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

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

\(=x^2-2xy+y^2+x^2+2xy+y^2\)

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

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

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

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

\(=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)

\(=4a^2+4ab+b^2-4a^2+4ab-b^2\)

\(=\left(4a^2-4a^2\right)+\left(4ab+4ab\right)+\left(b^2-b^2\right)\)

\(=8ab\)\

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

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

\(=x^2+2xy+y^2-x^2+2xy-y^2\)

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

\(=4xy\)

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

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

\(=4x^2-4x+1-8x^2+24x-18+4\)

\(=\left(4x^2-8x^2\right)-\left(4x-24x\right)+\left(1-18+4\right)\)

\(=-4x^2+20x-13\)

\(=-4x^2+20x-25+12\)

\(=-\left(4x^2-20x+25\right)-8\)

\(=-\left[\left(2x\right)^2-2.4x.5+5^2\right]-8\)

\(=-\left(2x-5\right)^2-8\)

câu 1:

a,x²+2x-4z²+1

=x²+2x.1+1²-(2z)²

=(x+1)²-(2z)²

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

bạn nên dùng hằng đẳng thức đã học

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

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

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

b) \(4x^2+4xy-16+y^2=\left(4x^2+4xy+y^2\right)-16\)

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

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

c)

\(A=\left(2x\right)^2-2.2x.5+5^2-4x.x+4x.6\)

\(=4x^2-20x+25-4x^2+24x=4x+25\)

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

\(=\left(7x-3y\right)\left(7x-3y-7x-3y\right)\)

\(=\left(7x-3y\right)\left(-6y\right)=18y^2-42xy\)

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

\(=9-2.3.2x+4x^2+9+2.3.2x+4x^2\)

\(=18+8x^2\)

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

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

bài 1.

a.\(A=x^2-2xy+y^2+x^2+2xy+y^2=2\left(x^2+y^2\right)\)

b.\(B=x^2+2xy+y^2-\left(x^2-2xy+y^2\right)=4xy\)

c.\(C=4a^2+4ab+b^2-\left(4a^2-4ab+b^2\right)=8ab\)

d.\(D=4x^2-4x+1-2\left(4x^2-12x+9\right)+4=-4x^2+20x-13\)

.bài 2

a.\(A=x^2+6x+9+x^2-9-2\left(x^2-2x-8\right)=10x+16;x=-\frac{1}{2}\Rightarrow A=9\)

b.\(B=9x^2+24x+16-x^2+16-10x=8x^2+14x+32\Rightarrow x=-\frac{1}{10}\Rightarrow B=\frac{767}{25}\)

c.\(C=x^2+2x+1-\left(4x^2-4x+1\right)+3\left(x^2-4\right)=6x-12\Rightarrow x=1\Rightarrow C=-6\)

d.\(D=x^2-9+x^2-4x+4-2x^2+8x=4x-5\Rightarrow x=-1\Rightarrow A=-9\)

Trả lời:

Bài 1: Rút gọn biểu thức:

a) A = ( x - y )² + ( x + y )²

= x² - 2xy + y² + x² + 2xy + y²

= 2x² + 2y² 

b) B = ( x + y )² - ( x - y )² 

= x² + 2xy + y² - ( x² - 2xy + y² )

= x² + 2xy + y² - x² + 2xy - y²

= 4xy

c) C = ( 2a + b )² - ( 2a - b )² 

= 4a² + 4ab + b² - ( 4a² - 4ab + b² )

= 4a² + 4ab + b² - 4a² + 4ab - b² 

= 8ab

d) D = ( 2x - 1 )² - 2 ( 2x - 3 )² + 4

= 4x² - 4x + 1 - 2 ( 4x² - 12x + 9 ) + 4

= 4x² - 4x + 1 - 8x² + 24x - 18 + 4

= - 4x² + 20x - 13

Bài 2: Rút gọn rồi tính giá trị biểu thức:

a) A = ( x + 3 )² + ( x - 3 )( x + 3 ) - 2 ( x + 2 )( x - 4 )

= x² + 6x + 9 + x² - 9 - 2 ( x² - 2x - 8 ) 

= 2x² + 6x - 2x² + 4x + 16

= 10x + 16

Thay x = 1/2 vào A, ta có:

