1. A: 8, B: 0

Câu 2

A:3

B:0

Bài 5:

\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}\\ \Rightarrow\dfrac{x-30}{40}=\dfrac{y-15}{20}=\dfrac{z-21}{28}=k\\ \Rightarrow\dfrac{x}{40}-\dfrac{3}{4}=\dfrac{y}{20}-\dfrac{3}{4}=\dfrac{z}{28}-\dfrac{3}{4}\\ \Rightarrow\dfrac{x}{40}=\dfrac{y}{20}=\dfrac{z}{28}=k\left(đặt.k\right)\\ \Rightarrow x=40k;y=20k;z=28k\)

Mà \(xyz=22400\Rightarrow22400k^3=22400\)

\(\Rightarrow k^3=1\Rightarrow k=1\Rightarrow\left\{{}\begin{matrix}x=40\\y=20\\z=28\end{matrix}\right.\)

Bài 5:

\(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\)

=>\(\frac{x-30}{40}=\frac{y-15}{20}=\frac{z-21}{28}\)

=>\(\frac{x}{40}-\frac34=\frac{y}{20}-\frac34=\frac{z}{28}-\frac34\)

=>\(\frac{x}{40}=\frac{y}{20}=\frac{z}{28}\)

=>\(\frac{x}{10}=\frac{y}{5}=\frac{z}{7}=k\)

=>x=10k; y=5k; z=7k

xyz=22400

=>\(10k\cdot5k\cdot7k=22400\)

=>\(350k^3=22400\)

=>\(k^3=64=4^3\)

=>k=4

=>\(\begin{cases}x=10\cdot4=40\\ y=5\cdot4=20\\ z=7\cdot4=28\end{cases}\)

Bài 6:

a: ĐKXĐ: x∉{0;1;-1}

Ta có: \(\frac{\left|x\right|+1}{\left|2x\right|-2}=\frac{2\left|x\right|-1}{4\left|x\right|}\)

=>\(4\left|x\right|\left(\left|x\right|+1\right)=\left(2\left|x\right|-2\right)\left(2\left|x\right|-1\right)\)

=>\(4x^2+4\left|x\right|=4x^2-2\left|x\right|-4\left|x\right|+2\)

=>\(10\left|x\right|=2\)

=>\(\left|x\right|=\frac15\)

=>\(\left[\begin{array}{l}x=\frac15\left(nhận\right)\\ x=-\frac15\left(nhận\right)\end{array}\right.\)

b: ĐKXĐ: x<>0

\(\frac{x-1}{x^2}=\frac{x}{x^2+x+1}\)

=>\(x^2\cdot x=\left(x-1\right)\left(x^2+x+1\right)\)

=>\(x^3=x^3-1\)

=>0=-1(vô lý)

=>x∈∅

Câu hỏi

10m=160000

(oh) hóa trị 1 mà zn hóa trị 2=> cthh la zn(oh)2

với lại ko có oh2 dau chi co OH hoac la H2O

phải viết là Zn(OH)₂ vì nhóm (OH) hóa trị I