a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)

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

b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)

\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)

c: \(C=x-4+\left|x-4\right|\)

=x-4+x-4

=2x-8

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

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

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

@seven 

phân tích thành nhân tử ở mẫu và tử sau đó ta rút gọn vậy là ra đáp số

a) \(=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(4x-5\right)}\)\(\)

\(=\frac{5x\cdot\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}\)

\(=\frac{5x\left(4x+5\right)}{x-3}\)

b) \(=\frac{3^2-\left(x+5\right)^2}{\left(x+2\right)^2}\)

\(=\frac{\left(3-x-5\right)\left(3+x+5\right)}{\left(x+2\right)^2}\)

\(=\frac{\left(x+2\right)\left(8+x\right)}{\left(x+2\right)^2}\)

\(=\frac{8+x}{x+2}\)

chịu

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

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

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

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

\(=y^2\)

\(A=\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\)

\(=\frac{\left(1-x\right)\left(1+x+x^2\right)-x+x^2}{1-x}.\frac{\left(1-x\right)-x^2\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}\)

\(=\frac{\left(1-x\right)\left(1+x+x^2\right)-x\left(1-x\right)}{1-x}.\frac{\left(1-x\right)\left(1-x^2\right)}{\left(1-x\right)\left(1+x\right)}\)

\(=\frac{\left(1-x\right)\left(1+x^2\right)}{1-x}.\frac{\left(1-x\right)\left(1-x\right)\left(1+x\right)}{\left(1-x\right)\left(1+x\right)}\)

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

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

Ta có : \(A=\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\)

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

\(=\left(\left(1+x+x^2\right)-x\right):\frac{\left(1-x\right)\left(1+x\right)}{\left(1-x\right)-x^2\left(x-1\right)}\)

\(=\left(1+x^2\right):\frac{\left(1-x\right)\left(1+x\right)}{\left(1-x\right)\left(1-x^2\right)}\)

\(=\left(1+x^2\right):\frac{\left(1-x\right)\left(1+x\right)}{\left(1-x\right)\left(1-x\right)\left(x+1\right)}\)

\(=\left(1+x^2\right):\frac{1}{1-x}\)

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

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

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

\(=4x^2\)

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