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

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

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

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

$x^2-8xy+16y^2$

$=x^2-2.x.4y+(4y)^2$

$=(x-4y)^2$

$=5^2=25$

$(2x+3y)^2+(2x-3y)^2-2(4x^2-9y^2)$

$=(2x-3y)^2-2(2x-3y)(2x+3y)+(2x+3y)^2$

$=[(2x-3y)-(2x+3y)]^2$

$=(2x-3y-2x-3y)^2$

$=(-6y)^2=36y^2$

a) \(x^2+y^2-4y+3=0\)

\(\Leftrightarrow x^2+\left(y-2\right)^2=1\)

Xét 2TH:

TH1: \(\left\{{}\begin{matrix}x=1\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x=0\\y-2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=3\end{matrix}\right.\)

Vậy có các cặp số nguyên \(\left(1;2\right),\left(3;0\right)\) thỏa mãn đề bài.

b) \(x^2+4y^2-2x+12y+1=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(2y+3\right)^2=9\)

Ta thấy \(2x+3\) là số lẻ nên ta chỉ có 1 TH duy nhất là 

\(\left\{{}\begin{matrix}2y+3=9\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=1\end{matrix}\right.\)

Vậy cặp số nguyên \(\left(1;3\right)\) thỏa mãn ycbt.

a: \(x^2+y^2-4y+3=0\)

=>\(x^2-1+\left(y^2-4y+4\right)=0\)

=>\(\left(x-1\right)\left(x+1\right)+\left(y-2\right)^2=0\)

=>\(\left\{{}\begin{matrix}\left(x-1\right)\left(x+1\right)=0\\\left(y-2\right)^2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\in\left\{1;-1\right\}\\y=2\end{matrix}\right.\)

b: \(x^2+4y^2-2x+12y+1=0\)

=>\(x^2-2x+1+4y^2+12y=0\)

=>\(\left(x-1\right)^2+4y\left(y+3\right)=0\)

=>\(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\4y\left(y+3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y\in\left\{0;-3\right\}\end{matrix}\right.\)

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