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1: \(1026-\left\lbrack\left(3^4+1\right):41\right\rbrack\)
\(=1026-82:41\)
=1026-2
=1024
\(2^{11}:\left\lbrace1026-\left\lbrack\left(3^4+1\right):41\right\rbrack\right\rbrace\)
\(=2^{11}:2^{10}=2\)
2: \(250:\left\lbrace1500:\left\lbrack4\cdot5^3-2^3\cdot25\right\rbrack\right\rbrace\)
\(=250:\left\lbrace1500:\left\lbrack4\cdot125-8\cdot25\right\rbrack\right\rbrace\)
\(=250:\left\lbrace1500:\left\lbrack500-200\right\rbrack\right\rbrace=250:\frac{1500}{3}=250:500=0,5\)
3: \(12+3\cdot\left\lbrace90:\left\lbrack39-\left(2^3-5\right)^2\right\rbrack\right\rbrace\)
\(=12+3\cdot\left\lbrace90:\left\lbrack39-\left(8-5\right)^2\right\rbrack\right\rbrace\)
\(=12+3\cdot\left\lbrace90:\left\lbrack39-3^2\right\rbrack\right\rbrace\)
\(=12+3\cdot\left\lbrace90:\left\lbrack39-9\right\rbrack\right\rbrace\)
\(=12+3\cdot\left\lbrace90:30\right\rbrace=12+3\cdot3=21\)
4: \(24:\left\lbrace390:\left\lbrack500-\left(5^3+49\cdot5\right)\right\rbrack\right\rbrace\)
\(=24:\left\lbrace390:\left\lbrack500-\left(125+245\right)\right\rbrack\right\rbrace\)
\(=24:\left\lbrace390:\left\lbrack500-125-245\right\rbrack\right\rbrace\)
\(=24:\left\lbrace390:\left\lbrack375-245\right\rbrack\right\rbrace\)
\(=24:\left\lbrace390:130\right\rbrace=\frac{24}{3}=8\)
5: \(117:\left\lbrace\left\lbrack79-3\cdot\left(3^3-17\right)\right\rbrack:7+2\right\rbrace\)
\(=117:\left\lbrace\left\lbrack79-3\cdot\left(27-17\right)\right\rbrack:7+2\right\rbrace\)
\(=117:\left\lbrace\left\lbrack79-3\cdot10\right\rbrack:7+2\right\rbrace\)
\(=117:\left\lbrace49:7+2\right\rbrace=\frac{117}{9}=13\)
6: \(514-4\cdot\left\lbrace\left\lbrack40+8\left(6-3\right)^2\right\rbrack-12\right\rbrace\)
\(=514-4\cdot\left\lbrace\left\lbrack40+8\cdot3^2\right\rbrack-12\right\rbrace\)
\(=514-4\cdot\left\lbrace\left\lbrack40+8\cdot9\right\rbrack-12\right\rbrace\)
\(=514-4\cdot\left\lbrace112-12\right\rbrace\)
\(=514-4\cdot100=514-400=114\)
7: \(25\cdot\left\lbrace32:\left\lbrack\left(12-4\right)+4\cdot\left(16:2^3\right)\right\rbrack\right\rbrace\)
\(=25\cdot\left\lbrace32:\left\lbrack8+4\cdot2\right\rbrack\right\rbrace\)
\(=25\cdot\left\lbrace32:16\right\rbrace=25\cdot2=50\)
8: \(30:\left\lbrace175:\left\lbrack355-\left(135+37\cdot5\right)\right\rbrack\right\rbrace\)
\(=30:\left\lbrace175:\left\lbrack355-\left(135+185\right)\right\rbrack\right\rbrace\)
