\(25< 5^{2x-1}< 5^5\)

\(5^2< 5^{2x-1}< 5^5\)

\(2< 2x-1< 5\)

\(2+1< 2x< 5+1\)

\(3< 2x< 6\)

\(\dfrac{3}{2}< x< 3\)

x=2

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\(A=\dfrac{6n-1}{2n+3}=\dfrac{3\left(2n+3\right)-10}{2n+3}\\ =3-\dfrac{10}{2n+3}\)

Để A nguyên thì: \(\dfrac{10}{2n+3}\) nguyên

\(\Rightarrow10⋮\left(2n+3\right)\)

\(\Rightarrow2n+3\inƯ\left(10\right)=\left\{1;-1;2;-2;5;-5;10;-10\right\}\\ \Rightarrow2n\in\left\{-2;-4;-1;-5;2;-8;7;-13\right\}\\ \Rightarrow n\in\left\{-1;-2;-\dfrac{1}{2};-\dfrac{5}{2};1;-4;\dfrac{7}{2};-\dfrac{13}{2}\right\}\)

\(\left(x-3\right)+\left(x-5\right)+\left(x-7\right)+...+\left(x-19\right)=3618\\ x-3+x-5+x-7+...+x-19=3618\\ 9x-\left(3+5+7+...+19\right)=3618\\ 9x-99=3618\\ 9x=3618+99\\ 9x=3717\\ x=3717:9\\ x=413\)

Vậy...

a: x+|-2|=0

=>x+2=0

=>x=-2

b: \(4x-20=2^5:2^3\)

=>\(4x-20=2^2=4\)

=>\(4x=20+4=24\)

=>\(x=\dfrac{24}{4}=6\)

a) x+|-2| = 0

⇒|x| = 0-(-2)

⇒|x| = 2⇒x= 2 hoặc x= -2

Vậy x = 2 hoặc x = -2 ϵ z

b) 4x - 20 = 2⁵ : 2³

⇒ 4x - 20 = 32 : 8

⇒ 4x - 20 = 4

⇒ 4x = 4+20

⇒4x = 24

⇒ x = 6

Vậy x = 6 ϵ z

Bài 5:

Tổng số tiền Lan phải trả khi mua đồ là:

\(2\cdot26500+5\cdot18000+2\cdot15000=173000\left(đ\right)\)

Số tiền mẹ Lan còn là:

\(200000-173000=27000\left(đ\right)\)

Bài 5:

Số tiền phải trả cho 2kg khoai tây là:

\(2\cdot26500=53000\left(đồng\right)\)

Số tiền phải trả cho 5kg gạo là:

\(5\cdot18000=90000\left(đồng\right)\)

Số tiền phải trả cho 2 nải chuối là:

\(2\cdot15000=30000\left(đồng\right)\)

Số tiền còn lại là:

200000-53000-90000-30000=27000(đồng)

Bài 6:

Đặt x=*

Đặt \(A=\overline{x63x}\)

A chia hết cho 5 và 2 nên x=0

=>\(A=\overline{0630}=630\)

Vì 630 chia hết cho cả 3 và 9

nên A=630 thỏa mãn yêu cầu đề bài

=>x=0

=>*=0

Bài 7:

a: \(126=2\cdot3^2\cdot7;210=2\cdot3\cdot5\cdot7\)

=>\(ƯCLN\left(126;210\right)=2\cdot3\cdot7=42\)

\(126⋮x;210⋮x\)

=>\(x\inƯC\left(126;210\right)\)

=>\(x\inƯ\left(42\right)\)

mà 15<x<30

nên x=21

b: \(12=2^2\cdot3;21=3\cdot7;28=2^2\cdot7\)

=>\(BCNN\left(12;21;28\right)=2^2\cdot3\cdot7=4\cdot3\cdot7=84\)

\(x⋮12;x⋮21;x⋮28\)

=>\(x\in B\left(84\right)\)

mà 150<x<300

nên \(x\in\left\{168;252\right\}\)

a: \(\left[\left(-21,8\right)+4,125\right]+\left[11,8+\left(-2,125\right)\right]\)

=-21,8+4,125+11,8-2,125

=(4,125-2,125)+(-21,8+11,8)

=2-10

=-8

b: \(\left(-124,5\right)+\left(-6,24+124,5\right)\)

\(=-124,5-6,24+124,5\)

=-6,24

\(a,\left[\left(-21,8\right)+4,125\right]+\left[11,8+\left(-2,125\right)\right]\)

\(=-21,8+4,125+11,8-2,125\)

\(=\left[\left(-21,8\right)+11,8\right]+\left(4,125-2,125\right)\)

\(=-10+2\)

\(=-8\)

\(b,\left(-124,5\right)+\left(-6,24+124,5\right)\)

\(=-124,5-6,24+124,5\)

\(=\left[\left(-124,5\right)+124,5\right]-6,24\)

