\(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}\)

a) ĐKXĐ : \(\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)

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

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

\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x^2-9}{\left(x-2\right)\left(x-3\right)}+\frac{2x^2-8}{\left(x-2\right)\left(x-3\right)}\)

\(=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}\)

\(=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\frac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}=\frac{x+4}{x-3}\)

b) Ta có : \(A=\frac{x+4}{x-3}=\frac{x-3+7}{x-3}=1+\frac{7}{x-3}\)

Để A đạt giá trị nguyên thì \(\frac{7}{x-3}\)đạt giá trị nguyên

=> 7 ⋮ x - 3

=> x - 3 ∈ Ư(7) = { ±1 ; ±7 }

So với ĐKXĐ ta thấy x = 4 , x = 10 , x = -4 thỏa mãn 

Vậy với x ∈ { ±4 ; 10 } thì A đạt giá trị nguyên

(....) dùng để nhìn được chữ số ở phân số cuối cùng thôi, ko dùng để làm gì.

( ác ) là từ ( các ) 

(gia strij) là từ ( giá trị )

Ta có:\(\text{​​}\text{​​}\frac{x+1}{58}+\frac{x+2}{57}+\frac{x+3}{56}+\frac{x+69}{5}=-1\)

\(\Leftrightarrow(\frac{x+1}{58}+1)+\left(\frac{x+2}{57}+1\right)+\left(\frac{x+3}{56}+1\right)+\left(\frac{x+69}{5}-2\right)=0\)

\(\Leftrightarrow\frac{x+59}{58}+\frac{x+59}{57}+\frac{x+59}{56}+\frac{x+59}{5}=0\)

\(\Leftrightarrow\left(x+59\right)\left(\frac{1}{59}+\frac{1}{57}+\frac{1}{56}+\frac{1}{5}\right)=0\)

\(\Leftrightarrow x+59=0\left(do\frac{1}{58}+\frac{1}{57}+\frac{1}{56}+\frac{1}{5}\ne0\right)\)

\(\Leftrightarrow x=-59\)

\(x^{41}\div x^2+1\)

Ta có:\(x^{41}=x^{41}-x+x=x\left(x^{40}-1\right)+x\)

Vì \(x^{40}-1=\left(x^4\right)^{10}-1^{10}⋮x^4-1\)

Mà \(x^4-1=\left(x^2-1\right)\left(x^2+1\right)⋮x^2+1\)

Nên \(x^{41}\)chia \(x^2-1\)dư \(x\)

ai giúp mk với!!! :((

có thêm thông tin gì nx ko bn ?

Kẻ đường cao \(AH\)

Xét \(\Delta AHD\)vuông tại \(H\), có: \(\widehat{D}=30^o\)

\(\Rightarrow DH=\frac{AD}{2}=\frac{8}{2}=4\left(cm\right)\)(Áp dụng tính chất trong tam giác vuông , cạnh đối diện với góc 30 độ thì bằng 1 nửa cạnh huyền)

Áp dụng định lý Py-ta-go cho \(\Delta AHD\)vuông tại \(H\), ta có:

\(AH^2=AD^2-DH^2=8^2-4^2=64-16=48\)

\(\Rightarrow AH=\sqrt{48}=4\sqrt{3}\left(cm\right)\)

Ta có : \(S_{ABCD}=\frac{\left(AB+CD\right)AH}{2}=\frac{\left(7+9\right).4\sqrt{3}}{2}=32\sqrt{3}\left(cm^2\right)\)

Vậy \(S_{ABCD}=32\sqrt{3}cm^2\)

Sửa đề : \(B=\frac{x+3}{x+1}-\frac{2x-1}{1-x}+\frac{x+7}{x^2-1}\)

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

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

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

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

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

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

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

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

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

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