Ta có : H(x)+Q(x)=P(x)H(x)+Q(x)=P(x)

<=>H(x)=P(x)−Q(x)<=>H(x)=P(x)−Q(x)

<=>H(x)=(4x3−32x2−x+10)−(10−12x−2x2+4x3)<=>H(x)=(4x3−32x2−x+10)−(10−12x−2x2+4x3)

<=>H(x)=(4x3−4x3)+(−32x2+2x2)+(−x+12x)+(10−10)<=>H(x)=(4x3−4x3)+(−32x2+2x2)+(−x+12x)+(10−10)

<=>H(x)=12x2−12x=(12x)(x−1)

HT

1.a,Q=x+32x+1−x−72x+1=x+32x+1+7−x2x+11.a,Q=x+32x+1−x−72x+1=x+32x+1+7−x2x+1

            =x+3+7−x2x+1=102x+1=x+3+7−x2x+1=102x+1

b,b, Vì x∈Z⇒(2x+1)∈Zx∈ℤ⇒(2x+1)∈ℤ

Q nhận giá trị nguyên ⇔102x+1⇔102x+1 nhận giá trị nguyên

                                ⇔10⋮2x+1⇔10⋮2x+1

                                ⇔2x+1∈Ư(10)={±1;±2;±5;±10}⇔2x+1∈Ư(10)={±1;±2;±5;±10}

Mà (2x+1):2(2x+1):2 dư 1 nên 2x+1=±1;±52x+1=±1;±5

⇒x=−1;0;−3;2⇒x=−1;0;−3;2

Vậy.......................

HT

\(A=\left(\frac{x^2-16}{x-4}+1\right):\left(\frac{x-2}{x-3}+\frac{x+3}{x+1}+\frac{x+2-x^2}{x^2-2x-3}\right)\)

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

\(=\left(x+5\right):\left(\frac{x^2+x-2x-2+x^2-9+x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)

\(=\left(x+5\right):\left(\frac{x^2-9}{\left(x-3\right)\left(x+1\right)}\right)\)

\(=\left(x+5\right):\left(\frac{x+3}{x+1}\right)=\frac{x+3}{\left(x+5\right)\left(x+1\right)}\)

Sai đề ở chỗ \(\left(\frac{x^2-16}{x-4}+1\right)\)thành -1

Cậu chỉ cần sửa x+5 thành x+3 là được

\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{8}{x^2-1}\right):\left(\frac{1}{x-1}-\frac{7x+3}{1-x^2}\right)\)

\(A=\left[\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x+1\right)\left(x-1\right)}+\frac{8}{\left(x+1\right)\left(x-1\right)}\right]:\left[\frac{x+1}{\left(x+1\right)\left(x-1\right)}-\frac{3-7x}{\left(x+1\right)\left(x-1\right)}\right]\)

\(A=\left[\frac{x^2+2x+1-x^2+2x-1+8}{\left(x+1\right)\left(x-1\right)}\right]:\frac{x+1-3+7x}{\left(x+1\right)\left(x-1\right)}\)

\(A=\frac{4x+8}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{8x-2}\)

...................... 

tìm giá trị x nguyên để A nguyên đi

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne-2\end{cases}}\)

\(D=\left(\frac{x}{x+2}+\frac{8x+8}{x^2+2x}-\frac{x+2}{x}\right):\left(\frac{x^2-x+3}{x^2+2x}+\frac{1}{x}\right)\)

\(\Leftrightarrow D=\left(\frac{x}{x+2}+\frac{8x+8}{x\left(x+2\right)}-\frac{x+2}{x}\right):\frac{x^2-x+3+x+2}{x\left(x+2\right)}\)

\(\Leftrightarrow D=\frac{x^2+8x+8-\left(x+2\right)^2}{x\left(x+2\right)}:\frac{x^2+5}{x\left(x+2\right)}\)

\(\Leftrightarrow D=\frac{\left(x^2+8x+8-x^2-4x-4\right)x\left(x+2\right)}{x\left(x+2\right)\left(x^2+5\right)}\)

\(\Leftrightarrow D=\frac{4x+4}{x^2+5}\)

Để \(D\inℤ\)

\(\Leftrightarrow4x+4⋮x^2+5\)

\(\Leftrightarrow4x^2+4x⋮x^2+5\)

\(\Leftrightarrow4\left(x^2+5\right)-16x⋮x^2+5\)

\(\Leftrightarrow16x⋮x^2+5\)

\(\Leftrightarrow256\left(x^2+5\right)-1280⋮x^2+5\)

\(\Leftrightarrow1280⋮x^2+5\)

\(\Leftrightarrow x^2+5\inƯ\left(1280\right)\)

Đoạn này bạn làm nốt nhé

bài mik sai từ đoạn \(4x^2+4x⋮x^2+5\)

k tương đương đc với \(4\left(x^2+5\right)-16x⋮x^2+5\)nhaaa !! 