\(=0-6,24\)

\(=-6,24\)

$\color{#90EE90}{\text{4}}$  $\color{#B0E0E6}{\text{56}}$

a) 

\(\left(-12,5\right)+3.4+12,5+\left(-3,4\right)\\ =\left(12,5-12,5\right)+\left(3,4-3,4\right)\\ =0+0=0\)

b) 

\(32,8+4,2+\left(-4,3\right)+\left(-32,8\right)+4,3\\ =\left(32,8-32,8\right)+\left(4,3-4,3\right)+4,2\\ =0+0+4,2\\ =4,2\)

c) 

\(-\left(42,5+150\right)\cdot2,5-7,5\cdot2,5\\ =2,5\left(-42,5-150-7,5\right)\\ =2,5\cdot\left(-50-150\right)\\ =2,5\cdot-200\\ =-500\) 

d) 

\(\left(-2,45\right)\cdot2,6+2,6\cdot\left(-7,55\right)\\ =2,6\cdot\left(-2,45-7,55\right)\\ =2,6\cdot-10\\ =-26\)

a: \(\left(-12,5\right)+3,4+12,5+\left(-3,4\right)\)

\(=\left(-12,5+12,5\right)+\left(3,4-3,4\right)\)

=0+0=0

b: \(32,8+4,2+\left(-4,3\right)+\left(-32,8\right)+4,3\)

\(=\left(32,8-32,8\right)+\left(4,3-4,3\right)+4,2\)

=0+0+4,2

=4,2

c: \(-\left(42,5+150\right)\cdot2,5-7,5\cdot2,5\)

\(=2,5\cdot\left(-42,5-150-7,5\right)\)

\(=2,5\cdot\left(-200\right)=-500\)

d: Sửa đề: \(\left(-2,45\right)\cdot2,6+2,6\cdot\left(-7,55\right)\)

\(=2,6\left(-2,45-7,55\right)\)

\(=2,6\cdot\left(-10\right)=-26\)

a: \(\left(-\dfrac{2}{3}-\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\left(-\dfrac{2}{3}-\dfrac{3}{7}-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\left(-1+\dfrac{1}{7}\right)\cdot\dfrac{5}{4}=\dfrac{-6}{7}\cdot\dfrac{5}{4}=\dfrac{-30}{28}=-\dfrac{15}{14}\)

b: \(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\)

\(=\dfrac{5}{9}:\dfrac{-3}{22}+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{10}{15}\right)\)

\(=\dfrac{5}{9}\cdot\dfrac{-22}{3}+\dfrac{5}{9}:\dfrac{-9}{15}\)

\(=\dfrac{5}{9}\cdot\dfrac{-22}{3}+\dfrac{5}{9}\cdot\dfrac{15}{-9}\)

\(=\dfrac{5}{9}\left(-\dfrac{22}{3}-\dfrac{5}{3}\right)=\dfrac{5}{9}\cdot\dfrac{-27}{3}=\dfrac{5}{9}\cdot\left(-9\right)=-5\)

c: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)

\(=\left(\dfrac{12}{12}+\dfrac{8}{12}-\dfrac{3}{12}\right)\cdot\left(\dfrac{16}{20}-\dfrac{15}{20}\right)^2\)

\(=\dfrac{17}{12}\cdot\left(\dfrac{1}{20}\right)^2=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)

d: \(2:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^2=2:\left(\dfrac{3}{6}-\dfrac{4}{6}\right)^2\)

\(=2:\left(-\dfrac{1}{6}\right)^2=2:\dfrac{1}{36}=72\)

a; (- \(\dfrac{2}{3}\) - \(\dfrac{3}{7}\)): \(\dfrac{4}{5}\) + (- \(\dfrac{1}{3}\) + \(\dfrac{4}{7}\)): \(\dfrac{4}{5}\)

= (- \(\dfrac{2}{3}\) - \(\dfrac{3}{7}\)) x \(\dfrac{5}{4}\) + (- \(\dfrac{1}{3}\) + \(\dfrac{4}{7}\)) x \(\dfrac{5}{4}\)

= (- \(\dfrac{2}{3}\) - \(\dfrac{3}{7}\) - \(\dfrac{1}{3}\) + \(\dfrac{4}{7}\)) x \(\dfrac{5}{4}\)

= [ (- \(\dfrac{2}{3}\) - \(\dfrac{1}{3}\)) - (\(\dfrac{3}{7}\) - \(\dfrac{4}{7}\))] x \(\dfrac{5}{4}\)

= [ - 1 + \(\dfrac{1}{7}\)] x \(\dfrac{5}{4}\)

= [- \(\dfrac{7}{7}\) + \(\dfrac{1}{7}\)] x \(\dfrac{5}{4}\)

= - \(\dfrac{6}{7}\) x \(\dfrac{5}{4}\)

= - \(\dfrac{15}{14}\)