\(=30:\left\lbrace175:\left\lbrack355-320\right\rbrack\right\rbrace=30:\left\lbrace175:35\right\rbrace=\frac{30}{5}=6\)
9: \(32:\left\lbrace160:\left\lbrack300-\left(175+21\cdot5\right)\right\rbrack\right\rbrace\)
\(=32:\left\lbrace160:\left\lbrack300-\left(175+105\right)\right\rbrack\right\rbrace\)
\(=32:\left\lbrace160:\left\lbrack300-280\right\rbrack\right\rbrace\)
\(=32:\left\lbrace160:20\right\rbrace=\frac{32}{8}=4\)
10: \(750:\left\lbrace130-\left\lbrack\left(5\cdot14-65\right)^3+3\right\rbrack\right\rbrace\)
\(=750:\left\lbrace130-\left\lbrack\left(70-65\right)^3+3\right\rbrack\right\rbrace\)
\(=750:\left\lbrace130-\left\lbrack5^3+3\right\rbrack\right\rbrace\)
\(=750:\left\lbrace130-128\right\rbrace=750:2=375\)
\(1)2^8:2^4+3^2\cdot3=2^4+3^3=16+27=43\)
\(2)3^{24}:3^{21}+2^2\cdot2^3=3^3+2^5=27+32=59\)
\(3)5^9:5^7+12\cdot3+7^0=5^2+36+1=25+37=62\)
\(4)5^6:5^4+3^2-2021^0=5^2+3^2-1=25+9-1=33\)
\(5)3^{19}:3^{16}+5^2\cdot2^3-1^{2021}=3^3+25\cdot8-1=27+200-1=226\)
\(6)3^6:3^5+2\cdot2^3+2021^0=3^1+2^4+1=3+16+1=20\)
Câu 6:
2\(xy\) - 5 = y - 2\(x\)
2\(xy\) + 2\(x\) = 5 + y
\(x\)(2y + 2) = 5 + y
\(x\) = \(\frac{5+y}{2y+2}\)
\(x\) ∈ Z ⇔ (5+ y) ⋮ (2y + 2)
(10 + 2y) ⋮ (2y + 2)
[(2y+ 2) + 8] ⋮ (2y+ 2)
8 ⋮ (2y+ 2)
(2y+ 2) ∈ Ư(8) = {-8; -4; -2; - 1; 1; 2; 4; 8}
Lập bảng ta có:
2y+2 | -8 | -4 | -2 | -1 | 1 | 2 | 4 | 8 |
y | -5 | -3 | -2 | -3/2 | 1/2 | 0 | 1 | 3 |
\(x=\frac{5+y}{2y+2}\) | 0 | -1/2 | -3/2 | 5/4 | 3/2 | 1 | ||
\(x;y\in Z\) | ktm | ktm | ktm | ktm | ktm | tm |
Theo bảng trên ta có: (\(x;y\)) = (0; -5); (1; 3)
2xy - 5 = y - 2x
2xy - y + 2x =5
2x( y+1 ) - y =5
2x( y+1 ) - ( y+1 ) = 5-1
(2x+1)(y+1) = 4
Vì x,y là số nguyên nên:
2x+1 là số nguyên lẻ
y+1 là số nguyên.
Tá có:
4= 1.4 = 4.1 =2.2 = (-1)(-4) = (-4)(-1) = (-2)(-2)
TH1:
2x+1=1
y+1=4
=> x=0 ; y=3 (Thoả mãn)
Tương tự với các TH khác...
1: \(3^2\cdot5^3+9^2\)
\(=9\cdot125+81\)
=1125+81
=1206
2: \(55+45:3^2\)
\(=55+45:9\)
=55+5
=60
3: \(8^3:4^2-5^2=64:16-25=4-25=-21\)
4: \(5\cdot3^2-32:2^2=5\cdot9-32:4=45-8=37\)
5: \(16:2^3+5^2\cdot4=16:8+25\cdot4\)
=2+100
=102
6: \(5\cdot2^2-18:3^2\)
\(=5\cdot4-18:9\)
=20-2
=18
7: \(3\cdot5^2-15\cdot2^2=3\cdot25-15\cdot4=75-60=15\)
8: \(2^3\cdot6-72:3^2=8\cdot6-72:9=48-8=40\)
9: \(5\cdot2^2-27:3^2\)
\(=5\cdot4-27:9\)
=20-3
=17
10: \(3\cdot2^4+81:3^2=3\cdot16+81:9=48+9=57\)
11: \(4\cdot5^3-32:2^5=4\cdot125-32:32=500-1=499\)
12: \(6\cdot5^2-32:2^4=6\cdot25-32:16=150-2=148\)
Ta có: \(F=5+5^3+5^5+\cdots+5^{101}\)
