\(\frac{1}{2}x^2\left(6x-3\right)-x\left(x^2+\frac{1}{2}\right)+\frac{1}{2}\left(x+4\right)\)

\(=\frac{1}{2}x^2.6x+\frac{1}{2}x^2.\left(-3\right)+\left(-x\right).x^2+\left(-x\right).\frac{1}{2}+\frac{1}{2}.x+\frac{1}{2}.4\)

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

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

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

\(a,\)\(\frac{1}{2}x^2\left(6x-3\right)-x\left(x^2+\frac{1}{2}\right)+\frac{1}{2}\left(x+4\right)\)

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

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

\(b,\)\(2x\left(3x^3-x\right)-4x^2\left(x-x^2+1\right)+\left(x-3x^2\right)x\)

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

\(=10x^4-7x^3-5x^2\)

\(\frac{1}{2}x^2.\left(6x-3\right)-x\left(x^2+\frac{1}{2}\right)+\frac{1}{2}\left(x+4\right)\)

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

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

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

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

\(=\frac{\sqrt{x}+1}{\sqrt{x}}\)

Lời giải:

a)

\(=\frac{(\sqrt{x}+1)\sqrt{x}(\sqrt{x}-\sqrt{y}))\sqrt{x}+\sqrt{y})}{(x-y)x(\sqrt{x}+1)}=\frac{(\sqrt{x}+1)\sqrt{x}(x-y)}{(x-y)x\sqrt{x}+1)}=\frac{1}{\sqrt{x}}\)

b)

\(=\frac{(2-\sqrt{x}-\sqrt{x}-3)(2-\sqrt{x}+\sqrt{x}+3)}{1+2\sqrt{x}}=\frac{(-1-2\sqrt{x}).5}{2\sqrt{x}+1}=\frac{-5(2\sqrt{x}+1)}{2\sqrt{x}+1}=-5\)

\(a,\frac{\left(\sqrt{x}+1\right)\cdot\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\sqrt{x}\left(x+1\right)}\)\(=\frac{\left(\sqrt{x}+1\right)\sqrt{x}\left(x-y\right)}{\left(x-y\right)\sqrt{x} \left(x+1\right)}\)\(=\frac{\sqrt{x}+1}{x+1}\)

\(b,\frac{\left(2-\sqrt{x}\right)^2-\sqrt{x}-3}{1+2\sqrt{x}}=\frac{4+x-4\sqrt{x}-\sqrt{x}-3}{1+2\sqrt{x}}=\frac{1+x-5\sqrt{x}}{1+2\sqrt{x}}\)

Lời giải:

ĐKXĐ: \(x\geq 0; x\neq 1\)

Ta có:

\(A=\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+2)}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}=\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+2)}+\frac{\sqrt{x}+2}{(\sqrt{x}-1)(\sqrt{x}+2)}+\frac{\sqrt{x}-1}{(\sqrt{x}+2)(\sqrt{x}-1)}\)

\(=\frac{x+\sqrt{x}+1+\sqrt{x}+2+\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{x+3\sqrt{x}+2}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{(\sqrt{x}+1)(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

a/ \(P=\left(\frac{3}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{x-1}\right):\left(\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)

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

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

\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)

\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)

\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)

\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{1}{\left(\sqrt{x}-1\right)}\right)\)

\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{\sqrt{x}-1}{1}\right)\)

=> \(P=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)}\)

b/ \(P=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)}=\sqrt{x}-1\)

<=> \(4\sqrt{x}=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)

<=> \(4\sqrt{x}=x-1\). Bình phương 2 vế, ta được:

<=> 16x=(x-1)²

<=> 16x=x²-2x+1

<=> x²-18x+1=0

\(\Delta'=81-1=80=>\sqrt{\Delta'}=4\sqrt{5}\)

=> \(x_1=9-4\sqrt{5}\)

\(x_2=9+4\sqrt{5}\)

Bài 2: 

a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)

\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)

b: Thay x=1/2 vào B, ta được:

\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)

Thay x=-1/2 vào B, ta được:

\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)

c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)

=>6(x-2)=-1/2

=>x-2=-1/12

hay x=23/12