b: -7<x<7

\(a,\left(-31\right).\left(x+7\right)=0\\ \Rightarrow x+7=0\\ \Rightarrow x=-7\\ b,\left(8-x\right).\left(x+13\right)=0\\ \Rightarrow\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\\ c,\left(x^2-25\right)\left(3-x\right)=0\\ \Rightarrow\left(x-5\right)\left(x+5\right)\left(3-x\right)=0\\\Rightarrow \left[{}\begin{matrix}x-5=0\\x+5=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\\x=3\end{matrix}\right.\\ d,\left(x-3\right)\left(x^2+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x^2=-4\left(loại\right)\end{matrix}\right.\\ \Rightarrow x=3\)

Mình cảm ơn nhaa!

a)\(\left(x2+7\right).\left(x2-49\right)< 0\)

\(\left(x2+7\right).\left(x2-49\right)< 0\) chứng tỏ hai vế \(\left(x2+7\right)\) và \(\left(x2-49\right)\) khác dấu nhau .

\(\left\{{}\begin{matrix}\left(x2+7\right)>0\\\left(x2-49\right)< 0\end{matrix}\right.\)

Vì \(\left(x2+7\right)\) > \(\left(x2-49\right)\)

Nên ta có:

\(\left\{{}\begin{matrix}\left(x2+7\right)>0\\\left(x2-49\right)< 0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{{}\begin{matrix}\left(x+7\right)=0\\\left(x-49\right)=0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{{}\begin{matrix}x=-7\\x=49\end{matrix}\right.\)

Vậy hai số nguyên đó là -7 và 49 .

Còn phần còn lại bạn làm tương tự nhé !

a) \(x\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b) \(\left(-7-x\right)\left(-x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)

c) \(\left(x+3\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)

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

d) \(\left(x-3\right)\left(x^2+12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)

\(\Rightarrow x=3\)

e) \(\left(x+1\right)\left(2-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)

\(\Rightarrow-1\le x\le2\)

f) \(\left(x-3\right)\left(x-5\right)\le0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow3\le x\le5\)

a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)

d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3

c: \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\\x=-5\\x=5\end{matrix}\right.\)

mik lớp 6 bạn

a, \(\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

b, \(3\left(x-2\right)+13⋮x-2\Rightarrow x-2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

c, \(x\left(x+7\right)+2⋮x+7\Rightarrow x+7\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Lời giải:
a. $(x^2-9)(5x+15)=0$

$\Rightarrow x^2-9=0$ hoặc $5x+15=0$
Nếu $x^2-9=0$

$\Rightarrow x^2=9=3^2=(-3)^2$

$\Rightarrow x=3$ hoặc $-3$
Nếu $5x+15=0$

$\Rightarrow x=-3$
b.

$x^2-8x=0$
$\Rightarrow x(x-8)=0$

$\Rightarrow x=0$ hoặc $x-8=0$

$\Rightarrow x=0$ hoặc $x=8$

c. 

$5+12(x-1)^2=53$

$12(x-1)^2=53-5=48$

$(x-1)^2=48:12=4=2^2=(-2)^2$

$\Rightarrow x-1=2$ hoặc $x-2=-2$
$\Rightarrow x=3$ hoặc $x=0$

d.

$(x-5)^2=36=6^2=(-6)^2$
$\Rightarrow x-5=6$ hoặc $x-5=-6$

$\Rightarrow x=11$ hoặc $x=-1$

e.

$(3x-5)^3=64=4^3$

$\Rightarrow 3x-5=4$

$\Rightarrow 3x=9$

$\Rightarrow x=3$

f.

$4^{2x}+2^{4x+3}=144$
$2^{4x}+2^{4x}.8=144$

$2^{4x}(1+8)=144$

$2^{4x}.9=144$

$2^{4x}=144:9=16=2^4$

$\Rightarrow 4x=4\Rightarrow x=1$

TK

=> x² - 7 và x² - 49 trái dấu

Nhận xét: x² - 7 > x² - 9 nên để x² - 7  và x² - 49 trái dấu thì x² - 7 > 0 và x² - 49 < 0

x² - 7 > 0  => x² > 7

x² - 49 < 0 => x² < 49

=> 7 < x² < 49. Vì x nguyên nên x² = 9; 16 ; 25; 36

x² = 9 => x = -3 hoặc x = 3

x² = 16 => x = -4 hoặc 4

x² = 25 => x = -5 ; 5

x² = 36 => x = 6;-6

Vậy ....

`(x^2+7)(x^2-49)<0`

Vì `x^2+7>=7>0`

`=>x^2-49<0`

`<=>x^2-7x+7x-49<0`

`<=>x(x-7)+7(x-7)<0`

`<=>(x-7)(x+7)<0`

Vì `x+7>x-7`

`=>` $\begin{cases}x+7>0\\x-7<0\\\end{cases}$

`=>` $\begin{cases}x>-7\\x<7\\\end{cases}$

`=>-7<x<7`

Vậy `-7<x<7`

Ta có: \(\left(x^2+7\right)\left(x^2-49\right)< 0\)

mà \(x^2+7>0\)

nên \(x^2-49< 0\)

\(\Leftrightarrow x^2< 49\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>-7\\x< 7\end{matrix}\right.\)

Vậy: -7<x<7

a) (x - 2).3⁵ = 3⁷

x - 2 = 3⁷ : 3⁵

x - 2 = 3²

x - 2 = 9

x = 9 + 2

x = 11

b) x² - 2x = 0

x(x - 2) = 0

⇒ x = 0 hoặc x - 2 = 0

*) x - 2 = 0

x = 2

Vậy x = 0; x = 2

c) (2x - 1)² = 49

⇒ 2x - 1 = 7 hoặc 2x - 1 = -7

*) 2x - 1 = 7

2x = 7 + 1

2x = 8

x = 8 : 2

x = 4

*) 2x - 1 = -7

2x = -7 + 1

2x = -6

x = -6 : 2

x = -3

Vậy x = -3; x = 4