Giaỉ các phương trình sau
a) x-3/x-2 -x-2/x-4 = 5/21
b)x+1/x-2 - x-1/x+2 = 2(x^2+2)/x^2-4
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12 tháng 8 2021
a: Ta có: \(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=\dfrac{-7}{x+2}\)
\(\Leftrightarrow3-\left(x+2\right)=-7\left(x-1\right)\)
\(\Leftrightarrow3-x-2+7x-7=0\)
\(\Leftrightarrow6x-6=0\)
hay x=1(loại
b: Ta có: \(\dfrac{2}{-x^2+6x-8}-\dfrac{x-1}{x-2}=\dfrac{x+3}{x-4}\)
\(\Leftrightarrow\dfrac{-2}{\left(x-2\right)\left(x-4\right)}-\dfrac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}\)
Suy ra: \(-2-x^2+5x-4=x^2+x-6\)
\(\Leftrightarrow-x^2+5x-6-x^2-x+6=0\)
\(\Leftrightarrow-2x^2+4x=0\)
\(\Leftrightarrow-2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)
12 tháng 8 2021
\(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{\left(x^2-x\right)+\left(2x-2\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{x\left(x-1\right)+2\left(x-1\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{1}{x-1}+\dfrac{7}{x+2}=0\)
\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{x+2}{\left(x+2\right)\left(x-1\right)}+\dfrac{7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Rightarrow\dfrac{3-\left(x+2\right)+7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Rightarrow3-x-2+7x-7=0\)
\(\Rightarrow6x-6=0\)
\(\Rightarrow x=1\)
ND
0
ND
1
12 tháng 8 2021
Ta có: \(\dfrac{4}{x^2+2x-3}=\dfrac{2x-5}{x+3}-\dfrac{2x}{x-1}\)
\(\Leftrightarrow\dfrac{\left(2x-5\right)\left(x-1\right)-2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=\dfrac{4}{\left(x+3\right)\left(x-1\right)}\)
Suy ra: \(2x^2-2x-5x+5-2x^2-6x=4\)
\(\Leftrightarrow13x=-1\)
hay \(x=-\dfrac{1}{13}\)
TT
1
15 tháng 1 2018
Mình làm mẫu câu a nha
a, pt <=> ( x-2/7 - 1 ) + ( x-1/8 - 1 ) = ( x-4/5 - 1 ) + ( x-3/6 - 1 )
<=> x-9/7 + x-9/8 = x-9/5 + x-9/6
<=> x-9/5 + x-9/6 - x-9/7 - x-9/8 = 0
<=> (x-9).(1/5+1/6-1/9-1/8) = 0
<=> x-9 = 0 ( vì 1/5+1/6-1/9-1/8 > 0 )
<=> x = 9
Vậy x = 9
Tk mk nha
26 tháng 8 2021
d: Ta có: \(\dfrac{2x+1}{3}-\dfrac{1-x}{2}\ge1-\dfrac{x}{4}\)
\(\Leftrightarrow8x+4-6+6x\ge12-3x\)
\(\Leftrightarrow14x+3x\ge12+2=14\)
\(\Leftrightarrow x\ge\dfrac{14}{17}\)
e: Ta có: \(\dfrac{x+1}{2}-\dfrac{2-x}{3}< \dfrac{2x-3}{4}\)
\(\Leftrightarrow6x+12+4x-8< 6x-9\)
\(\Leftrightarrow4x< -9+8-12=-13\)
hay \(x< -\dfrac{13}{4}\)
12 tháng 8 2021
Ta có: \(\dfrac{x-3}{x-2}-\dfrac{x-2}{x-4}=1\dfrac{5}{21}\)
\(\Leftrightarrow\dfrac{21\left(x-3\right)\left(x-4\right)}{21\left(x-2\right)\left(x-4\right)}-\dfrac{21\left(x-2\right)^2}{21\left(x-2\right)\left(x-4\right)}=\dfrac{26\left(x-2\right)\left(x-4\right)}{21\left(x-2\right)\left(x-4\right)}\)
\(\Leftrightarrow26\left(x^2-6x+8\right)=21\left(x^2-7x+12\right)-21\left(x^2-4x+4\right)\)
\(\Leftrightarrow26x^2-156x+208=21x^2-147x+252-21x^2+84x-84\)
\(\Leftrightarrow26x^2-156x+208+63x-168=0\)
\(\Leftrightarrow26x^2-93x+40=0\)
\(\text{Δ}=\left(-93\right)^2-4\cdot26\cdot40\)
\(=8649-4160\)
\(=4489\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{93-67}{52}=\dfrac{1}{2}\left(nhận\right)\\x_2=\dfrac{93+67}{52}=\dfrac{40}{13}\left(nhận\right)\end{matrix}\right.\)
21 tháng 12 2020
a, \(\dfrac{6-x}{4x-3}=\dfrac{2}{4x-3}\)
ĐKXĐ: \(x\ne\dfrac{3}{4}\)
PT đã cho \(\Leftrightarrow\)\(\dfrac{\left(6-x\right)\left(4x-3\right)}{4x-3}=\dfrac{2\left(4x-3\right)}{4x-3}\)
\(\Rightarrow6-x=2\)
\(\Leftrightarrow x=4\)(thỏa mãn ĐKXĐ)
21 tháng 12 2020
b, \(\dfrac{3-x}{2x-3}+x-1=\dfrac{-4}{2x-3}\)
ĐKXĐ: \(x\ne\dfrac{3}{2}\)
PT đã cho \(\Leftrightarrow\)\(\dfrac{\left(3-x\right)\left(2x-3\right)}{2x-3}+\left(x+1\right)\left(2x-3\right)=\dfrac{-4\left(2x-3\right)}{2x-3}\)
\(\Rightarrow3-x+2x-3x+2x-3=-8x+12\)
\(\Leftrightarrow8x=12\)
\(\Leftrightarrow x=\dfrac{3}{2}\)(không thỏa mãn ĐKXĐ)
Vậy \(x\in\varnothing\).
26 tháng 8 2021
a: Ta có: \(2x+3>1-x\)
\(\Leftrightarrow3x>-2\)
hay \(x>-\dfrac{2}{3}\)
b: Ta có: \(15-2\left(x-3\right)< -2x+5\)
\(\Leftrightarrow15-2x+6+2x-5< 0\)
\(\Leftrightarrow16< 0\left(vôlý\right)\)
c: Ta có: \(\left(x+1\right)\left(x-3\right)\le\left(x+4\right)\left(x-1\right)\)
\(\Leftrightarrow x^2-3x+x-3-x^2+x-4x+4\le0\)
\(\Leftrightarrow-5x\le-1\)
hay \(x\ge\dfrac{1}{5}\)
b: Ta có: \(\dfrac{x+1}{x-2}-\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow x^2+3x+2-x^2+3x-2-2x^2-4=0\)
\(\Leftrightarrow-2x^2+6x-4=0\)
a=-2; b=6; c=-4
Vì a+b+c=0 nên phương trình có hai nghiệm phân biệt là:
\(x_1=1\left(nhận\right);x_2=\dfrac{c}{a}=2\left(loại\right)\)