Gọi mẫu số là x

(ĐIều kiện: \(x\ne0\))

Vì phân số nhỏ hơn 1 nên mẫu số>tử số

=>Mẫu số>32/2=16

Tử số là 32-x

Mẫu số khi tăng thêm 10 đơn vị là x+10

Tử số khi giảm đi một nửa là \(\dfrac{32-x}{2}\)

Phân số mới là 2/17 nên \(\dfrac{32-x}{2}:\left(x+10\right)=\dfrac{2}{17}\)

=>\(\dfrac{32-x}{2x+20}=\dfrac{2}{17}\)

=>17(32-x)=2(2x+20)

=>544-17x=4x+40

=>-21x=40-544=-504

=>x=24

Tử số là 32-24=8

Vậy: Phân số cần tìm là \(\dfrac{8}{24}\)

1: cross

2: having

3: breaking

4: go

5: not believing

6: to believe

7: to phone

8: cheat

9: spend

1 cross

2 having

3 breaking

4 go

5 not believing

6 to believe

7 to phone

8 cheat

9 spend

\(4xy^2\cdot x-\left(-12x^2y^2\right)\)

\(=4x^2y^2+12x^2y^2\)

\(=16x^2y^2\)

#include <bits/stdc++.h>

using namespace std;

int main()

{

int n,i,t;

cin>>n;

t=0;

for (int i=1; i<=n; i++)

t=t+i;

cout<<t;

return 0;

}

12.B

13.C

14.C

15.D

16.A

17.A

18.B

hơi khó nha

a)

\(2^{2024}=2^{8.11.23}\)

\(2^8\equiv4\left(mod7\right)\)

\(2^{8.11}\equiv\left(2^8\right)^{11}\left(mod7\right)\equiv4^{11}\left(mod7\right)\equiv2\left(mod7\right)\)

\(\Rightarrow2^{8.11.23}\equiv\left(2^{8.11}\right)^{23}\left(mod7\right)\equiv2^{23}\left(mod7\right)\equiv4\left(mod7\right)\)

\(\Rightarrow2^{2024}\) chia 7 dư 4

\(41^{2023}=41.\left(41^2\right)^{1011}\)

\(41^2\equiv1\left(mod7\right)\)

\(\Rightarrow\left(41^2\right)^{1011}\equiv1^{1011}\left(mod7\right)\equiv1\left(mod7\right)\)

\(\Rightarrow41.\left(41^2\right)^{1011}\equiv41.1\left(mod7\right)\equiv6\left(mod7\right)\)

\(\Rightarrow2^{2024}+41^{2023}\equiv4+6\left(mod7\right)\equiv3\left(mod7\right)\)

Vậy \(2^{2024}+41^{2023}\) chia 7 dư 3

$16(x-1)^2-25=0$

$\Leftrightarrow (4x-4)^2-5^2=0$

$\Leftrightarrow (4x-4-5)(4x-4+5)=0$

$\Leftrightarrow (4x-9)(4x+1)=0$

$\Leftrightarrow \left[\begin{array}{} 4x-9=0\\4x+1=0 \end{array} \right. \Leftrightarrow \left[\begin{array}{} 4x=9\\4x=-1 \end{array} \right.$

$\Leftrightarrow \left[\begin{array}{} x=\frac94\\x=-\frac14 \end{array} \right.$

#$\mathtt{Toru}$

\(16\left(x-1\right)^2-25=0\)

\(16\left(x-1\right)^2=0+25\)

\(16\left(x-1\right)^2=25\)

\(\left(x-1\right)^2=\dfrac{25}{16}\)

\(x-1=\dfrac{5}{4};x-1=-\dfrac{5}{4}\)

*) \(x-1=\dfrac{5}{4}\)

\(x=\dfrac{5}{4}+1\)

\(x=\dfrac{9}{4}\)

*) \(x-1=-\dfrac{5}{4}\)

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

\(x=-\dfrac{1}{4}\)

Vậy \(x=-\dfrac{1}{4};x=\dfrac{9}{4}\)

$(2x-5)^2-x^2=0$

$\Leftrightarrow (2x-5-x)(2x-5+x)=0$

$\Leftrightarrow (x-5)(3x-5)=0$

$\Leftrightarrow \left[\begin{array}{} x-5=0\\ 3x-5=0 \end{array} \right.$

$\Leftrightarrow \left[\begin{array}{} x=5\\x=\frac53 \end{array} \right.$

(x-y)(x-2)=11

=>\(\left(x-y;x-2\right)\in\left\{\left(1;11\right);\left(11;1\right);\left(-1;-11\right);\left(-11;-1\right)\right\}\)
=>\(\left(x-2;x-y\right)\in\left\{\left(1;11\right);\left(11;1\right);\left(-1;-11\right);\left(-11;-1\right)\right\}\)

=>\(\left(x;x-y\right)\in\left\{\left(3;11\right);\left(13;1\right);\left(1;-11\right);\left(-9;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(3;-8\right);\left(13;12\right);\left(1;12\right);\left(-9;-8\right)\right\}\)