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5 tháng 9 2020

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

b) \(\left(3x-2y\right)^2+4.\left(3x-2y\right)+4\)

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

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

5 tháng 9 2020

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

b) \(\left(3x-2y\right)^2+4\left(3x-2y\right)+4=\left(3x-2y+2\right)^2\)

3 tháng 6 2016

\(A=x^2-2x+1-x^2+4=5-2x\)

\(B=27x^3+8-x^2+9=27x^3-x^2+17\)

\(C=3x^2y-6xy^2-2x\left(x^2-2x^2y+x^2y^2\right)=3x^2y-6xy^2-2x^3+4x^3y-2x^3y^2\)

Em chỉ cần nhớ hằng đẳng thức và áp dụng là biến đổi được ^^

9 tháng 6 2018

a) \(\left(3x-4\right)^2+2\left(3x-4\right)\left(2x+4\right)+\left(2x+4\right)^2\)\(=\left(3x-4+2x+4\right)^2=\left(5x\right)^2=25x^2\)

b)\(\left(3x+4\right)^2+\left(7+3x\right)^2-\left(6x+8\right)\left(3x+7\right)\)

\(=\left(3x+4\right)^2-2\left(3x+4\right)\left(7+3x\right)+\left(7+3x\right)^2\)

\(=\left[3x+4-\left(7+3x\right)\right]^2=\left(3x+4-7-3x\right)^2=\left(-3\right)^2=9\)

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

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

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

\(=\left(2x+1+2x-1\right)^2=\left(4x\right)^2=14x^2\)

xong rồi đấy,bạn k cho mình nhé

25 tháng 7 2018

\(a.\left(2x-3\right)\left(4x^2+6x+9\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\\ =\left(2x\right)^3-3^3-\left[\left(2x\right)^3+3^3\right]\\ =8x^3-9-\left(8x^3+9\right)\\ =8x^3-9-8x^3-9=-18\)

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

\(c.\left(3x-1\right)\left(3x+1\right)-\left(3x-2\right)^2\\ =9x^2-1-\left(9x^2-12x+4\right)\\ =9x^2-1-9x^2+12x-4\\ =12x-5\)

\(d.\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\\ =\left(2x-3\right)\cdot\left[\left(2x-3\right)-\left(2x+3\right)\right]\\ =\left(2x-3\right)\cdot\left(2x-3-2x-3\right)\\ =\left(2x-3\right)\cdot\left(-6\right)\\ =-12x\cdot18\)

\(e.\left(3x-4\right)^2-\left(2x+4\right)^2\\ =9x^2-24x+16-\left(4x^2+16x+16\right)\\ =9x^2-24x+16-4x^2-16x-16\\ =5x^2-40x\)

\(f.\left(3x-5\right)^3-\left(3x+5\right)^3\\ =27x^3-135x^2+225x-125-\left(27x^3+135x^2+225x+125\right)\\ =27x^3-135x^2+225x-125-27x^3-135x^2-225x-125\\ =-270x^2-250\)

\(g.\left(2x-1\right)^2-\left(3x-1\right)^2\\ =4x^2-4x+1-\left(9x^2-6x+1\right)\\ =4x^2-4x+1-9x^2+6x-1\\ =-5x^2+2x\)

\(h.\left(x-2y\right)\left(x^2+2xy+4y^2\right)+\left(x^3-6y^3\right)\\ =x^3-8y^3+x^3-6y^3\\ =2x^3-14y^3\)

24 tháng 7 2019

\(a,VT=\left(a+b+c\right)\left(a-b+c\right)\)

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

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

\(=a^2+2ac+c^2-b^2=VP\)

\(b,VT=\left(3x+2y\right)\left(3x-2y\right)-\left(4x-2y\right)\left(4x+2y\right)\)

\(=9x^2-4y^2-16x^2+4y^2=-7x^2=VP\)

\(c,VT=x^3-1-x^3-1=-2=VP\)

\(d,VT=8x^3+1-8x^3+1=2=VP\)

\(e,VT=\left(x^2+2xy+4y^2\right)\left(x-2y-2x+1\right)\)

\(=\left(x^2+2xy+4y^2\right)\left(-x-2y+1\right)\)

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

( bn kiểm tra lại đề nhé)

14 tháng 8 2015

a/ \(=3y^2-6y-2x+1\)

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

c/ \(=\left(2-x\right)^3\)

d/ \(=xy^2+x^2y+3xy+x^2y+x^3+3x^2-3xy-3x^2-9x\)

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

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

e/ \(=xy-x^2+2x-y^2+xy-2y\)

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

14 tháng 8 2015

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

b)=-(x-1)3

c)=-(x3-6x2+12x-8)=-(x-2)3

d)x3 + 2x2y + xy2 – 9x

    = x(x2 + 2xy + y2 -9)

    = x[(x2 + 2xy + y2) - 32]

    = x[(x + y)2 - 32]

    = x (x + y – 3)(x + y + 3)

e) 2x-2y-x2+2xy-y2=2(x-y)-(x-y)2=(x-y)(2-x+y)

3 tháng 12 2017

- Viết 7 hằng đẳng thức đáng nhớ :

\(\left(A+B\right)^2=A^2+2AB+B^2\)

\(\left(A-B\right)^2=A^2-2AB+B^2\)

\(A^2-B^2=\left(A-B\right)\left(A+B\right)\)

\(\left(A+B\right)^3=A^3+3A^2B+3AB^2+B^3\)

\(\left(A-B\right)^3=A^3-3A^2B+3AB^2-B^3\)

\(A^3-B^3=\left(A-B\right)\left(A^2+AB+B^2\right)\)

\(A^3+B^3=\left(A+B\right)\left(A^2-AB+B^2\right)\)

- Áp dụng :

\(a,\left(x+2y\right)^2=x^2+4xy+4y^2\)

\(b,\left(\dfrac{5x-1}{2}\right)^2=\dfrac{\left(5x-1\right)^2}{2^2}=\dfrac{25x^2-10x+1}{4}\)

\(c,\left(\dfrac{1}{3x-3}\right)\left(\dfrac{1}{3x+3}\right)=\dfrac{1.1}{\left(3x-3\right)\left(3x+3\right)}=\dfrac{1}{9x^2-9}\)

\(d,\left(2x+3\right)^3=8x^3+36x^2+54x+27\)

\(e,\left(\dfrac{1}{4y-2x}\right)^2=\dfrac{1}{\left(4y-2x\right)^2}=\dfrac{1}{16y^2-16xy+4x^2}\)

\(f,\left(2x-y\right)\left(4x^2+2xy+y^2\right)=\left(2x\right)^3-y^3=8x^3-y^3\)

\(g,\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)