Ta có \(-2x+3y\) \(=3\left(7x+y\right)-23x\), lại có \(7x+y⋮23\)  và \(23x⋮23\) nên \(3\left(7x+y\right)-23x⋮23\) hay \(-2x+3y⋮23\) (đpcm)

Theo đề ta có:

\(4a-8=3a+6\)

\(\Rightarrow4a-3a=6+8\)

\(\Rightarrow a=14\)

Vậy với a=14 thì f(a)=g(a)

Lời giải:
$x^2+5x+5=0$

$(x^2+2.2,5x+2,5^2)-1,25=0$

$(x+2,5)^2=1,25$

$\Rightarrow x+2,5=\pm \sqrt{1,25}$

$\Rightarrow x=\pm \sqrt{1,25}-2,5$

`@` `\text {Ans}`

`\downarrow`

\(\left(-3\right)^x\div81=-27\)

`=>`\(\left(-3\right)^x=-27\cdot81\)

`=>`\(\left(-3\right)^x=\left(-3\right)^3\cdot3^4\)

`=>`\(\left(-3\right)^x=\left(-3\right)^7\)

`=> x=7`

Vậy, `x=7.`

`@` `\text {Kaizuu lv u}`

(-3)^x . 81 = -27

<=> (-3)^x . (-3)⁴ = (-3)³

<=> (-3)^x+4 = (-3)³

=> x+4=3

<=>x= -1

a) Ta có: \(x\in B\left(6\right);15< x\le54\)

\(\Rightarrow x\in\left\{18;24;30;36;42;48;54\right\}\)

b) Ta có: \(x\in BC\left(5;8\right);40\le x\le200\)

\(\Rightarrow x\in\left\{40;80;120;160;200\right\}\)

c) Ta có: \(x\inƯC\left(36;48\right);4\le x\le20\)

\(\Rightarrow x\in\left\{4;6;9;12\right\}\)

d) Ta có: \(x\inƯC\left(18,36\right);2\le x\le30\)

\(\Rightarrow x\in\left\{2;3;6;9;18\right\}\)

Lời giải:
Đặt $x+\frac{2}{3}=\frac{a}{b}$ với $a,b$ là số nguyên, $b\neq 0$

$\Rightarrow x=\frac{a}{b}-\frac{2}{3}=\frac{3a-2b}{3b}$

Thấy rằng $3a-2b\in\mathbb{Z}$ với mọi $a,b$ nguyên, $3b\in\mathbb{Z}\neq 0$ với mọi số nguyên $b$ khác $0$

$\Rightarrow x$ là số hữu tỉ.

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1. 

$(3^2-2^3)x+3^2.2^2=4^2.3$

$\Leftrightarrow x+36=48$

$\Leftrightarrow x=48-36=12$

2.

$x^5-x^3=0$

$\Leftrightarrow x^3(x^2-1)=0$

$\Leftrightarrow x^3(x-1)(x+1)=0$

$\Leftrightarrow x^3=0$ hoặc $x-1=0$ hoặc $x+1=0$

$\Leftrightarrow x=0$ hoặc $x=\pm 1$
3.

$(x-1)^2+(-3)^2=5^2(-1)^{100}$

$\Leftrightarrow (x-1)^2+9=25$

$\Leftrightarrow (x-1)^2=25-9=16=4^2=(-4)^2$

$\Rightarrow x-1=4$ hoặc $x-1=-4$

$\Leftrightarrow x=5$ hoặc $x=-3$

4.

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

$\Leftrightarrow (2x-1)(2x-1-1)=0$

$\Leftrightarrow (2x-1)(2x-2)=0$

$\Leftrightarrow 2x-1=0$ hoặc $2x-2=0$

$\Leftrightarrow x=\frac{1}{2}$ hoặc $x=1$

$\Lef

`@` `\text {Ans}`

`\downarrow`

\((3^2-2^3)x+3^2.2^2=4^2.3\)

`=> x + (3*2)^2 = 48`

`=> x+6^2 = 48`

`=> x + 36 = 48`

`=> x = 48 - 36`

`=> x=12`

Vậy, `x=12`

\(x^5-x^3=0\)

`=> x^3(x^2 - 1)=0`

`=>`\(\left[{}\begin{matrix}x^3=0\\x^2-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

Vậy, `x \in {0; +- 1 }`

\(\left(x-1\right)^2+\left(-3\right)^2=5^2\cdot\left(-1\right)^{100}\)

`=> (x-1)^2 + 9 = 25*1`

`=> (x-1)^2 + 9 = 25`

`=> (x-1)^2 = 25 - 9`

`=> (x-1)^2 = 16`

`=> (x-1)^2 = (+-4)^2`

`=>`\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4+1\\x=-4+1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy, `x \in {5; -3}`

\((2x-1)^2-(2x-1)=0\)

`=> (2x-1)(2x-1) - (2x-1)=0`

`=> (2x-1)(2x-1-1)=0`

`=>`\(\left[{}\begin{matrix}2x-1=0\\2x-2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

Vậy, `x \in {1; 1/2}`