1. \(\text{6 + 2.(x - 19) = 16}.\)

\(\Leftrightarrow2.\left(x-19\right)=10.\)

\(\Leftrightarrow x-19=5.\)

\(\Leftrightarrow x=24.\)

Vậy \(x=24.\)

2. \(\text{(-240) : x – 16 = 64}.\)

\(\Leftrightarrow\left(-240\right):x=80.\)

\(\Leftrightarrow x=-3.\)

Vậy \(x=-3.\)

3. \(2x^3=16.\)

\(\Leftrightarrow x^3=8.\)

\(\Leftrightarrow x=2.\)

Vậy \(x=2.\)

bạn giúp mình câu này nhé !
(2x – 3)^2 = 49

6+x-16

=6+16-16

=22-16

=6

  6+x-16

=6+16-16

=22-16

=6

tick mik nha

a. 37500      b. x = 5600

ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow\left(x-6\sqrt{x}+9\right)-25=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)^2-5^2=0\\ \Leftrightarrow\left(\sqrt{x}-8\right)\left(\sqrt{x}+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=8\\\sqrt{x}=-2\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=64\)

x⁸ = (16x)⁶ = 16⁶.x⁶ => x² = 16⁶ = (16³)² => x = 16³

X⁸ = X⁶xX²

=>X²xX⁶=16xX⁶

=> X²=16

=>X = 4

a)\(\dfrac{4}{x}=\dfrac{x}{16}\)

<=>\(x^2=4.16=64\)

<=>\(x=\pm8\)

<=>x=-8(vì x<0)

b)\(\dfrac{x}{-24}=\dfrac{-6}{x}\)

<=>\(x^2=\left(-24\right)\left(-6\right)=144\)

<=>\(x=\pm12\)

<=>x=12(Vì x>0)

-x-13-16=-16+6-4

-x=-10-4+13+16

-x=15

=>x=-15

X = 15

AI TICK GIÙM MÌNH ĐI, LÀM ƠN ĐÓ !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\(\left|x^2+x+16\right|=x^2+\left|x+16\right|\)( vì \(x^2\ge0\))

\(\left|x^2+x-6\right|=x^2+\left|x-6\right|\)(vì \(x^2\ge0\))

\(\left|x+16\right|+\left|x-6\right|=\left|x+16\right|=\left|-x+6\right|\ge\left|22\right|=22\)

dấu = xảy ra khi và chỉ khi \(\left(x+16\right).\left(-x+6\right)\ge0\Rightarrow-16\le x\le6\)(1)

\(x^2\ge0\Rightarrow x^2+x^2\ge0\)

dấu = xảy ra khi và chỉ khi x=0 (2)

=> \(x^2+\left|x+16\right|+x^2+\left|x-6\right|\ge22+0=22\)

dấu = xảy ra khi dấu = ở (1) và (2) đồng thời xảy ra 

=> x=0

Vậy min A=22 khi và chỉ khi x=0

p/s: ko chắc lắm, sai sót bỏ qua :))

(x-3) tất cả mũ 2 à bạn

(x-3) tất cả mũ 2 à bạn

a: ĐKXĐ: x∉{2;-2;3}

\(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}\cdot\frac{4x^2-8x+16}{x^2-4}\right):\frac{16}{x^2-x-6}\)

\(=\left(\frac{4x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\frac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\right):\frac{16}{\left(x-3\right)\left(x+2\right)}\)

\(=\left(\frac{4x}{x+2}-\frac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right)\cdot\frac{\left(x-3\right)\cdot\left(x+2\right)}{16}\)

\(=\frac{4x\left(x+2\right)-4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\cdot\frac{\left(x-3\right)\left(x+2\right)}{16}=\frac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)}\cdot\frac{x-3}{16}\)

\(=\frac{-16\left(x-3\right)}{16\left(x+2\right)}=\frac{-x+3}{x+2}\)

b: A<0

=>\(\frac{-x+3}{x+2}<0\)

=>\(\frac{x-3}{x+2}>0\)

=>x>3 hoặc x<-2

c: A>=5

=>\(\frac{-x+3}{x+2}-5\ge0\)

=>\(\frac{-x+3-5x-10}{x+2}\ge0\)

=>\(\frac{-6x-7}{x+2}\ge0\)

=>\(\frac{6x+7}{x+2}\le0\)

=>-2<x<=-7/6