ĐKXĐ: \(x\ne-1\)

\(\dfrac{1}{x+1}-\dfrac{x}{x^2-x+1}=\dfrac{3}{x^3+1}\)

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

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

=>\(x^2-x+1-x^2-x=3\)

=>-2x+1=3

=>-2x=2

=>x=-1(loại)

vậy: \(x\in\varnothing\)

1) 100 - 7 x ( x - 5 ) = 65

    -7 x ( x - 5 ) = 65 - 100

    -7 x ( x - 5 ) = -35

      x - 5 = \(\dfrac{-35}{-7}\)

      x - 5 = 5 

      x = 10 

2) 7 + 2 x ( x - 3 ) =11 

    2 x ( x - 3 ) = 11 - 7 

    2 x ( x - 3 ) = 4

     x - 3  = \(\dfrac{4}{2}\)

    x - 3 = 2

    x = 5

\(2\cdot4\cdot6\cdot8\cdot10\cdot12⋮5\)

\(40⋮5\)

Do đó: \(A=2\cdot4\cdot6\cdot8\cdot10\cdot12+40⋮5\)

\(2\cdot4\cdot6\cdot8\cdot10\cdot12⋮8;40⋮8\)

Do đó: \(A=2\cdot4\cdot6\cdot8\cdot10\cdot12+40⋮8\)

`A = 2.4.6.8.10.12 + 40`

Ta có: 

`2.4.6.8.10.12` có thừa số `8` và `5 `

`=> 2.4.6.8.10.12⋮ 8` và `5`

`40 ⋮ 8` và `5`

`=> A = 2.4.6.8.10.12 + 40 ⋮ 8` và `5 (dpcm)`

-------------------------------

Nếu `a ⋮c` và `b ⋮c  => a + b  ⋮c `

không kể góc bẹt nha

\(A=n\left(2n-3\right)-2n\left(n+1\right)=2n^2-3n-2n^2-2n=-5n⋮5\)

`A = n(2n - 3) - 2n(n+1) `

`= 2n^2 - 3n - 2n^2 - 2n`

`= -5n `

Mà `-5 ⋮ 5`

`=> -5n  ⋮ 5 ∀n` thuộc `Z`

Hay `A ⋮ 5 ∀n` thuộc `Z`

\(\left(x+4\right)\left(x-4\right)-\left(x-3\right)^2\)

\(=x^2-16-\left(x^2-6x+9\right)\)

\(=x^2-16-x^2+6x-9\)

=6x-25

\(\left(4x+12\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}4x+12=0\\x+5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=-12\\x=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{-3;-5\right\}\)

(4x+12)(x+5)=0

=>4(x+3)(x+5)=0

=>(x+3)(x+5)=0

=>\(\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\\ \Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}+\dfrac{1}{3^{99}}\\ 3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow2B=1-\dfrac{1}{3^{100}}< 1\\ \Rightarrow B< \dfrac{1}{2}< 1\left(DPCM\right)\)

Ta có:

\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\\ \Rightarrow3B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{99}}\\ \Rightarrow3B-B=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\right)\\ \Rightarrow2B=1-\dfrac{1}{3^{100}}\\ \Rightarrow B=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)

Vì \(1-\dfrac{1}{3^{100}}< 1\) nên:

\(\dfrac{1-\dfrac{1}{3^{100}}}{2}< \dfrac{1}{2}< 1\) hay \(B< 1\)

Vậy...

5: a+b=7

=>a=7-b

b+c=9

=>c=9-b

c+a=8

=>7-b+9-b=8

=>16-2b=8

=>2b=16-8=8

=>b=4

=>a=7-4=3;c=9-4=5

1: a+b=17

a+b+c=20

=>c=20-17=3

a+c=15

=>a=15-c=15-3=1

b=17-a=17-1=16

2: \(\left\{{}\begin{matrix}a+b=5\\b+c=9\\a+c=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\c=9-b\\5-b+9-b=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}14-2b=6\\a=5-b\\c=9-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2b=8\\a=5-b\\c=9-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=4\\a=5-4=1\\c=9-4=5\end{matrix}\right.\)

3: \(c=\dfrac{abc}{ab}=\dfrac{288}{24}=12\)

bc=96

=>b=96/12=8

\(a=\dfrac{24}{b}=\dfrac{24}{8}=3\)

4: \(\left\{{}\begin{matrix}ab=36\\bc=45\\ca=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{36}{b}\\c=\dfrac{45}{b}\\\dfrac{36}{b}\cdot\dfrac{45}{b}=20\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}b^2=36\cdot45:20=81\\a=\dfrac{36}{b}\\c=\dfrac{45}{b}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b\in\left\{9;-9\right\}\\a=\dfrac{36}{b}\\c=\dfrac{45}{b}\end{matrix}\right.\)

TH1: b=9

\(a=\dfrac{36}{9}=4;c=\dfrac{45}{b}=\dfrac{45}{9}=5\)

TH2: b=-9

=>\(a=\dfrac{36}{-9}=-4;c=\dfrac{45}{-9}=-5\)