a: \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)

=>\(\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

b: \(\left|2x+1\right|+\dfrac{3}{2}=2\)

=>\(\left|2x+1\right|=\dfrac{1}{2}\)

=>\(\left[{}\begin{matrix}2x+1=\dfrac{1}{2}\\2x+1=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

c: (2x-3)²=36

=>\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

d: \(7^{x+2}+2\cdot7^x=357\)

=>\(7^x\cdot49+7^x\cdot2=357\)

=>\(7^x=7\)

=>x=1

a) \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

\(---\)

b) \(\left|2x+1\right| +\dfrac{2}{3}=2\)

\( \Rightarrow\left|2x+1\right|=2-\dfrac{2}{3}\)

\(\Rightarrow\left|2x+1\right|=\dfrac{4}{3}\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=\dfrac{4}{3}\\2x+1=-\dfrac{4}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}\\2x=-\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)

\(---\)

c) \(\left(2x-3\right)^2=36\)

\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(---\)

d) \(7^{x+2}+2\cdot7^x=357\)

\(\Rightarrow7^x\cdot7^2+2\cdot7^x=357\)

\(\Rightarrow7^x\cdot\left(7^2+2\right)=357\)

\(\Rightarrow7^x\cdot\left(49+2\right)=357\)

\(\Rightarrow7^x\cdot51=357\)

\(\Rightarrow7^x=357:51\)

\(\Rightarrow7^x=7\)

\(\Rightarrow x=1\)

Bài 2: 

b: =>x-1>-4 và x-1<4

=>-3<x<5

c: =>x-2011>2012 hoặc x-2011<-2012

=>x>4023 hoặc x<-1

d: \(\left(3x-1\right)^{2016}+\left(5y-3\right)^{2018}>=0\forall x,y\)

mà \(\left(3x-1\right)^{2016}+\left(5y-3\right)^{2018}< 0\)

nên \(\left(x,y\right)\in\varnothing\)

bài 1:

a. \((x+1)(x+3) - x(x+2)=7 \)

    \(x^2+ 3x +x +3 - x^2 -2x =7\)

    \(x^2+4x+3-x^2-2x=7\)

\(=> 2x+3=7\)

    \(2x=4\)

    \(x = 2\)

Bài 2:

a)

\((3x-5)(2x+11) -(2x+3)(3x+7) \)

\(= 6x^2 +33x-10x-55-6x^2-14x-9x-10\)

\(= (6x^2-6x^2)+(33x-10x-14x-9x)-(55+10)\)

\(=-65\)

\(\)

a) Ta có I 2x - 1 I = 2015

=> 2x-1=2015 hoặc 2x-1=2015

+,Th1: 2x-1=2015

             2x=2015+1

             2x=2016

              x=2016:2

             x=1008

+,Th2: 2x-1=-2015

              2x=-2015+1

             2x=-2014

              x=-2014:2

             x=-1007

Vậy x=1008, x=-1007

|x-1|<5

th1: x-1<5=> x<6

th2: x-1<-5=> x<-4

vậy x <6 hoặc<-4

|x-1|>5 cũng tương tự như thế

còn mấy câu khác Nguyễn Diệu Thảo làm thế chắc bạn cũng biết cách làm rồi

L_I_K_E CHO MÌNH NHA!!!

1: A>=5

Dấu '=' xảy ra khi x=0

2: A>=4

Dấu '=' xảy ra khi x=-1

3: A>=-7

Dấu '=' xảy ra khi x=3

4: A>=2015

Dấu '=' xảy ra khi x=5