a: Ta có: \(3x\left(2x+1\right)+\left(2x-3\right)\left(x+1\right)\)

\(=6x^2+3x+2x^2+2x-3x-3\)

\(=8x^2+2x-3\)

mình chẳng bt bạn hỏi j cả

câu b là : tính và rút gọn

câu c là : chứng ming đẳng thức

a: \(\left(2x-1\right)^5-\left(2x-1\right)^8=0\)

=>\(\left(2x-1\right)^5\cdot\left\lbrack1-\left(2x-1\right)^3\right\rbrack=0\)

=>\(\left[\begin{array}{l}\left(2x-1\right)^5=0\\ 1-\left(2x-1\right)^3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\left(2x-1\right)^5=0\\ \left(2x-1\right)^3=1\end{array}\right.\)

=>\(\left[\begin{array}{l}2x-1=0\\ 2x-1=1\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=1\\ 2x=2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac12\\ x=1\end{array}\right.\)

b: (2x+1)(2x-3)<0

TH1: \(\begin{cases}2x+1\ge0\\ 2x-3\le0\end{cases}\Rightarrow\begin{cases}x\ge-\frac12\\ x\le\frac32\end{cases}\)

=>\(-\frac12\le x\le\frac32\)

TH2: \(\begin{cases}2x+1\le0\\ 2x-3\ge0\end{cases}\Rightarrow\begin{cases}2x\le-1\\ 2x\ge3\end{cases}\Rightarrow\begin{cases}x\le-\frac12\\ x\ge\frac32\end{cases}\)

=>x∈∅

c: (x-1)(2x+3)>0

TH1: \(\begin{cases}x-1>0\\ 2x+3>0\end{cases}\Rightarrow\begin{cases}x>1\\ x>-\frac32\end{cases}\)

=>x>1

TH2: \(\begin{cases}x-1<0\\ 2x+3<0\end{cases}\Rightarrow\begin{cases}x<1\\ x<-\frac32\end{cases}\)

=>\(x<-\frac32\)

Bài 1

A= (x-2)(2x-1)-2x(x+3)=2x²-x-4x+2-2x²-6x=-11x+2

Bài 1:

a) \(A=\left(x-2\right)\left(2x-1\right)-2x\left(x+3\right)\)

\(A=2x^2-x-4x+2-2x^2-6x\)

\(A=-11x+2\)

b) \(B=\left(3x-2\right)\left(2x+1\right)-\left(6x-1\right)\left(x+2\right)\)

\(B=6x^2+3x-4x-2-6x^2-12x+x+2\)

\(B=-12x\)

c) \(C=6x\left(2x+3\right)-\left(4x-1\right)\left(3x-2\right)\)

\(C=12x^2+18x-12x^2+8x+3x-2\)

\(C=29x-2\)

d) \(D=\left(2x+3\right)\left(5x-2\right)+\left(x+4\right)\left(2x-1\right)-6x\left(2x-3\right)\)

\(D=10x^2-4x+15x-6+2x^2-x+8x-4-12x^2+18x\)

\(D=36x-10\)

bn cho mik xin lại cái đè bài câu a đc k mà bn lớp mấy thế

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

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

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

\(=4x^2-6x+2x^2-x-1\)

\(=6x^2-7x-1\)

a) (2x+1)^2-2(2x+1)(2x-1)+(2x-1)^2

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

=2^2=4

b)\(\left(2x^3-3x^2+6x-9\right)\left(2x-3\right)\)

\(=\left[x^2\left(2x-3\right)+x\left(2x-3\right)\right]\left(2x-3\right)\)

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

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

tự làm tiếp đi nha

tiếp câu b)

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

a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)

\(=6x^2-2x-6x^2-2x+18x+6\)

=14x+6

b: Ta có: \(\left(2x-3\right)^2-\left(2x+1\right)\left(2x-1\right)+3\left(2x-3\right)\)

\(=4x^2-12x+9-4x^2+1+6x-9\)

\(=-6x+1\)

c: Ta có: \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2\)

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

=1

a) \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)=6x^2-2x-6x^2-2x+18x+6=14x+6\)

b) \(\left(2x-3\right)^2-\left(1+2x\right)\left(2x-1\right)+3\left(2x-3\right)=4x^2-12x+9-4x^2+1+6x-9=-6x+1\)

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