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\(a,\left(x+3\right)\left(5-x\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
\(c,x+17⋮x+3\\ x+3+14⋮x+3\\ 14⋮x+3\\ x+3\inƯ\left(14\right)=\left\{\pm14;\pm7\pm2;\pm1\right\}\)
Từ đó bạn tìm những giá trị của x nha!
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Ta có: 23 + 3x = 56 : 53
23 + 3x = 5(6-3)
23 + 3x = 53
23 + 3x = 125
3x = 125 – 23
3x = 102
x = 102 : 3
x = 34
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a, 3 x + 1 : 3 4 = 81
3 x - 3 = 3 4
x – 3 = 4
x = 7
Vậy x = 7
b, 3 x + 3 . 3 x + 1 = 729
3 2 x + 4 = 3 6
2x + 4 = 6
x = 1
Vậy x = 1
c, 2 x + 3 . 2 x = 128
2 2 x + 3 = 2 7
2x + 3 = 7
x = 2
Vậy x = 2
d, 23 + 3 x = 5 6 : 5 3
23 + 3 x = 5 3
23 + 3x = 125
3x = 102
x = 34
Vậy x = 34
e, 2 x + 2 x + 4 = 272
2 x + 2 x . 2 4 = 272
2 x ( 1 + 2 4 ) = 272
2 x . 17 = 272
2 x = 16
2 x = 2 4
x = 4
Vậy x = 4
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a)23+3 𝑥 =125 | b) 70 – 5.(𝑥 - 3)= 45 | c) 20 + 5 𝑥 = 57:55
3 𝑥 +23=125 | 70−5 𝑥 +15=45 | 5 𝑥 +20=5755
𝑥=34 | 𝑥=8 | 𝑥=−104
275
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a) Đặt: \(A=1+2^2+2^3+...+2^{10}\)
\(\Rightarrow2A=2\left(1+2^2+2^3+...+2^9+2^{10}\right)\)
\(\Rightarrow2A=2+2^3+2^4+...+2^{10}+2^{11}\)
\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)
\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2-1\right)+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=0+0+...+1+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=1+2^{11}-2^2=1+2048-4=2045\)
Vậy: \(1+2^2+2^3+...+2^{10}=2045\)
b)
a] \(60-3\left(x-1\right)=2^3\cdot3\)
\(\Rightarrow60-3\left(x-1\right)=24\)
\(\Rightarrow3\left(x-1\right)=36\)
\(\Rightarrow x-1=12\)
\(\Rightarrow x=13\)
b] \(\left(3x-2\right)^3=2\cdot2^5\)
\(\Rightarrow\left(3x-2\right)^3=2^6\)
\(\Rightarrow\left(3x-2\right)^3=\left(2^2\right)^3\)
\(\Rightarrow3x-2=2^2\)
\(\Rightarrow3x=6\)
\(x=2\)
c] \(5^{x+1}-5^x=500\)
\(\Rightarrow5^x\left(5-1\right)=500\)
\(\Rightarrow5^x\cdot4=500\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
d] \(x^2=x^4\)
\(\Rightarrow x=x^2\)
\(\Rightarrow x-x^2=0\)
\(\Rightarrow x\left(1-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\1-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
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Bài 3:
a: Ta có: \(23\left(42-x\right)=23\)
\(\Leftrightarrow42-x=1\)
hay x=41
b: Ta có: 15(x-3)=30
nên x-3=2
hay x=5
Bài 1:
a: 32+89+68=100+89=189
b: 64+112+236=300+112=412
c: \(1350+360+650+40=2000+400=2400\)
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Câu 1:
a: =126+5=131
b: =-564-724+564+224=-500
c: =3x25+2-8
=75+2-8
=77-8
=69
d: =1024:32+140:70-49
=32-49+2
=-15