`|x(x-3)|=2/3 x`

TH1: `x(x-3)>=0`

`x(x-3)=2/3 x`

`x^2 -3x = 2/3 x`

` x^2- 11/3 x=0`

`x(x-11/3)=0`

`[(x=0),(x=11/3):}`

TH2: `x(x-3)<0`

`-x(x-3)=2/3 x`

`-x^2+3x=2/3x`

`-x^2+7/3 x=0`

`x(-x+7/3)=0`

`[(x=0(L)),(x=7/3):}`

Vây `x=0; x=11/3 ; x=7/3`

Bài 1:

\(f\left(-1\right)=1\)

\(f\left(\dfrac{1}{2}\right)=-\dfrac{1}{2}\)

\(f\left(0\right)=-1\)

\(f\left(-5\right)=49\)

Bài 2:

Bạn lấy toạ độ điểm A(1;3); B(-1;-3), C(0;0). Đồ thị y=3x một đường thẳng đi qua gốc toạ độ O nhé!

\(\left|x+1\right|+\left|x+1\right|=3x\)

<=> \(\left[{}\begin{matrix}x+1+x+1=3x\left(x\ge-1\right)\\x+1-x-1=3x\left(x< -1\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

x = 2 thôi nhé

\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2021}\\ \Rightarrow3A-A=3^2+3^3+...+3^{2021}-3-3^2-3^3-...-3^{2020}\\ \Rightarrow2A=3^{2021}-3\\ \Rightarrow2A+3=3^{2021}=3^x\\ \Rightarrow x=2021\)

b:

6x^2

Mọi người giải giúp mình bài tìm x với

x=-3, x=0

\(x^2+3x=x\left(x+3\right)\)

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∈∅