TL: 

Giải thích các bước giải:

 (2x-1) mũ 2 =9

(2x-1) mũ 2 =3 mũ 2

=> 2x-1= 3

2x =3+1

2x =4

x=4:2

x =2

vậy x = 2

\(\left(2x-1\right)^2=9\)

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

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

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

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

a.

\(x^2+x+3=y^2\)

\(\Leftrightarrow4x^2+4x+12=4y^2\)

\(\Leftrightarrow\left(2x+1\right)^2+11=\left(2y\right)^2\)

\(\Leftrightarrow\left(2y\right)^2-\left(2x+1\right)^2=11\)

\(\Leftrightarrow\left(2y+2x+1\right)\left(2y-2x-1\right)=11\)

Vậy pt có các cặp nghiệm:

\(\left(x;y\right)=\left(-3;3\right);\left(2;3\right);\left(2;-3\right);\left(-3;-3\right)\)

b.

\(A=\dfrac{a+b}{2022a-b}+\dfrac{b+c}{2022c-b}=\dfrac{\dfrac{1}{a}+\dfrac{1}{b}}{\dfrac{2022}{b}-\dfrac{1}{a}}+\dfrac{\dfrac{1}{b}+\dfrac{1}{c}}{\dfrac{2022}{b}-\dfrac{1}{c}}\)

\(=\dfrac{\dfrac{1}{a}+\dfrac{1}{b}}{\dfrac{1}{c}}+\dfrac{\dfrac{1}{b}+\dfrac{1}{c}}{\dfrac{1}{a}}=\dfrac{c}{a}+\dfrac{a}{c}+\dfrac{c+a}{b}\)

Do \(\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{2022}{b}\Rightarrow\dfrac{a+c}{2022ac}=\dfrac{1}{b}\Rightarrow\dfrac{c+a}{b}=\dfrac{\left(c+a\right)^2}{2022ac}\)

\(\Rightarrow A=\dfrac{c}{a}+\dfrac{a}{c}+\dfrac{\left(c+a\right)^2}{2022ac}\ge2\sqrt{\dfrac{ca}{ac}}+\dfrac{4ac}{2022ac}=\dfrac{2024}{1011}\)

Dấu "=" xảy ra khi \(a=c=\dfrac{b}{1011}\)

`Answer:`

Giải hệ phương trình à bạn?

\(\hept{\begin{cases}56x+65y=35,4\\x+y=0,6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}56x+65y=35,4\\56x+56y=33,6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}9y=1,8\\56x+56y=33,6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=0,2\\56x+56y=33,6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=0,2\\56x+56.0,2=33,6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=0,2\\56x=33,6-11,2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=0,2\\x=0,4\end{cases}}\)

`Answer:`

\(15x-3\left(4x-7\right)< 8\)

\(\Leftrightarrow15x-12x+21< 8\)

\(\Leftrightarrow3x< 8-21\)

\(\Leftrightarrow3x< -13\)

\(\Leftrightarrow x< -\frac{13}{3}\)

x∈(3, ∞)

`Answer:`

`-x+3<0`

`<=>-x+3-3<0-3`

`<=>-x<-3`

`<=>-x.(-1)>(-3).(-1)` (Do `-1<0`; BĐT mới ngược chiều BĐT ban đầu.)

`<=>x>3`

\(\left|x^2+x-1\right|=2x-3\)

- Với \(x< \dfrac{3}{2}\) vế trái không âm, vế phải âm nên pt vô nghiệm

- Với \(x\ge\dfrac{3}{2}\Rightarrow x^2+x-1\ge\left(\dfrac{3}{2}\right)^2+\dfrac{3}{2}-1>0\) nên pt trở thành:

\(x^2+x-1=2x-3\)

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

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\) (vô nghiệm do \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0;\forall x\))

Vậy pt đã cho luôn luôn vô nghiệm

\(\dfrac{x-1}{x+3}\le x\)

\(\dfrac{x-1}{x+3}-x\le0\)

\(\dfrac{x-1-x\left(x+3\right)}{x+3}\le0\)

\(\dfrac{-\left(x^2+2x+1\right)}{x+3}\le0\)

\(\dfrac{-\left(x+1\right)^2}{x+3}\le0\)

Vì \(-\left(x+1\right)^2\le0\forall x\)

\(\Rightarrow x+3>0\)

\(\Rightarrow x>-3\)