\(A=\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=\dfrac{3x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{3x}{2\left(x-3\right)}\)

Thay x=1 vào A ta được\(A=\dfrac{3x}{2\left(x-3\right)}=\dfrac{3.1}{2\left(1-3\right)}=\dfrac{3}{2.\left(-2\right)}=\dfrac{-3}{4}\)

Thay x=4 vào A ta được\(A=\dfrac{3x}{2\left(x-3\right)}=\dfrac{3.4}{2\left(4-3\right)}=\dfrac{12}{2.1}=\dfrac{12}{2}=6\)

\(A=\dfrac{3x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}=\dfrac{3x}{2\left(x-3\right)}\\ x=1\Leftrightarrow A=\dfrac{3}{2\left(-2\right)}=-\dfrac{3}{4}\\ x=4\Leftrightarrow A=\dfrac{12}{2}=6\)

\(\left[{}\begin{matrix}A=\dfrac{3\cdot1^2+3\cdot1}{\left(1+1\right)\left(2\cdot1-6\right)}=-\dfrac{3}{4}\\A=\dfrac{3\cdot4^2+3\cdot4}{\left(4+1\right)\left(2\cdot4-6\right)}=6\end{matrix}\right.\)

a: Bạn ghi lại đề nha bạn

b: \(30\left(x+2\right)-6\left(x-5\right)-24x=100\)

=>\(30x+60-6x+30-24x=100\)

=>\(\left(30x-6x-24x\right)+\left(60+30\right)=100\)

=>0x=100-90=10(vô lý)

c: \(\left(x-7\right)\left(x+3\right)< 0\)

TH1: \(\left\{{}\begin{matrix}x-7>0\\x+3< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>7\\x< -3\end{matrix}\right.\)

=>\(x\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}x-7< 0\\x+3>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\)

=>-3<x<7

mà x nguyên

nên \(x\in\left\{-2;-1;0;1;2;3;4;5;6\right\}\)

d: -1<2x-1<4

=>\(-1+1< 2x< 4+1\)

=>0<2x<5

=>0<x<2,5

mà x nguyên

nên \(x\in\left\{1;2\right\}\)

\(\dfrac{1}{2}x^3+4\)

\(=\dfrac{x^2+8}{2}\)

\(=\text{(x+2)(x^2 −2x+4)}\)

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(2x+1\right)}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2x}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2x+1}=\dfrac{9}{20}\)

\(\Leftrightarrow2x+1=\dfrac{20}{9}\Leftrightarrow x=\dfrac{11}{18}\)

Em giải như XYZ olm em nhé

Sau đó em thêm vào lập luận sau:

\(x\) = \(\dfrac{11}{18}\)

Vì \(\in\) N* 

Vậy \(x\in\) \(\varnothing\)

Cảm ơn mn ạ.

a) ( x-2) ( y+1) =7

=> x-2 \(\in\)Ư(7)= { 1,7}

Nếu x-2 = 1 => x= 1+2 => x= 3

Nếu x-2= 7 => x= 7+2 => x= 9

Nếu x= 3 thì ( x-2) ( y+1) = ( 3-2)(y+1)=7

=>  y+1 =7 => y= 7-1 => y = 6

Nếu x = 9 thì ( x- 2 )( y+1)= 7 => ( 9-2) ( y+1) =7

=> 7( y+1) =7 => y+1= 7:7 => y+1 = 1 => y= 1-1 => y=0

Vậy...

Trình bày có chỗ nào sao mong mn sửa hộ nhaaa

b) ( 2x-1)( x+3) =6

=> ( 2x-1) \(\in\)Ư( 6) = { 1: 2: 3: 6}

Mà 2x-1 là số lẻ nên 2x-1 \(\in\){ 1; 3}

Nếu 2x-1 = 1 thì 2x= 1+1 => 2x= 2 => x= 2:2 => x= 1

Nếu 2x-1 = 3 thì 2x= 1+3 => 2x=4 => x= 4:2 => x= 2

Phần còn lại làm như phần a nha :33333

a: \(\left|3x-2\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)

b: Ta có: \(\left|5x-3\right|=\left|x-7\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-3=x-7\\5x-3=7-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-4\\6x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{3}\end{matrix}\right.\)

Theo đề ta có : x + 1 chia hết cho 2, 4, 5  và x là số nhỏ nhất hay x + 1 thuộc BCNN(2, 4, 5)

Ta có: 2 = 2 ; 4 = 2²  ; 5 = 5

=> BCNN(2, 4, 5) = 2² . 5 = 20

=> x + 1 = 20 => x = 20 - 1= 19

Vậy x = 19 

x chia 2 dư 1; x chia 4 dư 3; x chia 5 dư 4

\(\Rightarrow x+1\in BC\left(2,4,5\right)=B\left(20\right)=\left\{20;40;...\right\}\)

Mà \(x\) nhỏ nhất nên \(x-1\) nhỏ nhất

\(\Rightarrow x+1=20\Rightarrow x=19\)

Bài 1: 

\(=\dfrac{-3-39}{32}+\dfrac{-6-11}{17}+\dfrac{-1}{6}=-\dfrac{21}{16}+\dfrac{-1}{6}-1=-\dfrac{119}{48}\)

Bài 2: 

\(\Leftrightarrow x:5=-\dfrac{13}{20}\)

hay x=-13/4

⇔x:5=−1320⇔x:5=−1320

hay x=-13/4