345(789)432):--:  ...

\(D=\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)

\(\Rightarrow D=x^6-3x^4+3x^2-1-\left(x^6-1\right)\)

\(\Rightarrow D=-3x^4+3x^2=3x^2\left(1-x^2\right)\)

\(1\dfrac{1}{2}\times1\dfrac{1}{3}\times...\times1\dfrac{1}{99}=\dfrac{3}{2}\times\dfrac{4}{3}\times...\times\dfrac{100}{99}=\dfrac{100}{2}=50\)

` a/`

` 2 - 1 5/6 + 2 2/3 = 2 - 11/6 - 8/3 = 1/6+ 8/3 = 1/6 + 16/6 = 17/6 `

`b/`

`5/9 xx ( 2 5/6 - 1 2/3 ) = 5/9 xx ( 17/6 - 5/3 ) = 5/9 xx 7/6 = 35/54 `

`c/` 

`1 1/3 : ( 2 + 1 1/6 : 2 5/6 ) `

`= 4/3 : ( 2 + 7/6 xx 6/17 ) `

`= 4/3 : ( 2 + 7/17 ) `

`= 4/3 : 41/17 `

`= 4/3 xx 17/41 `

`= 86/123`

` d/` 

` 2 3/5 : 3/4 xx 1 4/5 = 13/5 xx 4/3 xx 9/5 =52/15 xx 9/5 = 156/25`

\(a,2\dfrac{3}{4}-a+\dfrac{1}{4}=1\dfrac{1}{2}\\ \Leftrightarrow a=\dfrac{3}{2}\\ b,3\dfrac{1}{4}-a-1\dfrac{3}{4}=\dfrac{7}{9}\\ \Leftrightarrow a=\dfrac{13}{18}\\ c,2\dfrac{5}{6}-1\dfrac{1}{2}-a=\dfrac{1}{6}\\ \Leftrightarrow a=\dfrac{7}{6}\)

mình không bieets làm

\(10⋮\left(x+5\right)\)

\(\Rightarrow x+5\in\left\{-1;1;-2;2;-5;5;-10;10\right\}\)

\(\Rightarrow x\in\left\{-6;-4;-7;-3;-10;0;-15;5\right\}\left(x\in Z\right)\)

Để 10 ⋮ (x+5) thì (x+5) ∈ Ư(10)

⇒ (x+5) ∈ \(\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

⇒ x ∈ {5;0;-3;-4;-6;-7;-10;-15}

\(b,\left(5x-1\right)^2:2=8\\ \Leftrightarrow\left(5x-1\right)^2=16\\ \Leftrightarrow\left[{}\begin{matrix}5x-1=4\\5x-1=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{3}{5}\end{matrix}\right.\\ c,\left(1-3x\right)^3=-64\\ \Leftrightarrow1-3x=-4\\ \Leftrightarrow3x=5\\ \Leftrightarrow x=\dfrac{5}{3}\)

a) \(\left(2x-1\right)^3-4x^2\left(2x-3\right)=5\)

\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\)

\(\Leftrightarrow6x-1=5\Leftrightarrow6x=6\Leftrightarrow x=1\)

b) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x+1\right)^2=-10\)

\(\Leftrightarrow\left(x+1-x+1\right)\left[\left(x^2+2x+1+x^2-2x+1+\left(x^2-1\right)\right)\right]-6\left(x^2+2x+1\right)=-10\)

\(\Leftrightarrow2\left(3x^2+1\right)-6x^2-12x-6=-10\)

\(\Leftrightarrow6x^2+2-6x^2-12x-6=-10\)

\(\Leftrightarrow-12x-4=-10\Leftrightarrow12x=-6\Leftrightarrow x=\dfrac{1}{2}\)

a) \(\dfrac{11}{2}=\dfrac{10+1}{2}=5+\dfrac{1}{2}\)

\(\dfrac{32}{9}=\dfrac{27+5}{9}=3+\dfrac{5}{9}< 5+\dfrac{1}{2}\)

Vậy \(\dfrac{11}{2}>\dfrac{32}{9}\)

b)\(\dfrac{100}{23}=\dfrac{92+8}{23}=4+\dfrac{8}{23}\)

 \(\dfrac{302}{123}=\dfrac{246+56}{123}=2+\dfrac{56}{123}< 4+\dfrac{8}{23}\)

Vậy \(\dfrac{100}{23}>\dfrac{302}{123}\)

c) \(\dfrac{515}{605}< \dfrac{515+1}{605+1}=\dfrac{516}{606}\Rightarrow\dfrac{515}{605}< \dfrac{516}{606}\)