OK E

KẾT BẠN ĐI

1+3 =4

1+3+5=9

1+3+5+4 

Ta có :

4 = 2²

6 = 2 . 3

=> ƯCLN ( 4 ; 6 ) = 2 

Vậy ƯCNN ( 4 ; 6 ) = 2

đáp án

ucln của 4 và 6 là

2

Bài làm:

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\(1,\left(\frac{\left(x+1\right)^2.\left(y+1\right)^2}{\left(x+1\right)^2}+\frac{\left(x+1\right)^2\left(y+1\right)^2}{\left(y+1\right)^2}\right)\left(xy+1\right)\ge\left(x+1\right)^2\left(y+1\right)^2\)

\(\left[\left(y+1\right)^2+\left(x+1\right)^2\right]\left(xy+1\right)\ge\left(xy+y+x+1\right)^2\)

\(\left(y^2+2y+1+x^2+2x+1\right)\left(xy+1\right)\ge\left(xy+y+x+1\right)^2\)

\(\left(y^2+2y+1+x^2+2x+1\right)\left(xy+1\right)-\left(xy+y+x+1\right)^2\ge0\)

\(\left(y^2+2y+1+x^2+2x+1\right)\left(xy+1\right)-\left(x^2+2x+1\right)\left(y^2+2y+1\right)\ge0\)

\(xy\left(x-1\right)^2+\left(xy-1\right)^2\ge0\)

\(< =>BĐT\)luôn đúng

dấu "=" xảy ra khi \(x=y=1\)

mình ko chắc đã đúngg đâu

ĐK: \(x^2-3y^2+30\ge0\).

Phương trình thứ nhất tương đương với: 

\(\left(x-y+3\right)\left(x+2y-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=y-3\\x=1-2y\end{cases}}\)

Với \(x=y-3\)thế vào phương trình thứ hai ta được: 

\(\sqrt{\left(y-3\right)^2-3y^2+30}+y-3-2y-5=0\)

\(\Leftrightarrow\sqrt{-2y^2-6y+39}=y+8\)

\(\Rightarrow-2y^2-6y+39=y^2+16y+64\)

\(\Leftrightarrow\orbr{\begin{cases}y=\frac{-11+\sqrt{46}}{3}\Rightarrow x=\frac{-20+\sqrt{46}}{3}\\y=\frac{-11-\sqrt{46}}{3}\Rightarrow x=\frac{-20-\sqrt{46}}{3}\end{cases}}\)

Thử lại thỏa mãn. 

Với \(x=1-2y\)làm tương tự, thu được thêm một nghiệm là: \(x=\frac{17-2\sqrt{61}}{5},y=\frac{-6+\sqrt{61}}{5}\).

Trả lời :

B. Sydeny là thành phố của nước Úc

ĐK: \(1-sinx\ne0\Leftrightarrow x\ne\frac{\pi}{2}+l2\pi\left(l\inℤ\right)\).

\(\frac{cos3x}{1-sinx}=0\)

\(\Rightarrow cos3x=0\)

\(\Leftrightarrow3x=\frac{\pi}{2}+k\pi\left(k\inℤ\right)\)

\(\Leftrightarrow x=\frac{\pi}{6}+\frac{k\pi}{3}\left(k\inℤ\right)\)

Đối chiếu điều kiện: 

\(\frac{\pi}{6}+\frac{k\pi}{3}=\frac{\pi}{2}+l\pi\Leftrightarrow\frac{1}{3}=\frac{k}{3}-l\Leftrightarrow k=3l+1\).

Vậy nghiệm phương trình là \(x=\frac{\pi}{6}+\frac{k\pi}{3}\)với \(k\inℤ,k\ne3l+1\left(l\inℤ\right)\).