`9(x-y)^2-4(x+y)^2`

`=[3(x-y)]^2-[2(x+y)]^2`

`=(3x-3y)^2-(2x+2y)^2`

`=(3x-3y+2x+2y)(3x-3y-2x-2y)`

`=(5x-y)(x-5y)`

`cosy+siny. tgy=cosy+ siny. (siny)/(cosy)=(cos^2y+sin^2y)/(cosy)=1/(cosy)`

`sin2x-cos3x=0`

`<=>sin2x=cos3x `

`<=>cos(π/2-2x)=cos3x`

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\text{π}}{2}-2x=3x+k2\text{π}\\\dfrac{\text{π}}{2}-2x=-3x+k2\text{π}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{10}+\dfrac{k2\pi}{5}\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

`sinx+sin3x=0`

`<=>sinx=sin(-3x)`

`<=>[(x=-3x+k2π),(x=π+3x+k2π):}`

`<=>[(x=k π/2),(x=-π/2+kπ):}`

There isn't quiet enough for us to learn our lessons.

Tách 5 ra để có HĐT ạ.

\(4+\dfrac{1}{4}+\dfrac{3}{4}=5\)

`A=x^2-4x+y^2-y+5`

`=(x^2-4x+4)+(y^2-y+1/4)+3/4`

`=(x-2)^2+(y-1/2)^2+3/4`

`=>A_(min) = 3/4 <=> {(x=2),(y=-1/2):}`

Gọi `x,y,z` là số bi của 3 bạn Thanh, Hiếu, Nam. (`x,y,z \in NN^(**)`)

Vì `x,y,z` tỉ lệ với `2,3,4 => x:y:z=2:3:4 => x/2=y/3=z/4`

Áp dụng tính chất của dãy tỉ số bằng nhau:

`x/2=y/3=z/4=(x+y+z)/(2+3+4)=36/9=4`

`=>x=4.2=8`

`y=4.3=12`

`z=4.4=16`

Vậy...

`f'(x) = x^2 - 4x+m`

`f'(x) >=0 <=>x^2-4x+m>=0`

`<=> \Delta' >=0`

`<=> 2^2-1.m>=0`

`<=> m<=4`

Vậy....

`\lim (\sqrt(4n+3) -\sqrt(n+1))`

`=\lim \sqrtn (\sqrt(4+3/n)-\sqrt(1+1/n))`
`=+oo`
Vì `{(\limn=+oo),(\lim(\sqrt(4+3/n)-\sqrt(1+1/n))=1>0):}`

`(-8) .72 + (-8).19 - (-8)`

`=(-8) .72 + (-8).19 - (-8). 1`

`=(-8) . (72+19-1)`