$y=f(x)=3x+1$

a) $f(-2) = 3.(-2)+1 = -6+1 = -5$

$f(\dfrac12) = 3.\dfrac12 + 1 = \dfrac32 + 1 = \dfrac52$

b) $y=f(x)=5x$

$5x = 20$

$\to x= 4$

$7\;\vdots\; x-2$

$\to x-2 \in Ư(7)=\{\pm 1; \pm 7\}$

$\to x\in \{3;1;9;-5\}$

$\dfrac{x+7}{x+4} = \dfrac{x+4+3}{x+4} = 1+\dfrac3{x+4}$

Để $\dfrac{x+7}{x+4}$ nguyên thì $3\;\vdots\; x+4$

$\to x+4 \in Ư(3)=\{\pm 1;\pm 3\}$

$\to x\in \{-3;-5;-1;-7\}$

$A=x^2-2x+4$

$A=x^2-2x+1+3$

$A=(x-1)^2+3\ge 3$

Vậy $\min A = 3$ khi $x-1 = 0\leftrightarrow x=1$

$\sqrt{12x+8}.\sqrt{27x+18}+3=0$

$\leftrightarrow 2\sqrt{3x+2}.3\sqrt{3x+2}+3=0$

$\leftrightarrow 6|3x+2|+3=0$

$\leftrightarrow |3x+2|=-\dfrac12$

$\leftrightarrow x\in \varnothing$

`x^4-49a^2`

`= (x^2)^2-(7a)^2`

`= (x^2-7a)(x^2+7a)`

`2x^2:x`

`= 2x`

`2x^2+2xy+4x`

`= 2x(x+y+2)`

`1+3x+3x^2+x^3`

`= (1+x)^3`

$y=f(x)=2x$

$\bullet\; f(1)=2.1=2 \to A(1;2)$

$\bullet\; f(2)=2.2=4 \to B(2;4)$

Từ hai điểm trên ta có được đồ thị $y=2x$

$(x-5)^2=16$

$\leftrightarrow x-5=4$ hoặc $x-5=-4$

$\leftrightarrow x=9$ hoặc $x=1$

Vậy $x\in \{9;1\}$

$10-(x-3):2=7^2-1^{10}$

$\leftrightarrow 10-(x-3):2=49-1$

$\leftrightarrow 10-(x-3):2=48$

$\leftrightarrow (x-3):2=-38$

$\leftrightarrow x-3=-76$

$\leftrightarrow x=-73$

Vậy $x=-73$

Ta có: $\dfrac{x}{2} = \dfrac y5 = k$

$\to \begin{cases}x=2k\\ y=5k\end{cases}$

Mặt khác: $xy=100$

$\to 10k^2=100$

$\to k^2=10$

$\to k=\pm \sqrt{10}$

$\to \begin{cases}x=2\sqrt{10}\\ y=5\sqrt{10}\end{cases}$ hoặc $\begin{cases}x=-2\sqrt{10}\\y=-5\sqrt{10}\end{cases}$

a) `5x^2+10x+5-5y^2`

`= 5(x^2+2x+1-y^2)`

`= 5[(x+1)^2-y^2]`

`= 5(x+1-y)(x+1+y)`

b) `x^2-2022x+xy-2022y`

`= x(x-2022)+y(x-2022)`

`= (x+y)(x-2022)`

c) `9x^2+x-8`

`= 9x^2+9x-8x-8`

`= 9x(x+1)-8(x+1)`

`= (9x-8)(x+1)`

$\cot (-3x)=1$

$\leftrightarrow \cot (-3x)=\cot \left(\dfrac \pi 4\right)$

$\leftrightarrow -3x = \dfrac \pi 4 + k\pi$

$\leftrightarrow x=-\dfrac{\pi}{12}-\dfrac{k\pi}{3} \ (k\in \mathbb{Z})$

$\dfrac{x}{10}=\dfrac y{15}=\dfrac z{12}$

$\to \dfrac{2x}{20}=\dfrac y{15}=\dfrac{3z}{36}$

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

$\dfrac{2x}{20}=\dfrac y{15}=\dfrac{3z}{36} = \dfrac{2x+y-3z}{20+15-36}=\dfrac{-2}{-1}=2$

$\to \begin{cases}x=20\\y=30\\z=24\end{cases}$

`BCN N(5;7)=5.7=35`

`x^2+4xy-4z^2+4y^2`

`= (x+2y)^2-4z^2`

`= (x+2y-2z)(x+2y+2z)`

`x^2-1-2xy+y^2`

`= (x-y)^2-1`

`= (x-y-1)(x-y+1)`

$\dfrac{x+5}{x+1}=\dfrac{x+1+4}{x+1}=1+\dfrac4{x+1}$

Để $x+5\;\vdots\; x+1$ thì $4\;\vdots\; x+1$

$\to x+1\in Ư(4)=\{\pm 1;\pm 2;\pm 3\}$

$\to x\in \{0;-2;1;-3;2;-4\}$