a: \(\left(x+5\right)^2>=0\forall x\)

\(\left(2y-8\right)^2>=0\forall y\)

Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)

b: \(\left(x+3\right)\left(2y-1\right)=5\)

=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)

Lời giải:

a. $21-2(x+6)=15$

$2(x+6)=21-15=6$

$x+6=6:2=3$

$x=3-6=-3$

b.

$3^{x+1}=27=3^3$

$\Rightarrow x+1=3$

$\Rightarrow x=2$

c.

$14-x=36$

$x=14-36=-22$

d.

$-3x+25=-2$

$-3x=-2-25=-27$

$x=(-27):(-3)=9$

e.

$37-2(x+6)=29$

$2(x+6)=37-29=8$

$x+6=8:2=4$

$x=4-6=-2$

h.

$2^{x-1}=16=2^4$

$x-1=4$

$x=4+1=5$

Câu 30:

\(\dfrac{x+1}{8}=\dfrac{2}{x+1}\)

=>\(\left(x+1\right)^2=2\cdot8=16\)

=>\(\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)

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

=>Chọn D

Câu 31:

\(-\dfrac{7}{15}\cdot\dfrac{5}{8}\cdot\dfrac{15}{-7}\cdot\left(-16\right)\)

\(=\dfrac{5}{8}\cdot\left(-16\right)=-10\)

=>Chọn D

câu 31 : D 

giúp mik câu 30 luôn dc ko bạn

a: =>y/15=-2/3

hay y=-10

b: 2/x=x/18

nên \(x^2=36\)

hay \(x\in\left\{6;-6\right\}\)

c: x/9=16/x

nên \(x^2=144\)

hay \(x\in\left\{12;-12\right\}\)

a)x=-10
b)x=-6

c)x=-12 nhé

25 nha

a, ( x - 35) . 16 =0

    x - 35 = 0 : 16=0

    x = 0 + 35

    x = 35

b,19. (x - 7) = 19

   x - 7=19 : 19=1

   x =1 + 7

   x =8

c,37.x + 63.x = 200

  (37 + 63)x = 200

   100.x=200

   x =200 : 100

   x = 2

a) 35

b) 8

c)200/73

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