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a)
\(175\cdot19+38\cdot175+43\cdot175\\ =175\cdot19+175\cdot38+175\cdot43\\ =175\cdot\left(19+38+43\right)\\ =175\cdot100\\ =17500\)
b)
\(125\cdot75+125\cdot13-80\cdot125\\ =125\cdot75+125\cdot13-125\cdot80\\ =125\cdot\left(75+13-80\right)\\ =125\cdot10\\ =125\cdot8\\ =1000\)
a, 175. 19 + 38. 175 + 43. 175
= 175. 19 + 175. 38 + 175. 43
= 175.(19 + 38 + 43)
= 175. 100
= 17500


2/
Xét phân số \(\dfrac{2n-3}{n+1}=\dfrac{2n+2-5}{n+1}=\dfrac{2n+2}{n+1}-\dfrac{5}{n+1}=\dfrac{2\left(n+1\right)}{n+1}-\dfrac{5}{n+1}=2-\dfrac{5}{n+1}\)
\(n\in Z\Rightarrow2n-3\inƯ\left(5\right)=\left\{-1;-5;1;5\right\}\)
Ta có bảng:
2n - 3 | -1 | -5 | 1 | 5 |
n | 1 | -1 | 2 | 4 |
Vậy \(n\in\left\{-1;1;2;4\right\}\)
1/
(x + 1) + (x + 3) + (x + 5) + ... + (x + 999) = 500
<=> (x + x + x + ... + x) + (1 + 3 + 5 + ... + 999) = 500
Xét tổng A = 1 + 3 + 5 + ... + 999
Số số hạng của A là: (999 - 1) : 2 + 1 = 500
Tổng A là: (999 + 1) x 500 : 2 = 250 000
Do A có 500 số hạng nên có 500 ẩn x.
Vậy ta có: 500x + 250 000 = 500
=> 500x = -249 500
=> x = 499
Vậy x = 499

Ta có: \(\frac{A}{10^{10}}=\frac{10^{20}-6}{10^{20}-6\cdot10^{10}}=\frac{10^{20}-6\cdot10^{10}+6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}=1+\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}\)
\(\frac{B}{10^{10}}=\frac{10^{21}-6}{10^{21}-6\cdot10^{10}}=\frac{10^{21}-6\cdot10^{10}+6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}=1+\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}\)
Ta có: \(10^{20}<10^{21}\)
=>\(10^{20}-6\cdot10^{10}<10^{21}-6\cdot10^{10}\)
=>\(\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}>\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}\)
=>\(\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}+1>\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}+1\)
=>\(\frac{A}{10^{10}}>\frac{B}{10^{10}}\)
=>A>B

\(\dfrac{15}{34}+\dfrac{1}{3}+\dfrac{19}{34}-\dfrac{4}{3}+\dfrac{3}{7}=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{3}{7}=1-1+\dfrac{3}{7}=\dfrac{3}{7}\)

c: \(\left(x-1\right)^3=\left(-9\right)^3\)
=>x-1=-9
=>x=-9+1=-8
f: \(3x-2^3=7+\left(-9\right)\)
=>3x-8=7-9=-2
=>3x=-2+8=6
=>x=2

