a: ĐKXĐ: x=0; x<>1

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

\(=\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)=4-x\)

b: Sửa đề: P=1/M

P=1/4-x=-1/x-4

Để P nguyên thì x-4 thuộc {1;-1}

=>x thuộc {5;3}

a:

ĐKXĐ: x>=0; x<>1

 Sửa đề: \(M=x-\dfrac{2x-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x\sqrt{x}+1}{x-\sqrt{x}+1}+1\)

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

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

b: \(M=x-\sqrt{x}+2\)

\(=x-\sqrt{x}+\dfrac{1}{4}+\dfrac{7}{4}\)

\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\sqrt{x}-\dfrac{1}{2}=0\)

=>\(\sqrt{x}=\dfrac{1}{2}\)

=>x=1/4

a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b: \(M=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\)

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

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

c: Để M=-1/2 thì 2(x+1)=-2

=>x+1=-1

hay x=-2(nhận)

a, ĐKXĐ của M là :

\(\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)\ne0\\2\left(1-x\right)\left(1+x\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)

b, \(M=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\)

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

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

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

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

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

c, Để giá trị của biểu thức M bằng \(\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{1}{2x-2}=\dfrac{1}{2}\)

\(\Leftrightarrow2x-2=2\)

\(\Leftrightarrow2x=4\Leftrightarrow x=2\)

Vậy x = 2 thfi giá trị của M bằng 1/2

a.

M xác định \(\Leftrightarrow\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne\pm1\end{matrix}\right.\)

b.

\(M=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\\ =\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2\left(x^2-1\right)}\\ =\dfrac{x\left(x+1\right)}{2\left(x^2-1\right)}-\dfrac{x^2+1}{2\left(x^2-1\right)}\\ =\dfrac{x-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2x+2}\)

c.

M=1/2

\(\Leftrightarrow\dfrac{1}{2x+2}=\dfrac{1}{2}\\ \Leftrightarrow2x+2=2\\ \Leftrightarrow x=0\)

a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b: \(M=\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)

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

c: Để M=1/2 thì 2x+2=2

hay x=0(nhận)

A = 3x^3 +6x^2 + 3xy^3

x= 1 phần 2 ;  p = -1 phần 3

A=3.1 phần 2^3 . -1 phần 3     + 6.(1 phần 2)^2 . (-1 Phần 3)^2+3 1 phần 2 . (-1 phần 3)^3

=-1 phần 8      + -1 phần 2 - 1 phần 2

= -1 phần 4

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