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\(ĐKXĐ:x\ne1\)

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

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

\(\Leftrightarrow A=\frac{2\left(x^2-1\right)+3x^2-8x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\Leftrightarrow A=\frac{2x^2-2+3x^2-8x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\Leftrightarrow A=\frac{5x^2-8x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)

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

\(\Leftrightarrow A=\frac{5x-3}{x^2+x+1}\)

b) Để \(A=1\)

\(\Leftrightarrow5x-3=x^2+x+1\)

\(\Leftrightarrow x^2-4x+4=0\)

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

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy để \(A=1\Leftrightarrow x=2\)

Bạn có thể đăng từng bài k như thế nhìn đã sợ ai làm

1)đặt nhân tử chung quy đồng là xong

2)phân tích x+2cănx-3=(1-cănx)(3+cănx)

3)2a+căn a đặt căn a ra r rút gọn

Câu 1:

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

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

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

Câu 2: thay x vào A có :

\(A=\frac{-\frac{1}{2}}{\frac{1}{4}-1}=\frac{2}{3}\)

Câu c :

2A=1 => \(\frac{x}{x^4-1}=\frac{1}{2}\)ĐK \(\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)

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

\(\left(x+1\right)\left(x^2+1\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)loại do điều kiện  vậy ko có giá trị nào của x thỏa mãn

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 : \(x\ne0;x\ne3\)

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

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

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