Câu 8:
a:Sửa đề: \(4+4^2+\cdots+4^{2025}\)
Ta có: \(4+4^2+\cdots+4^{2025}\)
\(=\left(4+4^2+4^3\right)+\left(4^4+4^5+4^6\right)+\cdots+\left(4^{2023}+4^{2024}+4^{2025}\right)\)
\(=4\left(1+4+4^2\right)+4^4\left(1+4+4^2\right)+\cdots+4^{2023}\left(1+4+4^2\right)\)
\(=21\left(4+4^4+\cdots+4^{2023}\right)\) ⋮21
b: \(5+5^2+5^3+5^4+\cdots+5^{2024}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdots+\left(5^{2023}+5^{2024}\right)\)
\(=\left(5+5^2\right)+5^2\left(5+5^2\right)+\cdots+5^{2022}\left(5+5^2\right)\)
\(=30\left(1+5^2+\cdots+5^{2022}\right)\) ⋮30
Câu 7:
a: \(A=2+2^2+2^3+\cdots+2^{99}\)
=>\(2A=2^2+2^3+\cdots+2^{100}\)
=>\(2A-A=2^2+2^3+\cdots+2^{100}-2-2^2-\cdots-2^{99}\)
=>\(A=2^{100}-2\)
b: \(B=1-7+7^2-7^3+\cdots+7^{48}-7^{49}\)
=>\(7B=7-7^2+7^3-7^4+\cdots+7^{49}-7^{50}\)
=>\(7B+B=7-7^2+7^3-7^4+\cdots+7^{49}-7^{50}+1-7+7^2-7^3+\cdots+7^{48}-7^{49}\)
=>\(8B=-7^{50}+1\)
=>\(B=\frac{-7^{50}+1}{8}\)
Câu 4:
a: \(x^3=125\)
=>\(x^3=5^3\)
=>x=5
b: \(11^{x+1}=121\)
=>\(11^{x+1}=11^2\)
=>x+1=2
=>x=2-1=1
c: \(\left(x-5\right)^3=27\)
=>\(\left(x-5\right)^3=3^3\)
=>x-5=3
=>x=3+5=8
d: \(4^5:4^{x}=16\)
=>\(4^{x}=4^5:16=4^5:4^2=4^3\)
=>x=3
e: \(5^{x-1}\cdot8=1000\)
=>\(5^{x-1}=1000:8=125=5^3\)
=>x-1=3
=>x=3+1=4
f: \(2^{x}+2^{x+3}=72\)
=>\(2^{x}+2^{x}\cdot8=72\)
=>\(2^{x}\cdot9=72\)
=>\(2^{x}=\frac{72}{9}=8=2^3\)
=>x=3
g: \(\left(3x+1\right)^3=343\)
=>\(\left(3x+1\right)^3=7^3\)
=>3x+1=7
=>3x=6
=>x=2
h: \(3^{x}+3^{x+2}=270\)
=>\(3^{x}+3^{x}\cdot9=270\)
=>\(10\cdot3^{x}=270\)
=>\(3^{x}=\frac{270}{10}=27=3^3\)
=>x=3
i: \(25^{2x+4}=125^{x+3}\)
=>\(\left(5^2\right)^{2x+4}=\left(5^3\right)^{x+3}\)
=>\(5^{4x+8}=5^{3x+9}\)
=>4x+8=3x+9
=>x=1
Câu 6:
1 giờ=3600 giây
Số tế bào hồng cầu được tạo ra sau mỗi giờ là:
\(25\cdot10^5\cdot3600=25\cdot36\cdot10^7=900\cdot10^7=9\cdot10^9\) =9 tỉ (tế bào)
câu 5:
a. \(16^{16}=\left(2^4\right)^{16}=2^{64}\)
\(64^{11}=\left(2^6\right)^{11}=2^{66}\)
vì \(2^{66}>2^{64}\) nên \(64^{11}>16^{16}\)
b. \(625^5=\left(5^4\right)^5=5^{20}\)
\(125^7=\left(5^3\right)^7=5^{21}\)
\(5^{20}<5^{21}\Rightarrow625^5<125^7\)
c. \(3^{36}=\left(3^3\right)^{12}=27^{12}\)
\(5^{24}=\left(5^2\right)^{12}=25^{12}\)
\(27^{12}>25^{12}\Rightarrow3^{36}>5^{24}\)

Lời giải:
a.
$=\frac{3}{5}-\frac{7}{4}=\frac{12-35}{20}=\frac{-23}{20}$
b.
$=-(2+\frac{5}{8})=-\frac{21}{8}$
c.
$=-(\frac{1}{8}+\frac{5}{9})=-\frac{9+8.5}{8.9}=\frac{-49}{72}$
d.
$=\frac{6}{13}-\frac{14}{39}=\frac{18}{39}-\frac{14}{39}=\frac{4}{39}$
e.
$=\frac{-3}{4}+\frac{5}{7}=\frac{5}{7}-\frac{3}{4}$
$=\frac{20-21}{7.4}=\frac{-1}{28}$
Lời giải:
$\overline{ab9}+\overline{ab}=174$
$\overline{ab}.10+9+\overline{ab}=174$
$\overline{ab}(10+1)+9=174$
$\overline{ab}.11+9=174$
$\overline{ab}=(174-9):11=15$
$\Rightarrow a=1; b=5$
$\Rightarrow a+b=1+5=6$