mong mọi người giải giúp em vs 

a) <=> 3x-2=0 hoặc 4x+5=0

1) 3x-2=0 <=> 3x=2 <=> x=2/3

2) 4x+5=0 <=> 4x=-5 <=> x= -5/4

a) tách ra 2 cái rồi tính mỗi cái

b) phân tích ra ta đc:

 x² - 8x + 16 - x² + 6x -2x + 12=0

sau đó bạn tự giải ra

c) áp dụng hằng đẳng thức ta đc

 (2x+1)(2x-1)=(2x+1)(3x-5)

a/ \(\Leftrightarrow\left(x+2\right)^2-3\left|x+2\right|=0\)

\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x+2\right|=0\\\left|x+2\right|=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x+2=3\\x+2=-3\end{matrix}\right.\)

b/

\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|-4=0\)

\(\Leftrightarrow\left(\left|x+2\right|+1\right)\left(\left|x+2\right|-4\right)=0\)

\(\Leftrightarrow\left|x+2\right|-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\)

c/

\(\Leftrightarrow\left|x^2-3\right|^2-6\left|x^2-3\right|+5=0\)

\(\Leftrightarrow\left(\left|x^2-3\right|-1\right)\left(\left|x^2-3\right|-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x^2-3\right|=1\\\left|x^2-3\right|=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-3=1\\x^2-3=-1\\x^2-3=5\\x^2-3=-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=4\\x^2=2\\x^2=8\\x^2=-2\left(l\right)\end{matrix}\right.\)

d/ ĐKXĐ: ...

\(\Leftrightarrow\frac{\left|x-2\right|^2}{\left(x-1\right)^2}+\frac{2\left|x-4\right|}{x-1}=3\)

Đặt \(\frac{\left|x-2\right|}{x-1}=a\)

\(a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\\\left|x-2\right|=-3\left(x-1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\left(x\ge1\right)\\\left|x-2\right|=3-3x\left(x\le1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x-1\left(vn\right)\\x-2=1-x\\x-2=3-3x\\x-2=3x-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{4}{5}\\x=\frac{1}{2}\end{matrix}\right.\)

e/ ĐKXĐ: ...

Đặt \(\left|\frac{2x-1}{x+2}\right|=a>0\)

\(a-\frac{2}{a}=1\Leftrightarrow a^2-a-2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\) \(\Rightarrow\left|\frac{2x-1}{x+2}\right|=2\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=2\left(x+2\right)\\2x-1=-2\left(x+2\right)\end{matrix}\right.\)

f/ ĐKXĐ: ...

Đặt \(\left|x-\frac{1}{x}\right|=a\ge0\Rightarrow a^2=x^2+\frac{1}{x^2}-2\Rightarrow x^2+\frac{1}{x^2}=a^2+2\)

Phương trình trở thành:

\(a^2+2-10=2a\)

\(\Leftrightarrow a^2-2a-8=0\Rightarrow\left[{}\begin{matrix}a=4\\a=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left|x-\frac{1}{x}\right|=4\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-1=0\\x^2+4x-1=0\end{matrix}\right.\)

b) \(\dfrac{x-5}{2017}-1+\dfrac{x-2}{2020}-1=\dfrac{x-6}{2016}-1+\dfrac{x-68}{1954}-1\)

​\(\dfrac{x-2022}{2017}+\dfrac{x-2002}{2020}=\dfrac{x-2022}{2016}+\dfrac{x-2022}{1954}\)​

\(\Leftrightarrow\left(x-2022\right)\left(\dfrac{1}{2017}+\dfrac{1}{2020}-\dfrac{1}{2016}-\dfrac{1}{1954}\right)=0\)

\(\Leftrightarrow x-2022=0\left(\dfrac{1}{2017}+\dfrac{1}{2020}-\dfrac{1}{2016}-\dfrac{1}{1954}\ne0\right)\)

\(\Leftrightarrow x=2022\)

a) Đặt \(t=\left|2x-\dfrac{1}{x}\right|\Leftrightarrow t^2=\left(2x-\dfrac{1}{x}\right)^2=4x^2-4+\dfrac{1}{x^2}\Leftrightarrow t^2+4=4x^2+\dfrac{1}{x^2}\) ĐK \(t\ge0\)

từ có ta có pt theo biến t : \(t^2+4+t-6=0\)

\(\Leftrightarrow t^2+t-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\left(nh\right)\\t=-2\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left|2x-\dfrac{1}{x}\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{x}=1\\2x-\dfrac{1}{x}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x^2-x-1=0\\2x^2+x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)

c: TH1: x>0

Pt sẽ là \(\dfrac{x^2-1}{x\left(x-2\right)}=2\)

=>2x^2-4x=x^2-1

=>x^2-4x+1=0

hay \(x=2\pm\sqrt{3}\)

TH2: x<0

Pt sẽ là \(\dfrac{x^2-1}{-x\left(x-2\right)}=2\)

=>-2x(x-2)=x^2-1

=>-2x^2+4x=x^2-1

=>-3x^2+4x+1=0

hay \(x=\dfrac{2-\sqrt{7}}{3}\)

b:

TH1: 2x^3-x>=0

 \(4x^4+6x^2\left(2x^3-x\right)+1=0\)

=>4x^4+12x^5-6x^3+1=0

\(\Leftrightarrow x\simeq-0.95\left(loại\right)\)

TH2: 2x^3-x<0

Pt sẽ là \(4x^4+6x^2\left(x-2x^3\right)+1=0\)

=>4x^4+6x^3-12x^5+1=0

=>x=0,95(loại)

cho mk hỏi một chút là đây đích thực có phải lớp 1 ko ak?