Ta có : \(14^0-\left(4+x\right)^2=-80\)

\(\Leftrightarrow1+80=\left(4+x\right)^2\)

\(\Leftrightarrow\left(4+x\right)^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}4+x=9\\4+x=-9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-13\end{matrix}\right.\)

Vì (x²-15)(x²-8) <0

=> x²-15  và  x²-8 khác dấu

Mà x²-15 < x²-8 

=> x²-15 < 0 và x²-8 > 0

Ta có: x²-15 < 0

=> x² < 15 (1)

Ta có : x²-8 > 0 

=> x²  > 8 (2)

Từ (1) và (2) => 8 < x² < 15

=> x² ∈ {9;10;11;12;13;14}

=> x² = 9

=> x = 3

\(x^2+5x-14=0\)

\(\Rightarrow x^2+7x-2x-14=0\)

\(\left(x^2+7x\right)-\left(2x+14\right)=0\)

\(x\left(x+7\right)-2\left(x+7\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x+7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+7=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-7\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=2\\x=-7\end{cases}}\)

\(x^2-5x-14=0\)

\(\Leftrightarrow\)\(x^2+2x-7x-14=0\)

\(\Leftrightarrow\)\(x\left(x+2\right)-7\left(x+2\right)=0\)

\(\Leftrightarrow\)\(\left(x-7\right)\left(x+2\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-7=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-2\end{cases}}}\)

bài 2: (x-3).(y+2) = -5

    Vì x, y \(\in\)Z   => x-3 \(\in\)Ư(-5) = {5;-5;1;-1}

Ta có bảng: 

bài 3: a(a+2)<0

TH1 : \(\orbr{\begin{cases}a< 0\\a+2>0\end{cases}}\)=>\(\orbr{\begin{cases}a< 0\\a>-2\end{cases}}\)=> -2<a<0 ( TM)

TH2: \(\orbr{\begin{cases}a>0\\a+2< 0\end{cases}}\Rightarrow\orbr{\begin{cases}a>0\\a< -2\end{cases}}\Rightarrow loại\)
 

           Vậy -2<a<0

Bài 5: \(\left(x^2-1\right)\left(x^2-4\right)< 0\)

TH 1 : \(\hept{\begin{cases}x^2-1>0\\x^2-4< 0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2>1\\x^2< 4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x>1\\x< 2\end{cases}}\)\(\Rightarrow\)1 < a < 2

TH 2: \(\hept{\begin{cases}x^2-1< 0\\x^2-4>0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2< 1\\x^2>4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x< 1\\x>2\end{cases}}\)\(\Rightarrow\)loại

                         Vậy 1<a<2

Vì : \(\left|x+2017\right|\ge0\forall x\)

\(\left|y-2017\right|\ge0\forall y\)

\(\Rightarrow\left|x+2017\right|+\left|y-2017\right|\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+2017=0\\y-2017=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2017\\y=2017\end{matrix}\right.\)

Vậy x = -2017 ; y = 2017

b, \(\left|2x-1\right|=\left|x+8\right|\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=x+8\\2x-1=-\left(x+8\right)\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}2x-x=8+1\\2x+x=-8+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=9\\3x=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{-7}{3}\end{matrix}\right.\)

Vậy x = 9

c, \(\left|3x-2\right|-\left|x+14\right|=0\Rightarrow\left|3x-2\right|=\left|x+14\right|\)

\(\Rightarrow\left[{}\begin{matrix}3x-2=x+14\\3x-2=-\left(x+14\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x=14+2\\3x+x=-14+2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=16\\4x=-12\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

Vậy ...

Làm 1 câu thôi nha bạn,mỏi tay lắm:

\(\left|x+2017\right|+\left|y-2017\right|=0\)

\(\left|x+2017\right|\ge0\)

\(\left|x-2017\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left|x+2017\right|=0\Rightarrow x+2017=0\Rightarrow x=-2017\)

\(\left|y-2017\right|=0\Rightarrow y-2017=0\Rightarrow y=2017\)

c) | x - 3 | + x - 3 = 0

| x - 3 | + x = 0 + 3

| x - 3 | + x = 3

| x - 3 | = 3 - x

=> x < 3

=> x = { 3 ; 2 ; 1 ; 0 ; -1 ; - 2 ; -3 ; .... }

2)

10 + 9 + 8 +.....+ x = 10

9 + 8 + ... + x = 10 - 10

9 + 8 + .... + x = 0

tổng : 9 + 8 + ... + x = \(\frac{\left(9+x\right).n}{2}\)trong đó n là số số hạng

ta có : \(\frac{\left(9+x\right).n}{2}=0\)

    ( 9 + x ) . n = 0 . 2

     ( 9 + x ) . n = 0

     9 + x = 0 : n

      9 + x = 0

       x = 0 - 9

   => x = -9

a) | x | + 2 = 5

| x | = 5 - 2

| x | = 3

=> x = \(\orbr{\begin{cases}3\\-3\end{cases}}\)

b) | x + 2 | - x = 2

| x + 2 | = 2 + x

=> x + 2 \(\in\orbr{\begin{cases}Z^-\\Z^+\end{cases}}\)

=> x = { 0 ; 1 ; 2 ; 3 ; 4; .... }

hoặc x = { -1 ; -2 ; -3 ; -4 ; ... }

làm ko nổi  nữa

a. \(\left|x-1\right|-x+1=0\)

\(\Leftrightarrow\left|x-1\right|=x-1\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=x-1\\x-1=1-x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}0x=2\left(loại\right)\\2x=2\end{cases}}\)

\(\Leftrightarrow x=1\)

Vậy ...

b/ \(\left(x^2+1\right)\left(81-x^2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\81-x^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\left(loại\right)\\x^2=81\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=9\\x=-9\end{cases}}\)

Vậy ..