a) \(\left(m+n\right)^2-\left(m-n\right)^2+\left(m+n\right)\left(m-n\right)\)

\(=\left(m+n+m-n\right)\left(m+n-m+n\right)+m^2-n^2\)

\(=m^2-n^2+4mn\)

b) \(\left(a+b\right)^3+\left(a-b\right)^3-2a^3\)

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

\(=2b\left[a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right]-2a^3\)

\(=2b\left(a^2+3b^2\right)-2a^3\)

\(=2a^2b+6b^3-2a^3.\)

Tương tự áp dụng các HĐT.

a) \(\left(m+n\right)^2-\left(m-n\right)^2=\left[\left(m+n\right)-\left(m-n\right)\right]\left[\left(m+n\right)+\left(m-n\right)\right]=\left(2n\right)\left(2m\right)=4mn\)\(\left(m+n\right)\left(m-n\right)=m^2-n^2\)

A=\(4mn+m^2-n^2\) tối giản rồi

b)

\(\left(a+b\right)^3+\left(a-b\right)^3=\left[\left(a+b\right)+\left(a-b\right)\right]^3-3\left(a+b\right)\left(a-b\right)\left[\left(a+b\right)+\left(a-b\right)\right]=8a^3-3.2a.\left(a^2-b^2\right)\)B=\(8a^3-3.2a.\left(a^2-b^2\right)-2a^3=6a\left[a^2-\left(a^2-b^2\right)\right]=6ab^2\)

Câu c :

\(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2=\left[\left(2x+1\right)+\left(2x-1\right)\right]^2=\left(2x+1+2x-1\right)^2\)

Câu d :

\(\left(a+b+c\right)^2-2\left(a+b+c\right)\left(b+c\right)+\left(b+c\right)^2=\left[\left(a+b+c\right)-\left(b+c\right)\right]^2=\left(a+b+c-b-c\right)^2\)

a) \(\cdot\left(m+n\right)^2-\left(m-n\right)^2+\left(m+n\right)\left(m-n\right)\)

\(=\left(m+n+m-n\right)\left(m+n-m+n\right)+\left(m+n\right)\left(m-n\right)\)

\(=\left(2m\cdot2n\right)+m^2-n^2\)

\(=4mn+m^2-n^2\)

b) \(\left(a+b\right)^2-\left(a-b\right)^2-2a^3\)

\(=\left(a+b+a-b\right)\left(a+b-a+b\right)-2a^3\)

\(=2ab-2a^3\)

c) \(\left(2x+1\right)^2+\left(2x-1\right)^2+2\left(4x^2-1\right)\)

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

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

\(=\left(4x\right)^2=16x^2\)

d) \(\left(a+b+c\right)^2-2\left(a+b+c\right)\left(b+c\right)+\left(b+c\right)^2\)

\(=\left(a+b+c-b-c\right)^2=a^2\)

xin lỗi mk ghi sai đề ở bài :d) (a+b+c)^2-2(a+b+c)(b+c)+(b+c)^2

1)a)=>x²+y²+2xy-4(x²-y²-2xy)

=>x²+y²+2xy-4.x²+4y²+8xy

=>-3.x²+5y²+10xy

a:

ĐKXĐ: x<>2

|2x-3|=1

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

=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

Thay x=1 vào A, ta được:

\(A=\dfrac{1+1^2}{2-1}=\dfrac{2}{1}=2\)

b: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)

\(B=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{x^2-x-2}\)

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

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

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

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

c: \(P=A\cdot B=\dfrac{-1}{x+1}\cdot\dfrac{x\left(x+1\right)}{2-x}=\dfrac{x}{x-2}\)

\(=\dfrac{x-2+2}{x-2}=1+\dfrac{2}{x-2}\)

Để P lớn nhất thì \(\dfrac{2}{x-2}\) max

=>x-2=1

=>x=3(nhận)

a: \(=y^2-9\)

b: \(=m^3+n^3\)

c: \(=8-a^3\)

d: \(=\left(a-b-c-a+b-c\right)\left(a-b-c+a-b+c\right)\)

\(=-2c\cdot\left(2a-2b\right)\)

\(=-4ac+4bc\)

f: \(=\left(1-x^3\right)\left(1+x^3\right)=1-x^6\)

em 2k6, đọc phần lí thuyết r lm, nên có lỗi j sai mong mn thông cảm

bài 1,

a, \(3xy\left(4xy^2-5x^2y-4xy\right)\)

= \(3xy.4xy^2-3xy.5x^2y-3xy.4xy\)

=\(12x^2y^3-15x^3y^2-12x^2y^2\)

b, \(\left(2x-1\right)\left(4x^2+2x+1\right)\)

=\(\left(2x.4x^2+2x.2x+2x.1\right)-\left(1.4x^2+1.2x+1.1\right)\)

=\(8x^3+4x^2+2x-4x^2-2x-1\)

=\(8x^3+\left(4x^2-4x^2\right)+\left(2x-2x\right)-1\)

=\(8x^3-1\)

a: \(P=\left(\dfrac{-\left(x+1\right)}{x-1}+\dfrac{x-1}{x+1}-\dfrac{4x^2}{\left(x-1\right)\left(x-1\right)}\right)\cdot\dfrac{\left(x-1\right)^2}{4\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-x^2-2x-1+x^2-2x+1-4x^2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x-1}{4\left(x+1\right)}\)

\(=\dfrac{-4x^2-4x}{x+1}\cdot\dfrac{1}{4\left(x+1\right)}\)

\(=\dfrac{-4x\left(x+1\right)}{x+1}\cdot\dfrac{1}{4\left(x+1\right)}=\dfrac{-x}{x+1}\)

b: khi x=5/8 thì \(P=\left(-\dfrac{5}{8}\right):\dfrac{13}{8}=\dfrac{-5}{13}\)

c: Để P là số nguyên thì \(-x-1+1⋮x+1\)

\(\Leftrightarrow x+1\in\left\{1;-1\right\}\)

hay \(x\in\left\{0;-2\right\}\)