\(45=3^2.5\\ 180=2^2.3^2.5\\ 25=5^2\)

\(BCNN\left(45;180;25\right)=3^2.5^2.2^2=900\)

\(\dfrac{16^3-64^2}{8^3}=\dfrac{2^{12}-2^{12}}{2^9}=\dfrac{0}{2^9}=0\)

gọi số cần tìm là x(x thuộc Z)

theo đề bài, ta có:

\(x+3276=6737-1234\\ \Leftrightarrow x=5503-3276=2227\)

\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\\ \Leftrightarrow\dfrac{x+1}{65}+\dfrac{x+3}{63}+2=\dfrac{x+5}{61}+\dfrac{x+7}{59}+2\\ \Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}=\dfrac{x+66}{61}+\dfrac{x+66}{59}\\ \Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\\ \Leftrightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\\ \Rightarrow x+66=0\Rightarrow x=-66\)

Là 324m2  bạn ạ!!!

Mjk làm r

\(\dfrac{3}{11}+\dfrac{a}{22}=\dfrac{b}{11}\Rightarrow\dfrac{6}{22}+\dfrac{a}{22}=\dfrac{2b}{22}\\ \Rightarrow6+a=2b\)

a và b nguyên khi a là số chẵn.

Gọi thương của phép chia đa thức \(f\left(x\right)\)cho \(x-1\)và cho \(x+2\), theo thứ tự là \(A\left(x\right),B\left(x\right)\)và dư theo thứ tự là  \(4\) và  \(1\)

Ta có:

\(f\left(x\right)=\left(x-1\right).A\left(x\right)+4\)

nên \(\left(x+2\right)f\left(x\right)=\left(x-1\right)\left(x+2\right).A\left(x\right)+4\left(x+2\right)\) \(\left(1\right)\)

\(f\left(x\right)=\left(x+2\right).B\left(x\right)+1\) 

nên \(\left(x-1\right)f\left(x\right)=\left(x+2\right)\left(x-1\right).B\left(x\right)+1\left(x-1\right)\) \(\left(2\right)\)

Lấy \(\left(1\right)\)trừ \(\left(2\right)\) vế theo vế, ta có:

\(\left[\left(x+2\right)-\left(x-1\right)\right]f\left(x\right)=\left(x-1\right)\left(x+2\right)\left[A\left(x\right)-B\left(x\right)+4\left(x+2\right)-1\left(x-1\right)\right]\)

\(\Leftrightarrow3f\left(x\right)=\left(x-1\right)\left(x+2\right)\left[A\left(x\right)-B\left(x\right)\right]+3x+9\)

Do đó: \(f\left(x\right)=\left(x-1\right)\left(x+2\right)\frac{A\left(x\right)-B\left(x\right)}{3}+\left(x+3\right)\)

\(\Leftrightarrow f\left(x\right)=5x^2\left(x-1\right)\left(x+2\right)+\left(x+3\right)\)

trong đó, bậc của \(x+3\) nhỏ hơn bậc của \(\left(x-1\right)\left(x+2\right)\)

Vậy, dư của phép chia \(f\left(x\right)\) cho \(\left(x-1\right)\left(x+2\right)\)là  \(x+3\)

\(\dfrac{1}{\sqrt{a}+1}-\dfrac{-2\sqrt{a}}{a\sqrt{a}+\sqrt{a}+a+1}=\dfrac{1}{\sqrt{a}+1}-\dfrac{-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}\\ =\dfrac{a+1+2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(a+1\right)\left(\sqrt{a}+1\right)}\\ =\dfrac{\sqrt{a}+1}{a+1}\)

bn rút gọn lại dc bao nhiêu
mình lười rút gọn quá