th1\(\hept{\begin{cases}5x-\frac{1}{2}>0\\1,25-3x>0\end{cases}}=>\hept{\begin{cases}x>\frac{1}{10}\\x< \frac{5}{12}\end{cases}}\)=>1/10<x<5/12

còn th2 vô lí

cảm ơn bạn nha!!!!

ok, nhưng đầy môn lắm

me too :>

\(\frac{x+\frac{3}{2}}{x-\frac{2}{3}}\)VÌ \(x-\frac{2}{3}< x+\frac{3}{2}\)=> \(x-\frac{2}{3}< 0;x+\frac{3}{2}>0\)

      => \(\frac{-3}{2}< x< \frac{2}{3}\)=> \(x=\left\{-\frac{8}{6};-\frac{7}{6};....;\frac{3}{6}\right\}\)

                      HỌC TỐT NHA

cảm ơn bạn nha!!!!!!!!!!!!1

Đặt \(A=4+4^2+4^3+\cdots+4^{24}\)

\(A=\left(4+4^2+\ldots+4^6\right)+\left(4^7+4^8+\cdots+4^{12}\right)+\cdots+\left(4^{19}+4^{20}+\cdots+4^{24}\right)\)

\(A=\left(4+4^2+\ldots+4^6\right)+4^6\cdot\left(4+4^2+\ldots+4^6\right)+4^{12}\cdot\left(4+4^2+\ldots+4^6\right)+4^{18}\cdot\left(4+4^2+\ldots+4^6\right)\)

\(A=5460\cdot\left(1+4^6+4^{12}+4^{18}\right)\)

\(A=420\cdot13\cdot\left(1+4^6+4^{12}+4^{18}\right)\)

⇒ A ⋮ 420

Vậy A ⋮ 420

số hơi to một chút, nếu có gì sai sót thì cho mk xin lỗi! ^^

15/90 × 94 + 15/94 × 98 + 15/98 × 102 + ... + 15/146 × 150

= 15/4 × (4/90×94 + 4/94×98 + 4/98×102 + ... + 4/146×150)

= 15/4 × (1/90 - 1/94 + 1/94 - 1/98 + 1/98 - 1/102 + ... + 1/146 - 1/150)

= 15/4 × (1/90 - 1/150)

= 15/4 × 1/30 × (1/3 - 1/5)

= 1/8 × 2/15

= 1/60

Ta có : \(x^2+1>0\)

Vậy để \(\frac{x^2+1}{x-5}< 0\) thì \(x-5< 0\Rightarrow x< 5\)

a: \(5x\left(x-3\right)-x\left(5x+1\right)=16\)

=>\(5x^2-15x-5x^2-x=16\)

=>-16x=16

=>x=-1

b: \(4x\left(x-1\right)+x\left(3-4x\right)=5\)

=>\(4x^2-4x+3x-4x^2=5\)

=>-x=5

=>x=-5

c: \(5\left(x^2+4x-3\right)-x\left(5x+3\right)=19\)

=>\(5x^2+20x-15-5x^2-3x=19\)

=>17x=19+15=34

=>x=2