\(x\left(x-\frac{1}{3}\right)< 0\)

Để \(x\left(x-\frac{1}{3}\right)< 0\)thì x và \(x-\frac{1}{3}\)trái dấu nhau

Thấy \(x>x-\frac{1}{3}\)\(\Rightarrow\hept{\begin{cases}x>0\\x-\frac{1}{3}< 0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x< \frac{1}{3}\end{cases}\Leftrightarrow}0< x< \frac{1}{3}}\)

x^2+1>=1

=>(x^2+1)^2>=1

y^2+2>=2

=>(y^2+2)^4>=16

=>(x^2+1)^2+(y^2+2)^4>=17

=>(x^2+1)^2+(y^2+2)^4-2>=15

Dấu = xảy ra khi x=y=0

Bài 4: 

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

nên \(\dfrac{a}{c}=\dfrac{b}{d}\)

\(\Leftrightarrow\dfrac{c}{a}=\dfrac{d}{b}\)

hay \(\dfrac{a+c}{a}=\dfrac{b+d}{b}\)

15×2121/4343+15×222222/434343
= 15 x (21x101)/(101x43) + 15x(22x 10101)/(43x10101)
= 15 x 21/43 + 15 x 22/43
= 15x(21/43 + 22/43)
= 15x1 = 15

=15 nha

Đáp án 15

mình làm cho câu dưới nha

\(x+y+xy+2=x^2+y^2\)

\(=>x^2+y^2-x-y-xy=2\)

=>\(2x^2+2y^2-2x-2y-2xy=4\\\)( chỗ này nhân mõi zế zs 2 á)

=>\(\left(x-y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2=4\)

ta lại có\(4=0+1+3=tgtựra\)

mình nghĩ thế

<=> x² -4+3x²= 4x²+4x+1+2x

<=> 4x^2 - 4= 4x^2 +6x +1

<=> - 4=6x +1

<=> 6x= -5

<=> x= \(-\frac{5}{6}\)

\(A=1\cdot4+2\cdot5+3\cdot6+...+99\cdot102\)

   \(=1\cdot\left(2+2\right)+2\cdot\left(2+3\right)+3\cdot\left(2+4\right)+...+99\cdot\left(2+100\right)\)

   \(=\left(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\right)+\left(2+4+6+...+198\right)\)

  Ta thấy : \(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)nhân với 3 được :

    \(1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\)

\(=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+99\cdot100\cdot101-98\cdot99\cdot100\)

\(=99\cdot100\cdot101\)

\(=999900\)

\(\Rightarrow1\cdot2+2\cdot3+3\cdot4+...+99\cdot100=999900:3=333300\)

\(2+4+6+...+198=\left(198-2\right):2+1=99\)( số hạng )

Tổng của \(2+4+6+...+198\)bằng : \(\left(198+2\right)\cdot99:2=9900\)

\(\Rightarrow A=333300+9900=343200\)

Vậy \(A=343200\)