a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)

b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)

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

Câu 3 : 

\(a,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\)  ĐKXđ : \(x\ne\pm1\)

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

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

\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{5\left(x-1\right)}{2x}\)

\(A=\frac{10}{x+1}\)

\(B=\left(\frac{x}{3x-9}+\frac{2x-3}{3x-x^2}\right).\frac{3x^2-9x}{x^2-6x+9}.\)

ĐKXđ : \(x\ne0;x\ne3\)

\(B=\left(\frac{x}{3\left(x-3\right)}+\frac{2x-3}{x\left(3-x\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)

\(B=\left(\frac{x^2}{3x\left(x-3\right)}+\frac{9-6x}{3x\left(x-3\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)

\(B=\frac{x^2-6x+9}{3x\left(x-3\right)}.\frac{3x\left(x-3\right)}{x^2-6x+9}=1\)

câu a, phân tích từng mẫu thành nhân tử (nếu cần)

rồi tìm mtc, ở đây, nhân chia cũng như cộng trừ, nên phân tích hết rồi ra mtc, đkxđ là cái mtc ấy khác 0

câu b với c tự làm

câu d thì lấy cái rút gọn rồi của câu b, rồi giải ra, để nguyên thì mẫu là ước của tử, thế thôi

bài1   A=\(\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

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

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

b)  thế \(x=-\frac{1}{2}\)vào biểu thức A

 \(-\frac{1}{3\cdot\left(-\frac{1}{2}\right)}=\frac{2}{3}\)

c)  A=\(-\frac{1}{3x}< 0\)

VÌ (-1) <0  nên  3x>0

                        x >0

a

\(ĐKXĐ:x\ne3;x\ne-3;x\ne0\)

b

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

\(=\left[\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right]:\left[\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right]\)

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

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

c

Với \(x=4\Rightarrow A=-3\)

d

Để A nguyên thì \(\frac{3}{x-3}\) nguyên

\(\Rightarrow3⋮x-3\)

 Làm nốt.

toi moi lop 5

a, ĐKXĐ: \(x\ne-3\) và \(x\ne\pm1\)

b, \(P=\frac{x\left(x+3\right)-11+x^2-3x+9}{x^3+27}:\frac{x^2-1}{x+3}\)

\(P=\frac{2x^2-2}{x^3+27}.\frac{x+3}{x^2-1}\)

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

\(=\frac{2}{x^2-3x+9}\)

c, \(P=\frac{2}{x^2-3x+9}==\frac{2}{\left(x-\frac{3}{2}\right)^2+\frac{27}{4}}\le\frac{2}{\frac{27}{4}}=\frac{8}{27}\)

Dấu "=" xảy ra khi: \(x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)

Vậy P lớn nhất bằng \(\frac{8}{27}\) \(\Leftrightarrow x=\frac{3}{2}\)

\(P=\left(\frac{x}{x^2-3x+9}-\frac{11}{x^3+27}+\frac{1}{x+3}\right):\frac{x^2-1}{x+3}.\)

ĐKXĐ : \(x\ne-3;x\ne0\)

\(P=\left(\frac{x\left(x+3\right)}{\left(x+3\right)\left(x^2-3x+9\right)}-\frac{11}{\left(x+3\right)\left(x^2-3x+9\right)}+\frac{x^2-3x+9}{\left(x+3\right)\left(x^2-3x+9\right)}\right).\frac{x+3}{x^2-1}\)

\(P=\left(\frac{x^2+3x-11+x^2-3x+9}{\left(x+3\right)\left(x^2-3x+9\right)}\right).\frac{x+3}{x^2-1}\)

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

\(P=\frac{2}{x^2-3x+9}\)

Nếu có câu "d" tìm giá trị nguyên của x để P nguyên thì sao ?

Ai giải được giải giúp với !!!

\(a,x\ne2;x\ne-2;x\ne0\)

\(b,A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\frac{6}{x+2}\)

\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)

\(=\frac{1}{2-x}\)

\(c,\)Để A > 0 thi \(\frac{1}{2-x}>0\Leftrightarrow2-x>0\Leftrightarrow x< 2\)