Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Việc tìm ĐKXĐ hơi phức tạp nên bạn có thể dùng cách sau:
\(=>\dfrac{x+5}{2x^2-5x+2}+\dfrac{x+2}{2x^2-7x+3}-\dfrac{2x+7}{2x^2-7x+3}=0\)
\(=>\dfrac{x+5}{2x^2-5x+2}-\dfrac{x+5}{2x^2-7x+3}=0\)
\(=>\left(x+5\right)\left(\dfrac{1}{2x^2-5x+2}-\dfrac{1}{2x^2-7x+3}\right)=0\)
\(=>\dfrac{\left(x+5\right)\left(1-2x\right)}{\left(2x^2-5x+2\right)\left(2x^2-7x+3\right)}=0\)
Xét tử số bằng 0 ta có:
\(\left\{{}\begin{matrix}x+5=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=\dfrac{1}{2}\end{matrix}\right.\)
Với x=2 thì \(2x^2-5x+2=0\) , phương trình đã cho không xác định;
- Với \(x\ne-5\) thì : \(\left(2x^2-5x+2\right)\left(2x^2-5x+2\right)\ne0\)
Vậy nghiệm của phương trình là -5;
\CHÚC BẠN HỌC TỐT....
Lời giải:
ĐKXĐ:.....
Ta có: \(\frac{x+5}{x(2x-5)+2}+\frac{x+2}{x(2x-7)+3}=\frac{2x+7}{2x^2-7x+3}\)
\(\Leftrightarrow \frac{x+5}{x(2x-5)+2}=\frac{2x+7}{2x^2-7x+3}-\frac{x+2}{2x^2-7x+3}\)
\(\Leftrightarrow \frac{x+5}{x(2x-5)+2}=\frac{2x+7-(x+2)}{2x^2-7x+3}\)
\(\Leftrightarrow \frac{x+5}{x(2x-5)+2}=\frac{x+5}{2x^2-7x+3}\)
\(\Leftrightarrow (x+5)\left(\frac{1}{x(2x-5)+2}-\frac{1}{2x^2-7x+3}\right)=0\)
TH1: \(x+5=0\Leftrightarrow x=-5\) (thỏa mãn)
TH2: \(\frac{1}{x(2x-5)+2}-\frac{1}{2x^2-7x+3}=0\)
\(\Leftrightarrow x(2x-5)+2=2x^2-7x+3\)
\(\Leftrightarrow -5x+2=-7x+3\)
\(\Leftrightarrow 2x=1\Leftrightarrow x=\frac{1}{2}\). Thử lại thây mẫu số bằng 0 (không thỏa mãn ĐKXĐ)
Vậy PT có nghiệm duy nhất \(x=-5\)
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
4.
\(\dfrac{x+1}{99}+\dfrac{x+3}{97}+\dfrac{x+5}{95}=\dfrac{x+7}{93}+\dfrac{x+9}{91}+\dfrac{x+11}{89}\\ \Rightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+5}{95}+1\right)=\left(\dfrac{x+7}{93}+1\right)+\left(\dfrac{x+9}{91}+1\right)+\left(\dfrac{x+11}{89}+1\right)\\ \Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{97}++\dfrac{x+100}{95}=\dfrac{x+100}{93}+\dfrac{x+100}{91}+\dfrac{x+100}{89}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{93}-\dfrac{1}{91}-\dfrac{1}{89}\right)=0\\ \Leftrightarrow x+100=0\Leftrightarrow x=-100\)
\(\text{1) }\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}=\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\\ \Leftrightarrow\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}\cdot24=\left[\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\right]24\\ \Leftrightarrow3\left(4x^2-9\right)=4\left(x^2-8x+16\right)+8\left(x^2-4x+4\right)\\ \Leftrightarrow12x^2-27=4x^2-32x+64+8x^2-32x+32\\ \Leftrightarrow12x^2-27=12x^2-64x+96\\ \Leftrightarrow12x^2-12x^2+64x=96+27\\ \Leftrightarrow64x=123\\ \Leftrightarrow x=\dfrac{123}{64}\\ \text{Vậy }S=\left\{\dfrac{123}{64}\right\}\\ \)
\(\text{2) }x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}=\dfrac{7x-\dfrac{x-3}{2}}{5}\\ \Leftrightarrow\left(x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}\right)15=\dfrac{7x-\dfrac{x-3}{2}}{5}\cdot15\\ \Leftrightarrow15x+30-2x-\dfrac{2x-5}{6}=21x-\dfrac{3x-9}{2}\\ \Leftrightarrow15x-2x-\dfrac{2x-5}{6}-21x+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow\left(-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}\right)6=-30\cdot6\\ \Leftrightarrow-48x-2x+5+9x-27=-180\\ \Leftrightarrow-41x==-158\\ \Leftrightarrow x=\dfrac{158}{41}\\ \text{Vậy }S=\left\{\dfrac{158}{41}\right\}\)
\(\text{3) }1-\dfrac{x-\dfrac{1+x}{3}}{3}=\dfrac{x}{2}-\dfrac{2x-\dfrac{10-7}{3}}{2}\\ \Leftrightarrow\left(1-\dfrac{x-1-x}{3}\right)6=\left(\dfrac{x}{2}-\dfrac{2x-1}{2}\right)6\\ \Leftrightarrow6+2=-3x+3\\ \Leftrightarrow-3x=8-3\\ \Leftrightarrow-3x=5\\ \Leftrightarrow x=-\dfrac{5}{3}\\ \\ \text{Vậy }S=\left\{-\dfrac{5}{3}\right\}\)
\(\Leftrightarrow\dfrac{x+5}{2x^2-5x+2}=\dfrac{2x+7-x-2}{2x^2-7x+3}\)
\(\Leftrightarrow\left(x+5\right)\left(\dfrac{1}{\left(2x-1\right)\left(x-2\right)}-\dfrac{1}{\left(x-3\right)\left(2x-1\right)}\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-3-x+2\right)=0\)
=>x+5=0
hay x=-5