MIk rút gọn đc D thôi :)) Phần còn lại chắc cậu tự làm nha

Kết quả rút gọn của bạn Minh đúng rồi nhé, mình làm tiếp nha !

Để D là số nguyên

\(\Leftrightarrow4x+4⋮x^2+5\)

\(\Rightarrow\left(4x+4\right)\left(4x-4\right)⋮x^2+5\)

\(\Leftrightarrow16x^2-16⋮x^2+5\)

\(\Leftrightarrow16\left(x^2+5\right)-96⋮x^2+5\)

\(\Leftrightarrow96⋮x^2+5\)

\(\Leftrightarrow x^2+5\inƯ\left(96\right)\)

\(\Leftrightarrow x^2+5\in\left\{\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12,\pm16,\pm24,\pm32,\pm48,\pm96\right\}\)

Lại có : \(x^2+5\ge5>0\)

Do đó \(x^2+5\in\left\{6,8,12,16,24,32,48,96\right\}\)

\(\Leftrightarrow x^2\in\left\{1,3,7,11,19,27,43,91\right\}\)

Mà \(x^2\) là số chính phương và x là số nguyên

\(\Rightarrow x^2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\) ( thỏa mãn ĐKXĐ )

Thử lại ta thấy \(x=-1\) thỏa mãn D là số nguyên.

Vậy : \(x=-1\) để D nhận giá trị nguyên.

\(A=\left(\frac{x^2-16}{x-4}-1\right):\left(\frac{x-2}{x-3}+\frac{x+3}{x+1}+\frac{x+2-x^2}{x^2-2x-3}\right)\)ĐK : \(x\ne3;-1;4\)

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

\(=\left(x-3\right):\left(\frac{x^2-x-2+x^2-9+x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)=\left(x-3\right):\left(\frac{x^2-9}{\left(x-3\right)\left(x-1\right)}\right)\)thơm thế :))

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

1) đk: \(x\ne\left\{-1;3;4\right\}\)

Ta có:

\(A=\left(\frac{x^2-16}{x-4}-1\right)\div\left(\frac{x-2}{x-3}+\frac{x+3}{x+1}+\frac{x+2-x^2}{x^2-2x-3}\right)\)

\(A=\left[\frac{\left(x-4\right)\left(x+4\right)}{x-4}-1\right]\div\frac{\left(x-2\right)\left(x+1\right)+\left(x+3\right)\left(x-3\right)+x+2-x^2}{\left(x+1\right)\left(x-3\right)}\)

\(A=\left(x+4-1\right)\div\frac{x^2-x-2+x^2-9-x^2+x+2}{\left(x+1\right)\left(x-3\right)}\)

\(A=\left(x+3\right)\div\frac{x^2-9}{\left(x+1\right)\left(x-3\right)}\)

\(A=\left(x+3\right)\cdot\frac{\left(x+1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(A=x+1\)

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

Để \(\frac{A}{x^2+x+1}\) nguyên thì \(\left(x+1\right)⋮\left(x^2+x+1\right)\Leftrightarrow\left(x+1\right)^2⋮\left(x^2+x+1\right)\)

\(\Rightarrow\left(x+1\right)^2-\left(x^2+x+1\right)⋮\left(x^2+x+1\right)\)

\(\Rightarrow x⋮\left(x^2+x+1\right)\Rightarrow1⋮x^2+x+1\)

\(\Rightarrow x^2+x+1\in\left\{-1;1\right\}\Rightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Rightarrow\orbr{\begin{cases}x=-1\left(ktm\right)\\x=0\left(tm\right)\end{cases}}\)

Vậy x = 0

ĐKXĐ : \(\hept{\begin{cases}x\ne4\\x\ne3\\x\ne-1\end{cases}}\)

\(A=\left(\frac{x^2-16}{x-4}-\frac{x-4}{x-4}\right)\div\left(\frac{\left(x-2\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}+\frac{x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)

\(=\left(\frac{x^2-x-12}{x-4}\right)\div\left(\frac{x^2-x-2+x^2-9+x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)

\(=\frac{\left(x-4\right)\left(x+3\right)}{x-4}\div\frac{x^2-9}{\left(x-3\right)\left(x+1\right)}\)

\(=\frac{\left(x-4\right)\left(x+3\right)}{x-4}\times\frac{\left(x-3\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}=x+1\)