\(A=10.\left(-\frac{1}{2}\right)+16=-5+16=11\)

b) B = ( 3x + 4 )² - ( x - 4 )( x + 4 ) - 10x

= 9x² + 24x + 16 - x² + 16 - 10x 

= 8x² + 14x + 32

Thay x = - 1/10 vào B, ta có:

\(B=8.\left(-\frac{1}{10}\right)^2+14.\left(-\frac{1}{10}\right)+32=\frac{767}{25}\)

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

= x² + 2x + 1 - 4x² + 4x - 1 + 3 ( x² - 4 )

= - 3x² + 6x + 3x² - 12

= 6x - 12

Thay x = 1 vào C, ta có:

\(C=6.1-12=-6\)

d) D = ( x - 3 )( x + 3 ) + ( x - 2 )² - 2x ( x - 4 ) 

= x² - 9 + x² - 4x + 4 - 2x² + 8x

= 4x - 5

Thay x = - 1 vào D, ta có:

\(D=4.\left(-1\right)-5=-9\)

a) Ta có: \(\left(3-xy^2\right)^2-\left(2+xy^2\right)^2\)

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

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

\(=5\cdot\left(1-2xy^2\right)\)

\(=5-10xy^2\)

b) Ta có: \(9x^2-\left(3x-4\right)^2\)

\(=\left[3x-\left(3x-4\right)\right]\left[3x+\left(3x-4\right)\right]\)

\(=\left(3x-3x+4\right)\cdot\left(3x+3x-4\right)\)

\(=4\cdot\left(6x-4\right)\)

\(=24x-16\)

c) Ta có: \(\left(a-b^2\right)\left(a+b^2\right)\)

\(=a^2-b^4\)

d) Ta có: \(\left(a^2+2a+3\right)\left(a^2+2a-3\right)\)

\(=\left(a^2+2a\right)^2-9\)

\(=a^4+4a^3+4a^2-9\)

e) Ta có: \(\left(x-y+6\right)\left(x+y-6\right)\)

\(=x^2+xy-6x-yx-y^2+6y+6x+6y-36\)

\(=x^2-y^2+12y-36\)

f) Ta có: \(\left(y+2z-3\right)\left(y-2z-3\right)\)

\(=\left(y-3\right)^2-\left(2z\right)^2\)

\(=y^2-6y+9-4z^2\)

g) Ta có: \(\left(2y-5\right)\left(4y^2+10y+25\right)\)

\(=\left(2y\right)^3-5^3\)

\(=8y^3-125\)

h) Ta có: \(\left(3y+4\right)\left(9y^2-12y+16\right)\)

\(=\left(3y\right)^3+4^3\)

\(=27y^3+64\)

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

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

\(=x^3-9x^2+27x-27-\left(x^3-6x^2+12x-8\right)\)

\(=x^3-9x^2+27x-27-x^3+6x^2-12x+8\)

\(=-3x^2+15x-19\)

j) Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

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

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

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

a) \(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)

\(A=20x^3-10x^2+5x-20x^3+10x^2+4x\)

\(A=9x\)

Thay x = 15 vào, ta có: 

\(A=9.15=135\)

b) \(B=5x\left(x-4y\right)-4y\left(y-5x\right)\)

\(B=5x^2-20xy-4y^2+20xy\)

\(B=5x^2-4y\)

Thay \(x=-\frac{1}{5};y=-\frac{1}{2}\) vào, ta có: 

\(B=5.\left(-\frac{1}{5}\right)^2-4.\left(-\frac{1}{2}\right)=\frac{11}{5}\)

c) \(C=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)-5y^2\left(x^2-xy\right)\)

\(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)

\(C=9x^2y^2-xy^3-8x^3\)

Thay \(x=\frac{1}{2};y=2\) vào, ta có:

\(C=9.\left(\frac{1}{2}\right)^2.2^2-\frac{1}{2}.2^3-8.\left(\frac{1}{2}\right)^3=4\)

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

\(D=6x^2-3x+10x-5+12x^2+8x-3x-2\)

\(D=18x^2+12x-7\)

Ta có: \(\left|2\right|=\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)

+) Với x = -2

\(D=18.\left(-2\right)^2+12.\left(-2\right)-7=41\)

+) Với x = 2

\(D=18.2^2+12.2-7=89\)