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

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

\(=36x^2\)(1)

Thay \(x=-\dfrac{1}{3}\) vào biểu thức (1), ta được:

\(36\cdot\left(-\dfrac{1}{3}\right)^2=36\cdot\dfrac{1}{9}=4\)

b) Sửa đề: \(\left(x+y-7\right)^2-2\cdot\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)

Ta có: \(\left(x+y-7\right)^2-2\cdot\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)

\(=\left(x+y-7-y+6\right)^2\)

\(=\left(x-1\right)^2=100^2=10000\)

a) \(=5\left|a\right|+3a=5a+3a=8a\)

b) \(=3\left|a^2\right|+3a^2=3a^2+3a^2=6a^2\)

c) \(=5.2\left|a^3\right|-3a^3=-10a^3-3a^3=-13a^3\)

a) \(P=\dfrac{\sqrt{3}+\sqrt{6}}{1+\sqrt{2}}=\dfrac{\left(\sqrt{3}+\sqrt{6}\right)\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}\)

                             \(=\dfrac{\sqrt{3}-\sqrt{6}+\sqrt{6}-\sqrt{12}}{1-2}=\sqrt{12}-\sqrt{3}\)

b) \(Q=\left(\sqrt{75}-\dfrac{3}{2}:\sqrt{3}-\sqrt{48}\right)\cdot\sqrt{\dfrac{16}{3}}\)

        \(=\left(5\sqrt{3}-\dfrac{3}{2}\cdot\dfrac{1}{\sqrt{3}}-4\sqrt{3}\right)\cdot\dfrac{4}{\sqrt{3}}\)

        \(=\sqrt{3}\left(5-\dfrac{1}{2}-4\right)\cdot\dfrac{4}{\sqrt{3}}\)

        \(=\left(1-\dfrac{1}{2}\right)\cdot4=2\)

a: \(=\sqrt{3}-1-\sqrt{3}=-1\)

b: \(=2\sqrt{3}-10\sqrt{3}+4\sqrt{3}=-4\sqrt{3}\)

a: \(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)

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

\(=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\)

\(=x^3-16x^2+25x\)

Ta có: \(A=\left(x-y-1\right)^3-\left(x-y+1\right)^3+6\left(x-y\right)^2\)

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

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

\(=-6\left(x-y\right)^2+6\left(x-y\right)^2-2\)

=-2

a: \(\left(2x+3y\right)\left(x-2y\right)-\dfrac{\left(4x^3y-6x^2y^2-3xy^3\right)}{2xy}\)

\(=2x^2-4xy+3xy-6y^2-\dfrac{2xy\cdot\left(2x^2-3xy-1,5y^2\right)}{2xy}\)
\(=2x^2-xy-6y^2-2x^2+3xy+1,5y^2\)

\(=2xy-4,5y^2\)

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

\(=x^3-6x^2+12x-8-x\left(x^2-1\right)-\left(9x^2-6x-3x+2\right)\)

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

\(=-15x^2+22x-10\)

(5x-1)²+(5x+1)²-2(1-25x²)=25x²-10x+1+25x²+10x+1-2-50x²

                                          = 0

Lời giải:

$(5x-1)^2+(5x+1)^2-2(1-25x^2)$

$=(5x-1)^2+(5x+1)^2-2(1-5x)(1+5x)$

$=(5x-1)^2+(5x+1)^2+2(5x-1)(5x+1)$

$=(5x-1+5x+1)^2$

$=(10x)^2=100x^2$

Ta có : \(\left(5x-1\right)^2-2\left(1-25x^2\right)+\left(5x+1\right)^2\)

\(=\left(5x-1\right)^2+2\left(25x^2-1\right)+\left(5x+1\right)^2\)

\(=\left(5x-1\right)^2+2\left(5x-1\right)\left(5x+1\right)+\left(5x+1\right)^2\)

\(=\left(5x-1+5x+1\right)^2=\left(10x\right)^2=100x^2\)

\(a,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\\ =\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|\\ =\sqrt{3}+\sqrt{2}-\left(\sqrt{3}-\sqrt{2}\right)\\ =\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\\=2\sqrt{2} \)

\(b,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1}+\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}+1+\sqrt{3}-1\\ =2\sqrt{3}\)

\(c,=x-4+\sqrt{\left(4^2-2.4.x+x^2\right)}\\ =x-4+\sqrt{\left(4-x\right)^2}\\ =x-4+\left|4-x\right|\\ =x-4+x-4=2x-8\)    (vì \(x>4\) )

@seven 

a: \(A=\dfrac{x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{2}}+y^{\dfrac{1}{3}}\cdot x^{\dfrac{1}{2}}}{x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}}=\dfrac{x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{3}}\left(x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}\right)}{x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}}=x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{3}}=\left(xy\right)^{\dfrac{1}{3}}\)

b: \(B=\dfrac{x^{3+\sqrt{3}}}{y^2}\cdot\dfrac{x^{-\sqrt{3}-1}}{y^{-2}}=\dfrac{x^{3+\sqrt{3}-\sqrt{3}-1}}{y^{2-2}}=x^2\)