Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(\Leftrightarrow\dfrac{x+1}{2x+1}=\dfrac{x+4}{2x+6}\)
=>(x+1)(2x+6)=(2x+1)(x+4)
\(\Leftrightarrow2x^2+6x+2x+6=2x^2+8x+x+4\)
=>9x+4=8x+6
=>x=2
b: \(x^2+5x=0\)
=>x(x+5)=0
=>x=0 hoặc x=-5
a: \(\dfrac{31-2x}{x+23}=\dfrac{9}{4}\)
=>121-8x=9x+207
=>-17x=86
hay x=-86/17
b: \(\dfrac{\left|2x-1\right|}{\dfrac{1}{2}}=\dfrac{18}{5}\)
=>|2x-1|=9/5
=>2x-1=9/5 hoặc 2x-1=-9/5
=>2x=14/5 hoặc 2x=-4/5
=>x=7/5 hoặc x=-2/5
a,
\(\left(\dfrac{3}{5}x-\dfrac{2}{3}x-x\right)\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)
\(\Rightarrow\dfrac{-16}{15}x\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)
\(\Rightarrow\dfrac{-16}{15}x=\dfrac{-\dfrac{5}{21}}{\dfrac{1}{7}}=-\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{-\dfrac{5}{3}}{-\dfrac{16}{15}}=\dfrac{25}{16}\)
b,
\(\left(5x-1\right)\left(2x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\2x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{6}\end{matrix}\right.\)
c,
\(\dfrac{5\left|x+1\right|}{2}=\dfrac{90}{\left|x+1\right|}\)
\(\Rightarrow5\left|x+1\right|^2=180\)
\(\Rightarrow\left|x+1\right|^2=36\)
Mà \(\left|x+1\right|\ge0\)
=> x + 1 = 6 <=> x = 7
a) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)
\(20x^2-16x-34=10x^2+3x-34\)
\(10x^2-19x=0\)
\(x\left(10x-19\right)=0\)
\(\Leftrightarrow x=0\)
hoặc \(10x-19=0\)
\(\Leftrightarrow x=\dfrac{19}{10}\)
Vạy ..............
b) \(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)
\(\Leftrightarrow1-\dfrac{x-1}{x+5}=1-\dfrac{6}{7}\)
\(\Leftrightarrow\dfrac{x+5}{x+5}-\dfrac{x-1}{x+5}=\dfrac{7}{7}-\dfrac{6}{7}\)
\(\Leftrightarrow\dfrac{\left(x+5\right)-\left(x-1\right)}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{x+5-x+1}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{\left(x-x\right)+\left(5+1\right)}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{6}{x+5}=\dfrac{1}{7}\)
\(\Leftrightarrow x+5=42\)
\(\Leftrightarrow x=37\)
a/ đk: x khác -7/5 ; x khác -1/5
pt <=> \(\dfrac{\left(3x+2\right)\left(5x+1\right)}{\left(5x+7\right)\left(5x+1\right)}=\dfrac{\left(3x-1\right)\left(5x+7\right)}{\left(5x+7\right)\left(5x+1\right)}\)
\(\Rightarrow15x^2+13x+2=15x^2+16x-7\)
\(\Leftrightarrow15x^2+13x-15x^2-16x^2=-7-2\)
\(\Leftrightarrow-3x=-9\Leftrightarrow x=3\left(tm\right)\)
vậy x = 3
b/ đk: x khác -1/2; x khác -3
pt <=> \(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(2x+1\right)\left(x+3\right)}=\dfrac{\left(0,5x+2\right)\left(2x+1\right)}{\left(2x+1\right)\left(x+3\right)}\)
\(\Rightarrow x^2+4x+3=x^2+4,5x+2\)
\(\Leftrightarrow x^2+4x-x^2-4,5x=2-3\)
\(\Leftrightarrow-0,5x=-1\Leftrightarrow x=2\left(tm\right)\)
vậy x = 2
a) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}=\dfrac{\left(3x+2\right)-\left(3x-1\right)}{\left(5x+7\right)-\left(5x+1\right)}=\dfrac{3}{6}=\dfrac{1}{2}\)
\(\Rightarrow2\left(3x+2\right)=5x+7\)
\(\Rightarrow6x+4=5x+7\)
\(\Leftrightarrow x=3\)
Vậy x = 3
b) Ta có: \(\dfrac{0,5x+2}{x+3}=\dfrac{2\left(0,5x+2\right)}{2\left(x+3\right)}=\dfrac{x+4}{2x+6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x+1}{2x+1}=\dfrac{0,5x+2}{x+3}=\dfrac{x+4}{2x+6}=\dfrac{\left(x+4\right)-\left(x+1\right)}{\left(2x+6\right)-\left(2x+1\right)}=\dfrac{3}{5}\)
\(\Rightarrow5\left(x+1\right)=3\left(2x+1\right)\)
\(\Rightarrow5x+5=6x+3\)
\(\Leftrightarrow x=2\)
1: \(\Leftrightarrow\left(x+1\right)^2=4\)
=>x+1=2 hoặc x+1=-2
=>x=1 hoặc x=-3
2: \(\Leftrightarrow7x-21=5x+25\)
=>2x=46
=>x=23
3: \(\Leftrightarrow x^2+4x+3=x^2+0.5x+4x+2\)
=>4,5x+2=4x+3
=>x=1
\(\dfrac{3x+2}{5x-7}=\dfrac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x-7\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2-21x-5x-7\)
\(\Leftrightarrow15x^2+13x+2=15x^2-26x-7\)
\(\Leftrightarrow15x^2-15x^2-13x-2=26x-7\)
\(\Leftrightarrow-13x-2=26x-7\)
\(\Leftrightarrow26x+13x=7+2\)
\(\Leftrightarrow39x=9\Leftrightarrow x=\dfrac{3}{13}\)
b tương tự
a/ \(-x-\dfrac{2}{3}=-\dfrac{6}{7}\)
\(\Leftrightarrow-x=-\dfrac{6}{7}+\dfrac{2}{3}\)
\(\Leftrightarrow-x=-\dfrac{18}{21}+\dfrac{14}{21}\)
\(\Leftrightarrow-x=-\dfrac{4}{21}\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
Vậy ....
b/ \(\left(5x-1\right)\left(2x-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy ..