Cách 2: Do \(\left|x\right|\ge0\forall x\) nên \(\left|x+1\right|+\left|x-2\right|+\left|x+7\right|\ge0\)

\(\Rightarrow5x-10\ge0\Rightarrow x\ge2\)

Với \(x\ge2\), ta có : \(x+7>0;x+1>0;x-2\ge0\)

Suy ra \(x+1+x-2+x+7=5x-10\)

\(\Leftrightarrow-2x=-16\Leftrightarrow x=8\left(tm\right)\)

Vậy x = 8.

Cách 1: Với \(x\le-7\), ta có : \(x+7\le0;x+1< 0;x-2< 0\)

Suy ra \(-x-1-x+2-x-7=5x-10\)

\(\Leftrightarrow-8x=-4\Leftrightarrow x=\frac{1}{2}\left(l\right)\)

Với \(-7< x\le-1\), ta có : \(x+7>0;x+1\le0;x-2< 0\)

Suy ra \(-x-1-x+2+x+7=5x-10\)

\(\Leftrightarrow-6x=-18\Leftrightarrow x=3\left(l\right)\)

Với \(-1< x\le2\), ta có : \(x+7>0;x+1>0;x-2\le0\)

Suy ra \(x+1-x+2+x+7=5x-10\)

\(\Leftrightarrow-6x=-20\Leftrightarrow x=\frac{10}{3}\left(l\right)\)

Với \(x>2\), ta có : \(x+7>0;x+1>0;x-2>0\)

Suy ra \(x+1+x-2+x+7=5x-10\)

\(\Leftrightarrow-2x=-16\Leftrightarrow x=8\left(tm\right)\)

Vậy x = 8.

(x + 20)⁴ + (2y - 1)²⁰²⁴ ≤ 0

⇒ (x + 20)⁴ = 0 và (2y - 1)²⁰²⁴ = 0

*) (x + 20)⁴ = 0

x + 20 = 0

x = 0 - 20

x = -20

*) (2y - 1)²⁰²⁴ = 0

2y - 1 = 0

2y = 1

y = 1/2

M = 5.(-20)².1/2 - 4.(-2).(1/2)²

= 1000 + 2

= 1002

Ta có: \(\hept{\begin{cases}\left(5x-y\right)^{2016}\ge0\\\left|x^2-4\right|^{2017}\ge0\end{cases}\Rightarrow\left(5x-y\right)^{2016}+\left|x^2-4\right|\ge}0\)

Mà \(\left(5x-y\right)^{2016}+\left|x^2-4\right|^{2017}\le0\)

\(\Rightarrow\hept{\begin{cases}\left(5x-y\right)^{2016}=0\\\left|x^2-4\right|^{2017}=0\end{cases}\Rightarrow\hept{\begin{cases}5x-y=0\\x^2-4=0\end{cases}}\Rightarrow\hept{\begin{cases}y=\pm10\\x=\pm2\end{cases}}}\)

Vậy các cặp (x;y) là (2;10);(-2;-10)

cảm ơn

1,

Vì \(\left|2x-27\right|^{2007}\ge0;\left(3y+10\right)^{2008}\ge0\)

\(\Rightarrow\left|2x-27\right|^{2007}+\left(3y+10\right)^{2008}\ge0\)

Mà \(\left|2x-27\right|^{2007}+\left(3y+10\right)^{2008}=0\)

\(\Rightarrow\hept{\begin{cases}\left|2x-27\right|^{2007}=0\\\left(3y+10\right)^{2008}=0\end{cases}\Rightarrow\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{27}{2}\\y=\frac{-10}{3}\end{cases}}}\)

2,

TH1: \(x\ge\frac{3}{5}\)

<=> 2(5x-3)-2x=14

<=> 10x-6-2x=14

<=>8x-6=14

<=>8x=20

<=>x=5/2 (thỏa mãn)

TH2: x < 3/5

<=> 2(3-5x)-2x=14

<=>6-10x-2x=14

<=>6-12x=14

<=>12x=-8

<=>x=-2/3 (thỏa mãn)

Vậy \(x\in\left\{\frac{5}{2};\frac{-2}{3}\right\}\)

1 x=13,5 ;y=-10/3

2 kết quả x =-2/3

ko bieet

d) Ta có: \(n^2+5n+9⋮n+3\)

\(\Leftrightarrow n^2+3n+2n+6+3⋮n+3\)

\(\Leftrightarrow n\left(n+3\right)+2\left(n+3\right)+3⋮n+3\)

mà \(n\left(n+3\right)+2\left(n+3\right)⋮n+3\)

nên \(3⋮n+3\)

\(\Leftrightarrow n+3\inƯ\left(3\right)\)

\(\Leftrightarrow n+3\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{-2;-4;0;-6\right\}\)

Vậy: \(n\in\left\{-2;-4;0;-6\right\}\)

d) Ta có: n2+5n+9⋮n+3n2+5n+9⋮n+3

⇔n2+3n+2n+6+3⋮n+3⇔n2+3n+2n+6+3⋮n+3

⇔n(n+3)+2(n+3)+3⋮n+3⇔n(n+3)+2(n+3)+3⋮n+3

mà n(n+3)+2(n+3)⋮n+3n(n+3)+2(n+3)⋮n+3

nên 3⋮n+33⋮n+3

⇔n+3∈Ư(3)⇔n+3∈Ư(3)

`b)` - Ta thấy : `|x+1|+|x-2|+|x+7|>=0`

`-> 5x-10>=0`

`-> 5x>=10`

`-> x>=2`

`-> |x+1|=x+1;|x-2|=x-2;|x+7|=x+7`

- Vậy ta có :

`(x+1)+(x-2)+(x+7)=5x-10`

`<=> x+1+x-2+x+7=5x-10`

`<=> 3x+6=5x-10`

`<=> 3x-5x=-10-6`

`<=> -2x=-16`

`<=> x=8`

\(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)=7\)

\(25x^2-10x+1-25x^2+16=7\)

\(17-10x=7\)

\(10x=10\)

\(x=1\)