\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)

\(a,Sửa:2021x-1+2022x\left(1-2021x\right)=0\\ \Leftrightarrow\left(2021x-1\right)\left(1-2022x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2021}\\x=\dfrac{1}{2022}\end{matrix}\right.\)

a) \(3\left(x-1\right)^2\cdot3x\left(x-5\right)=0\)

\(\Rightarrow9x\left(x-1\right)^2\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=5\end{matrix}\right.\)

b) \(\left(x+3\right)^2-5x-15=0\)

\(\Rightarrow\left(x+3\right)^2-5\left(x+3\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x+3-5\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)

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

c) \(2x^5-4x^3+2x=0\)

\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)

\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)

\(\Rightarrow2x\left(x^2-1\right)^2=0\)

\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)

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

\(\text{#}Toru\)

a. -3.(x + 2) > 0

Mà -3 < 0

=> x + 2  < 0

=> x < -2

b. 5.(1 - x) < 0

Mà 5 > 0

=> 1 - x < 0

=> x > 1

c. (x² + 2).(5 - x) < 0

+) x² + 2 < 0 (vô lí, loại); 5 - x > 0

+) x² + 2 > 0 (luôn đúng); 5 - x < 0

=> chỉ cần 5 - x < 0

=> x > 5

a) \(\Leftrightarrow\left[{}\begin{matrix}x-3,5=7,5\\x-3,5=-7,5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-4\end{matrix}\right.\)

b) \(\Leftrightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{1}{2}\\x+\dfrac{4}{5}=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{10}\\x=-\dfrac{13}{10}\end{matrix}\right.\)

c) \(\Leftrightarrow\left|x-0,4\right|=3,6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-0,4=3,6\\x-0,4=-3,6\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3,2\end{matrix}\right.\)

d) \(\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\4,5-x=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=4,5\end{matrix}\right.\)(vô lý)

Vậy \(S=\varnothing\)

Nói chung cả 3 câu :

Vì GTTĐ luôn lớn hơn hoặc bằng 0

=> tất cả các số hạng đều bằng 0 

sau đó tính ra là xong

a) | x - 1| + | y - 3| = 0

=> |x -1| = 0 => x = 1

|y-3| = 0 => y  = 3

KL:...

b) | x - 1 | + |x-3| + |x-5| = 0

Ta thấy: \(\left|x-1\right|;\left|x-3\right|;\left|x-5\right|\ge0.\)

=> | x - 1 | = 0 => x = 1 mà | 1-3| không bằng 0 (Loại)

...

ko tìm được x

c) \(\left|x-2018y\right|+\left|x-2018\right|\le0\)

mà  \(\left|x-2018y\right|;\left|x-2018\right|\ge0\)

=> | x - 2018y| + |x-2018| = 0

=> | x - 2018| = 0 => x = 2018

=> |x-2018y| = 0 => |2018-2018y| = 0 => y = 1

KL:...

Bài 2:

a, |x-1| -x +1=0

|x-1| = 0-1+x

|x-1| = -1 + x

 \(\orbr{\begin{cases}x-1=-1+x\\x-1=1-x\end{cases}}\)

 \(\orbr{\begin{cases}x=-1+x+1\\x=1-x+1\end{cases}}\)

 \(\orbr{\begin{cases}x=x\\x=2-x\end{cases}}\)

x = 2-x

2x = 2

x = 2:2

x=1

b, |2-x| -2 = x

|2-x| = x+2

\(\orbr{\begin{cases}2-x=x+2\\2-x=2-x\end{cases}}\)

2-x = x+2

x+x = 2-2

2x = 0

x = 0

kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

a. x= 1;2

b. x= 1;2;3;4;5;6

c. x= 6;7;8;9;...

d. x= 6;7;8;9;...

e. x= 1;2;3

a) x thuộc{1;-1;2;-2}

b)x thuộc {1;-1;2;-2;3;-3;4;-4;5;-5;6;-6}

c) x thuộc {6;-6;7;-7;...}

d) x thuộc {6:-6:7:-7;...}

f) x thuộc { 2;3;4;5;...}

e) x thuộc {0;1;2;3}

g) x thuộc {0;1}