Sửa đề: \(\sqrt{5+\sqrt{x-1}}=6-x\)

ĐK: \(x\ge1\)   ( 1)

pt <=> \(\sqrt{5+\sqrt{x-1}}=6-x\)

<=> \(\hept{\begin{cases}6-x\ge0\\5+\sqrt{x-1}=\left(6-x\right)^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le6\left(2\right)\\\sqrt{x-1}=31-12x+x^2\left(3\right)\end{cases}}\)

(3) <=> \(4\sqrt{x-1}=124-48x+4x^2\)

<=> \(4x-4+2.2\sqrt{x-1}.1+1=4x^2-44x+121\)

<=> \(\left(2\sqrt{x-1}+1\right)^2=\left(2x-11\right)^2\)

Em tự làm tiếp nhé!

\(y=\frac{x^3-x^2+2x+7}{x^2+1}=x-1+\frac{x+8}{x^2+1}\)

Đặt 

\(A=\frac{x+8}{x^2+1}\)

\(\Leftrightarrow\left(x-8\right)A=\frac{x^2-64}{x^2+1}=1-\frac{65}{x^2+1}\)

Để A nguyên thì \(x^2+1\)phải là ước của 65. Làm nốt

\(b.=\frac{1\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{1\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{1\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\frac{1c-1a+1a-1b+1b-1c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=-\frac{2b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Sr nha

Kq mik nhầm

Ko phải -2b đâu mà = 0

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

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

\(\Leftrightarrow-4\left(x-2\right)=0\)

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

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

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

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

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)