a) Phân thức xác định được \(\Leftrightarrow\hept{\begin{cases}2x+10\ne0\\x\ne0\\2x\left(x+5\right)\ne0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x+5\ne0\end{cases}}\)

Vậy...

b) \(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)

=> \(P=\frac{x\left(x^2+2x\right)+2\left(x-5\right)\left(x+5\right)+50-5x}{2x\left(x+5\right)}\)

=> \(P=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)

=> \(P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}=\frac{x\left(x-1\right)\left(x+5\right)}{2x\left(x+5\right)}=\frac{\left(x-1\right)}{2}\)

\(P=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)

\(a)x\ne\pm\frac{4}{3}\)

\(b)x\ne2\)

\(c)x\ne\pm1\)

\(d)x\ne0;x\ne\frac{1}{2}\)

\(e)x\ne\pm1\)

\(f)x\ne-1;x\ne3\)

\(g)x\ne3;x\ne2\)

Mình Không Biết !

\(a.\frac{x}{2x-6}+\frac{x}{2x+2}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=\)\(0\)

\(\Leftrightarrow\frac{x}{2.\left(x-3\right)}+\frac{x}{2.\left(x+1\right)}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\frac{x^2+x+x^2-3x-4x}{2.\left(x+1\right).\left(x-3\right)}=0\)

\(\Leftrightarrow2x^2-6=0\)

\(\Leftrightarrow2x^2=6\)

\(\Leftrightarrow x^2=3\)

\(\Leftrightarrow x=\sqrt{3}\)

\(b.2x^3-5x^2+3x=0\)

\(\Leftrightarrow x.\left(2x^2-5x+3\right)=0\)

\(\Leftrightarrow x.\left(2x^2-2x-3x+3\right)=0\)

\(\Leftrightarrow x.\left[2x.\left(x-1\right)-3.\left(x-1\right)\right]=0\)

\(\Leftrightarrow x.\left(x-1\right).\left(2x-3\right)=0\)

Đến đây tự làm nhé có việc bận

câu a sai dzoii

Bài 6

\(\left(a-b\right)^2=a^2-2ab+b^2\)

\(=\left(a^2+2ab+b^2\right)-4ab\)

\(=\left(a+b\right)^2-4ab\)

Bài 5 :

\(a,16x^2-\left(4x-5\right)^2=15\)

\(16x^2-16x^2+40x-25-15=0\)

\(40x-40=0\)

\(40x=40\)

\(x=1\)

\(b,\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

\(4x^2+12x+9-4x^2+4=49\)

\(12x=36\)

\(x=3\)

\(c,\left(2x+1\right)\left(2x-1\right)+\left(1-2x\right)^2=18\)

\(4x^2-1+1-4x+4x^2=18\)

\(8x^2-4x-18=0\)

\(2\left(4x^2-2x-9\right)=0\)

\(x=\frac{1-\sqrt{37}}{4}\)

\(d,2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)

\(2x^2+4x+2-x^2+9-x^2+8x-16=0\)

\(12x=4\)

\(x=\frac{1}{3}\)

1.

a. $A=\frac{x^3-x+2}{x-2}=\frac{x^2(x-2)+2x(x-2)+4(x-2)+10}{x-2}$

$=x^2+2x+4+\frac{10}{x-2}$

Với $x$ nguyên, để $A$ nguyên thì $\frac{10}{x-2}$ là số nguyên. 

Khi $x$ nguyên, điều này xảy ra khi $10\vdots x-2$

$\Rightarrow x-2\in \left\{\pm 1; \pm 2; \pm 5; \pm 10\right\}$

$\Rightarrow x\in \left\{3; 1; 4; 0; 7; -3; 12; -8\right\}$

b.

\(B=\frac{2x^2+5x+8}{2x+1}=\frac{x(2x+1)+3x+8}{2x+1}=x+\frac{3x+8}{2x+1}\)

Với $x$ nguyên, để $B$ nguyên thì $3x+8\vdots 2x+1$

$\Rightarrow 2(3x+8)\vdots 2x+1$
$\Rightarrow 3(2x+1)+13\vdots 2x+1$

$\Rightarrow 13\vdots 2x+1$
$\Rightarrow 2x+1\in \left\{\pm 1; \pm 13\right\}$

$\Rightarrow x\in \left\{0; -1; 6; -7\right\}$

Bài 2:

$P=\frac{8x^3-12x^2+6x-1}{4x^2-4x+1}=\frac{(2x-1)^3}{(2x-1)^2}=2x-1$
Với $x$ nguyên thì $2x-1$ cũng là số nguyên.

$\Rightarrow P$ nguyên với mọi $x$ nguyên.