\(\left(-38\right).63+\left(-37\right).38\\ =\left(-38\right).63+37.\left(-38\right)\\ =\left(-38\right)\left(63+37\right)\\ =-38.100\\ =-3800\)

(−38).63+(−37).38=(−38).63+37.(−38)=(−38)(63+37)=−38.100=−3800

\((-38).63 + (-37) . 38\)
\(=38.(-63) + (-37) . 38\)
\(=38.[(-63) + (-37)] \)
\(=38.(-100) = -3800\)

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

\(4x-12+5=2x+1\)

\(4x-2x=1-5+12\)

\(2x=8\)

\(x=4\)

vay \(x=4\)

4.(x-3)+5=2x+1

\(\Rightarrow\)4x-12+5=2x+1

\(\Rightarrow\)4x-12=2x+1-5

\(\Rightarrow\)4x-12=2x+(-4)

\(\Rightarrow\)4x=2x+(-4)+12

\(\Rightarrow\)4x=2x+8

\(\Rightarrow\)4x-2x=8

\(\Rightarrow\)2x=8

\(\Rightarrow\) x=4

4.(x-3)+5=2x+1

⇒4x-12+5=2x+1

⇒4x-12=2x+1-5

⇒4x-12=2x+(-4)

⇒4x=2x+(-4)+12

⇒4x=2x+8

⇒4x-2x=8

⇒2x=8

⇒ x=4

ra 34 và - 20 

mk làm rùi đó ,bài dưới ế ,rõ ràng rành mạch lên pạn mk nha 

|7- x|=(-13)-5.(-8)

|7- x|=144

7- x=144

      x=7-144

        x=(-137)

Vậy : x=(-137); x=137

ọilmk

\(A=\frac{x^3-3x^2-7x-15}{x^5-x^4-10x^3-38x^2-51x-45}\)

\(=\frac{x^2\left(x-5\right)+2x\left(x-5\right)+3\left(x-5\right)}{x^4\left(x-5\right)+4x^3\left(x-5\right)+10x^2\left(x-5\right)+12x\left(x-5\right)+9\left(x-5\right)}\)

\(=\frac{\left(x-5\right)\left(x^2+2x+3\right)}{\left(x-5\right)\left(x^4+4x^3+10x^2+12x+9\right)}\)

\(=\frac{x^2+2x+3}{x^4+4x^3+10x^2+12x+9}\)

\(=\frac{x^2+2x+3}{\left(x^2\right)^2+2.x^2.2x+\left(2x\right)^2+6x^2+12x+9}\)

\(=\frac{x^2+2x+3}{\left(x^2+2x\right)^2+2.\left(x^2+2x\right).3+3^2}\)

\(=\frac{\left(x^2+2x+3\right)}{\left(x^2+2x+3\right)^2}=\frac{1}{x^2+2x+3}\)

b, \(A=\frac{1}{x^2+2x+3}=\frac{1}{\left(x+1\right)^2+2}\le\frac{1}{2}\forall x\)

Dấu "=" xảy ra khi: \(x+1=0\Rightarrow x=-1\)

Vậy GTLN của A là \(\frac{1}{2}\) khi x = -1

Đề ko rõ ràng \(\sqrt{x^2}+x+\dfrac{1}{4}\) hay \(\sqrt{x^2+x+\dfrac{1}{4}}\)??

m??