Đề thiếu rồi phải là $30n+2$

Gọi $ƯCLN(12n+1,30n+2)=d(d>0)(d \in N)$

$\to \begin{cases}12n+1 \vdots d\\30n+2 \vdots d\\\end{cases}$

$\to \begin{cases}60n+5 \vdots d\\60n+4 \vdots d\\\end{cases}$

$\to 60n+5-60n-4 \vdots d$

$\to 1 \vdots d$

$\to d=1$

Vậy ƯCLN(12n+1,30n+2)

`(x-3)(3-4x)+(x^2-6x+9)=0`

`=>(x-3)(3-4x)+(x-3)^2=0`

`=>(x-3)(3-4x+x-3)=0`

`=>(x-3).=0`

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

`=>` $\left[ \begin{array}{l}x=3\\x=0\end{array} \right.$

Vậy phương trình có tập nghiệm `S={3,0}`

`(x-3)(3-4x)+(x^2-6x+9)=0`

`=>(x-3)(3-4x)+(x-3)^2=0`

`=>(x-3)(3-4x+x+3)=0`

`=>(x-3)(6-3x)=0`

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

`=>` $\left[ \begin{array}{l}x-3=0\\2-x=0\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=3\\x=2\end{array} \right.$

Vậy phương trình có tập nghiệm `S={3,2}`

`-2/10=-3/15`

`10/(-2)=15/(-3)`

`15/10=(-3)/(-2)`

Lấy 4 số `-5;-3;10;15`

`10/15=(-2)/(-3)`

`4(2x+7)^2-9(x+3)^2 = 0`

`=>(4x+14)^2-(3x+9)^2=0`

`=>(4x+14-3x-9)(4x+14+3x+9)=0`

`=>(x+5)(7x+23)`

`=>` $\left[ \begin{array}{l}x=-5\\x=-\dfrac{23}{7}\end{array} \right.$

Vậy `S={-5,-23/7}`

`f(x)=x+2-4\sqrt{x-1}(x>=1)`

`=x-1-4\sqrt{x-1}+4-1`

`=(\sqrt{x-1}-2)^2-1>=-1`

Dấu "=" xảy ra khi `\sqrt{x-1}=2<=>x=5`

=>We sometimes play marbles

`\sqrt{4x^2+5x+1}-2\sqrt{x^2-x+1}=6-18x`

`<=>\sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}=6-18x`

`<=>(9x-3)/(\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4})+6(3x-1)=0`

`<=>(3x-1)(3/(\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4})+6)=0`

Ta thấy `3/(\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4})+6>0`

`=>3x-1=0`

`=>3x=1`

`=>x=1/3`

Vậy `S={1/3}`

`1/(x^2+9x+20)=1/15-1/(x^2+5x+4)(x ne -1,-4,-5)`

`=>1/((x+4)(x+5))=1/15-1/((x+1)(x+4))`

`=>1/(x+4)-1/(x+5)=1/15-1/((x+1)(x+4))`

`=>3/(x+4)-3/(x+5)=3/15-3/((x+1)(x+4))`

`=>3/(x+4)-3/(x+5)=3/15-1/(x+1)+1/(x+4)`

`=>2/(x+4)-3/(x+5)+1/(x+1)=3/15`

`=>30(x+1)(x+5)-45(x+1)(x+4)+15(x+4)(x+5)=3(x+1)(x+4)(x+5)`

`=>30(x^2+6x+5)-45(x^2+5x+4)+15(x^2+9x+20)=3(x^2+5x+4)(x+5)`

`<=>90x+270=3(x^3+8x^2+29x+20)`

`<=>x^3+24x^2-3x-210=0`

`=>x=-23\or\x=2,85\or\x=-3`

`A=(10^50+2)/(10^50-1)`

`=1+3/(10^50-1)`

Tương tự:

`B=1+3/(10^50-3)`

`10^50-1>10^50-3>0`

`=>3/(10^50-1)<3/(10^50-3)`

`=>A<B`

`20.2^x+1=10.4^2+1`

`=>20.2^x=10.4^2`

`=>2^x=4^2/2=2^3`

`=>x=3`

Vậy x=3