giải pt :\(\frac{2x-5}{2-x}>-1\)
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NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 2 2020
\(\left(2x+1\right)\left(x-1\right)>0\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -\frac{1}{2}\end{matrix}\right.\)
\(\left(3x+1\right)\left(x-5\right)\left(-4x+5\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-\frac{1}{3}\\\frac{5}{4}\le x\le5\end{matrix}\right.\)
\(\frac{x+2}{x-2}\le\frac{3x+1}{2x-1}\Leftrightarrow\frac{3x+1}{2x-1}-\frac{x+2}{x-2}\ge0\)
\(\Leftrightarrow\frac{x^2-8x}{\left(2x-1\right)\left(x-2\right)}\ge0\Leftrightarrow\frac{x\left(x-8\right)}{\left(2x-1\right)\left(x-2\right)}\ge0\Leftrightarrow\left[{}\begin{matrix}x\le0\\\frac{1}{2}< x< 2\\x\ge8\end{matrix}\right.\)
VH
1
1 tháng 4 2020
\(\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}ĐKXĐ:x\ne-1;-3\)
\(\frac{2x}{x+1}+\frac{18}{\left(x-1\right)\left(x+3\right)}=\frac{2x-5}{x+3}\)
\(2x\left(x-1\right)\left(x+3\right)+18\left(x+1\right)=\left(2x-5\right)\left(x+1\right)\left(x-1\right)\)
\(4x^2+12x+18=-2x-5x^2+5\)
\(4x^2+12x+18+2x+5x^2-5=0\)
\(9x^2-14x+13=0\)
=> vô nghiệm
TB
1
LM
21 tháng 6 2017
\(\frac{3\text{x}-1}{x-1}-\frac{2\text{x}+5}{x+3}=1-\)\(\frac{4}{x^2+2\text{x}-3}\) \(\left(\text{Đ}K\text{X}\text{Đ}:x\ne1;x\ne-3\right)\)
\(\Leftrightarrow\frac{\left(3\text{x}-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2\text{x}+5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Rightarrow\left(3\text{x}-1\right)\left(x+3\right)-\left(2\text{x}+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)
\(\Leftrightarrow3\text{x}^2+8\text{x}-3-2\text{x}^2-3\text{x}+5=x^2+2\text{x}-3-4\)
\(\Leftrightarrow3\text{x}^2-2\text{x}^2-x^2+8\text{x}-3\text{x}-2\text{x}=-3-4+3-5\Leftrightarrow3\text{x}=-9\Leftrightarrow x=-3\)(không thỏa mãn ĐKXĐ)
Vậy pt vô nghiệm
NN
1
KT
18 tháng 11 2018
mk lm k chắc đúng, sai đâu ib mk nhé
DKXD: \(x\ge-\frac{1}{2};\)\(x\ne0\)
Dat: \(\sqrt{2x+1}=a\) \(\left(a\ge0;a\ne1\right)\)
Khi đó bpt đã cho trở thành:
\(\frac{a^2-1}{a-1}>a^2+1\)
<=> \(a+1>a^2+1\)
<=> \(a\left(1-a\right)>0\)
<=> \(1-a>0\)
<=> \(a< 1\)
Khi đó: \(\sqrt{2x+1}< 1\)
<=> \(2x+1< 1\)
<=> \(x< 0\)
Vay: \(-\frac{1}{2}\le x< 0\)
N
1 tháng 4 2019
a.\(\Leftrightarrow\left(x+3\right)\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-x^2+2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-1\end{matrix}\right.\)
1 tháng 4 2019
(x-2)(x+1)(x+3)=(x+3)(x+1)(2x-58)
\(x^3+2x^2-5x-6\)=\(2x^3+3x^2-14x-15\)
\(-x^3-x^2+9x+9=0\)
\(-x^2\left(x+1\right)+9\left(x+1\right)=0\)
\(\left(x+1\right)\left(9-x^2\right)\)=0
(x+1)(3-x)(3+x)=0
*x+1=0 =>x=-1
*3-x=0=>x=3
*3+x=0=>x=-3
NA
0
BC
0
\(ĐKXĐ:x\ne2\)
\(pt\Leftrightarrow2x-5>x-2\)
\(\Leftrightarrow2x-x>5-2\)
\(\Leftrightarrow x>3\)
Vậy x > 3
\(\frac{2x-5}{2-x}>-1\)
=> \(2x-5>-\left(2-x\right)\)
=> \(2x-5>x-2\)
=> \(2x-x>5-2\)
=> \(x>3\)