=>\(25F=5^3+5^5+5^7+\cdots+5^{103}\)
=>\(25F-F=5^3+5^5+5^7+\cdots+5^{103}-5-5^3-5^5-\cdots-5^{101}\)
=>\(24F=5^{103}-5\)
=>\(F=\frac{5^{103}-5}{24}\)
Ta có: \(5^{103}+1>5^{103}-5\)
=>\(\frac{5^{103}+1}{24}>\frac{5^{103}-5}{24}\)
=>E>F
Bài 8:
a: \(5^3=125;3^5=243\)
mà 125<243
nên \(5^3<3^5\)
b: \(7\cdot2^{13}<8\cdot2^{13}=2^3\cdot2^{13}=2^{16}\)
c: \(27^5=\left(3^3\right)^5=3^{3\cdot5}=3^{15}\)
\(243^3=\left(3^5\right)^3=3^{5\cdot3}=3^{15}\)
Do đó: \(27^5=243^5\)
d: \(625^5=\left(5^4\right)^5=5^{4\cdot5}=5^{20}\)
\(125^7=\left(5^3\right)^7=5^{3\cdot7}=5^{21}\)
mà 20<21
nên \(625^5<125^7\)
Bài 9:
a: \(3^{x}\cdot5=135\)
=>\(3^{x}=\frac{135}{5}=27=3^3\)
=>x=3(nhận)
b: \(\left(x-3\right)^3=\left(x-3\right)^2\)
=>\(\left(x-3\right)^3-\left(x-3\right)^2=0\)
=>\(\left(x-3\right)^2\cdot\left\lbrack\left(x-3\right)-1\right\rbrack=0\)
=>\(\left(x-3\right)^2\cdot\left(x-4\right)=0\)
=>\(\left[\begin{array}{l}x-3=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\left(nhận\right)\\ x=4\left(nhận\right)\end{array}\right.\)
c: \(\left(2x-1\right)^4=81\)
=>\(\left[\begin{array}{l}2x-1=3\\ 2x-1=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=4\\ 2x=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\left(nhận\right)\\ x=-1\left(loại\right)\end{array}\right.\)
d: \(\left(5x+1\right)^2=3^2\cdot5+76\)
=>\(\left(5x+1\right)^2=9\cdot5+76=45+76=121\)
=>\(\left[\begin{array}{l}5x+1=11\\ 5x+1=-11\end{array}\right.\Rightarrow\left[\begin{array}{l}5x=10\\ 5x=-12\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\left(nhận\right)\\ x=-\frac{12}{5}\left(loại\right)\end{array}\right.\)
e: \(5+2^{x-3}=29-\left\lbrack4^2-\left(3^2-1\right)\right\rbrack\)
=>\(2^{x-3}+5=29-\left\lbrack16-9+1\right\rbrack\)
=>\(2^{x-3}+5=29-8=21\)
=>\(2^{x-3}=16=2^4\)
=>x-3=4
=>x=4+3=7(nhận)
f: \(3+2^{x-1}=24-\left\lbrack4^2-\left(2^2-1\right)\right\rbrack\)
=>\(2^{x-1}+3=24-\left\lbrack16-4+1\right\rbrack=24-13=11\)
=>\(2^{x-1}=11-3=8=2^3\)
=>x-1=3
=>x=4(nhận)
Bài 6:
a: \(5\cdot5\cdot5\cdot5\cdot5\cdot5=5^6\)
b: \(27\cdot14\cdot7\cdot2=27\cdot14\cdot14=3^3\cdot14^2\)
c: \(x\cdot x\cdot x\cdot y=x^3\cdot y\)
d: \(5^3\cdot5^4=5^{3+4}=5^7\)
e: \(7^8:7^2=7^{8-2}=7^6\)
f: \(42^7:6^7\cdot49=7^7\cdot49=7^7\cdot7^2=7^{7+2}=7^9\)







Câu 6:
a;
b;
Câu 7:
Các bộ ba điểm có trong hình vẽ trên là:
(I; M; R); (I; N; S); (I; P; T); (M; N; P); (R; S; T)
Câu 8:
Sơ đồ trồng cây tổ 1: với 7 cây trồng thành 5 hàng, mỗi hàng có 3 cây: