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

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

B = 6x + 4y

chiu thui

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mk moi hoclop 6

chuc bn hoc tot!

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\(1,\left(2x+1\right)^2+2\left(2x+1\right)+1\\ =\left(2x+1\right)^2+2.\left(2x+1\right).1+1^2\\ =\left[\left(2x+1\right)+1\right]^2\\ b,\left(3x-2y\right)^2+4\left(3x-2y\right)+4\\ =\left(3x-2y\right)^2+2.\left(3x-2y\right).2+2^2\\ =\left[\left(3x-2y\right)+2\right]^2\)

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

\(=\left(2x+1\right)^2+2\left(2x+1\right)\cdot1+1^2\)

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

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

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

\(=\left(3x+2y\right)^2+2\cdot\left(3x+2y\right)\cdot2+2^2\)

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

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

\(\left(x+2y\right)^2-\left(x-2y\right)^2\\ =\left[\left(x+2y\right)-\left(x-2y\right)\right]\left[\left(x+2y\right)+\left(x-2y\right)\right]\\ =\left(x+2y-x+2y\right)\left(x+2y+x-2y\right)\\ =4y.\left(2x\right)\\ =8xy\)

\(\left(3x+y\right)^2+\left(x-y\right)^2\\ =\left[\left(3x\right)^2+2.3x.y+y^2\right]+\left(x^2-2xy+y^2\right)\\ =6x^2+6xy+y^2+x^2-2xy-y^2\\ =7x^2+4xy\)

\(-\left(x+5\right)^2-\left(x-3\right)^2\\ =-\left(x^2+10x+25\right)-\left(x^2-6x+9\right)\\ =-x^2-10x-25-x^2+6x-9\\ =-2x^2-4x-34\)

\(\left(3x-2\right)^2-\left(3x-1\right)^2\\ =\left[\left(3x-2\right)-\left(3x-1\right)\right]\left[\left(3x-2\right)+\left(3x-1\right)\right]\\ =\left(3x-2-3x+1\right)\left(3x-2+3x-1\right)\\ =-1.\left(6x-3\right)\\ =-6x+3\)

Ta có: \(A^2=\dfrac{\left(3x-2y\right)^2}{\left(3x+2y\right)^2}\)

\(=\dfrac{9x^2+4x^2-12xy}{9x^2+4x^2+12xy}\)

\(=\dfrac{20xy-12xy}{20x^2+12xy}\)

\(=\dfrac{8xy}{32xy}=\dfrac{1}{4}\)

\(\Leftrightarrow A\in\left\{\dfrac{1}{2};-\dfrac{1}{2}\right\}\)(1)

Vì 2y<3x<0 nên 3x-2y>0 và 3x+2y<0

hay \(A=\dfrac{3x-2y}{3x+2y}< 0\)(2)

Từ (1) và (2) suy ra \(A=-\dfrac{1}{2}\)

Vậy: \(A=-\dfrac{1}{2}\)

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

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

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

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

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

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

Sửa đề: \(5x^2y^2\cdot\left(3x^2\right)^3\)

\(=5x^2y^2\cdot27x^6\)

\(=5\cdot27\cdot x^2\cdot x^6\cdot y^2\)
\(=135x^8y^2\)

\((5x^2y^2)\cdot(3x^2y^3)\) (sửa đề)

\(=\left(5\cdot3\right)\left(x^2\cdot x^2\right)\left(y^2\cdot y^3\right)\)

\(=15x^4y^5\)

Bài 1: 

a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)

= \(x^2\) -  16 - 5\(x\) - 5 + \(x^2\) + \(x\) 

= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)

= 2\(x^2\) - 4\(x\) - 21

b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)

=  3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7

= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)

= - 3\(x^2\) + 3\(xy\) - 3