Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-5\right).2=\left(x+2\right).3\)
\(\Rightarrow4x-10=3x+6\)
\(\Rightarrow x=16\)
\(\dfrac{2x+5}{3}=\dfrac{x+2}{2}\)
MTC : 6
Quy đồng mẫu thức :
\(\Rightarrow\) \(\dfrac{2\left(2x+5\right)}{6}=\)\(\dfrac{3\left(x+2\right)}{6}\)
Suy ra : 2(2x + 5) = 3(x + 2)
\(\Leftrightarrow\) 4x + 10 = 3x + 6
\(\Leftrightarrow\) 4x + 10 - 3x - 6 = 0
\(\Leftrightarrow\) x + 4 = 0
\(\Leftrightarrow\) x = - 4
Vậy S = \(\left\{-4\right\}\)
Chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
`(3x+2)/3 <= (x-4)/7`
`<=>7(3x+2) <= 3(x-4)`
`<=> 21x+14<=3x-12`
`<=>18x <= -26`
`<=> x <=-13/9`
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`(3x+2)/3 <= (x-4)/7`
`<=>7(3x+2) <= 3(x-4)`
`<=>21x+14 <= 3x-12`
`<=> 18x <=-26`
`<=>x <= -13/9`
Vậy `x<=-13/9`.
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi phân thức cần tìm là \(A\)
Ta có:
\(\dfrac{1}{x}.\dfrac{x}{x+1}.\dfrac{x+1}{x+2}.\dfrac{x+2}{x+3}.\dfrac{x+3}{x+4}.\dfrac{x+4}{x+5}.\dfrac{x+5}{x+6}.\dfrac{x+6}{x+7}.\dfrac{x+7}{x+8}.\dfrac{x+8}{x+9}.\dfrac{x+9}{x+10}\)
\(=\dfrac{x\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)\left(x+8\right)\left(x+9\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)\left(x+8\right)\left(x+9\right)\left(x+10\right)}\)\(=\dfrac{x}{x+10}\)
Suy ra:
\(\dfrac{1}{x}.\dfrac{x}{x+1}.\dfrac{x+1}{x+2}.\dfrac{x+2}{x+3}.\dfrac{x+3}{x+4}.\dfrac{x+4}{x+5}.\dfrac{x+5}{x+6}.\dfrac{x+6}{x+7}.\dfrac{x+7}{x+8}.\dfrac{x+8}{x+9}.\dfrac{x+9}{x+10}.A=1\)
\(\Leftrightarrow\dfrac{x}{x+10}.A=1\)
\(\Leftrightarrow A=\dfrac{x+10}{x}\)
Vậy phân thức cần điền vào chỗ trống là \(\dfrac{x+10}{x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow4\left(6-x\right)-3x=6\left(2x+3\right)-12\)
=>24-4x-3x=12x+18-12
=>12x+6=-7x+24
=>19x=18
=>x=18/19
b: \(\Leftrightarrow-210x-6\left(x-3\right)-15x=30x+10\left(2x+1\right)\)
=>-225x-6x+18=30x+20x+10
=>-231x+18-50x-10=0
=>-281x=-8
=>x=8/281
c: \(\Leftrightarrow36-2\left(x+3\right)=-4x+1-x\)
=>36-2x-6=-5x+1
=>3x=1+6-36=5-36=-31
=>x=-31/3
d: \(\Leftrightarrow-30\left(x-3\right)+10\left(2x-7\right)=6\left(6-x\right)\)
=>-30x+90+20x-70=36-6x
=>-10x+20=36-6x
=>-4x=16
=>x=-4
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{x^2-x+x+1+2x}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}\)
b: \(=\dfrac{x^2+2x-4x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)
c: \(=\dfrac{2x^2-3x-9-x^2+3x+x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x^2+6x}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x}{x-3}\)
ĐK: ` x\ne \pm 3`
`(x+1)/(x-3)+(x-1)/(x+3)=(x+6)/(x^2-9)`
`<=>(x+1)(x+3)+(x-1)(x-3)=x+6`
`<=>x^2+4x+3+x^2-4x+3=x+6`
`<=>2x^2+6=x+6`
`<=>2x^2-x=0`
`<=>x(2x-1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy `S={0; 1/2}`.
ĐKXĐ: x ≠ -3, x ≠ 3
\(\dfrac{x+1}{x-3}+\dfrac{x-1}{x+3}=\dfrac{x+6}{x^2-9}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+3\right)+\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow x^2+4x+3+x^2-4x+3=x+6\)
\(\Leftrightarrow2x^2-x=0\)
\(\Leftrightarrow x\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\)
Vậy...