow\left(\sqrt[]{x}+3\right)\in\left\{-1;1;-5;5\right\}\)
\(\Rightarrow x\in\left\{\varnothing;\varnothing;\varnothing;4\right\}\Rightarrow x\in\left\{4\right\}\left(x\in Z\right)\)
Ta có: \(\dfrac{\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-5}{\sqrt{x}+3}=1-\dfrac{5}{\sqrt{x}+3}\) nguyên khi:
5 ⋮ \(\sqrt{x}+3\)
\(\Rightarrow\sqrt{x}+3\inƯ\left(5\right)\)
Mà: \(Ư\left(5\right)=\left\{1;-1;5;-5\right\}\)
Và \(x\ge0\) nên \(\sqrt{x}+3\in\left\{5\right\}\)
Ta có bảng sau:
| \(\sqrt{x}+3\) | 5 |
| \(x\) | 4 |
Vậy biểu thức nguyên khi x=4
643 x 48 x 164
= (26)3 \(\times\) (22)8 \(\times\) (24)4
= 218 \(\times\) 216 \(\times\) 216
= 250
giúp mình v mn
a, 24.( 5 - 178 ) + 178 .(10 + 24)
= 24.5 - 24 .178 + 178 .10 + 178.24
= (24.5 + 178.10) -( 24.178 - 178.24)
=(120 + 1780) - 0
= 1900
b ,29.(-101)
= 29.( -100 -1)
= - 2900 - 29
= -2929
c, (-56 + 130) - (43 - 56) - (-20 - 43)
= -56 + 130 - 43 + 56 + 20 + 43
= (-56 + 56) - (43 - 43) + ( 130 + 20)
= 0 - 0 + 150
= 150