Giaỉ các phương trình sau ;
a) 2/x^3-x^2-x+1 = 3/1-x^2 - 1/x+1
b)x+2/x-2 - 2/x^2-2x = 1/x
Giaỉ các phương trình sau
a) 3/x^2+x-2 - 1/x-1 = -7/x+2
b) 2/-x^2+6x-8 - x-1/x-2 = x+3/x-4
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: \(-2-x^2+5x-4=x^2+x-6\)
\(\Leftrightarrow-x^2+5x-6-x^2-x+6=0\)
\(\Leftrightarrow-2x^2+4x=0\)
\(\Leftrightarrow-2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)
\(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{\left(x^2-x\right)+\left(2x-2\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{x\left(x-1\right)+2\left(x-1\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{1}{x-1}+\dfrac{7}{x+2}=0\)
\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{x+2}{\left(x+2\right)\left(x-1\right)}+\dfrac{7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Rightarrow\dfrac{3-\left(x+2\right)+7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Rightarrow3-x-2+7x-7=0\)
\(\Rightarrow6x-6=0\)
\(\Rightarrow x=1\)
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
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:
\(x_1=1\left(nhận\right);x_2=\dfrac{c}{a}=2\left(loại\right)\)
Giaỉ các phương trình sau:
a, \(\dfrac{6-x}{4x-3}\)=\(\dfrac{2}{4x-3}\)
b, \(\dfrac{3-x}{2x-3}\)+x-1=\(\dfrac{-4}{2x-3}\)
c, \(\dfrac{2x-4}{x-3}\)=2x+1
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)
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 \(x\in\varnothing\).
a) ĐK: \(x\ne\dfrac{3}{4}\)
PT \(\Rightarrow27x-18-4x^2=8x-6\)
\(\Leftrightarrow4x^2-19x+12=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=\dfrac{3}{4}\left(loại\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm \(x=4\)
b) ĐK: \(x\ne\dfrac{3}{2}\)
PT \(\Rightarrow3-x+2x^2-5x+3=-4\)
\(\Leftrightarrow x^2-3x+5=0\) (Vô nghiệm)
Vậy phương trình vô nghiệm
c) ĐK: \(x\ne3\)
PT \(\Rightarrow2x^2-5x-3=2x-4\)
\(\Leftrightarrow2x^2-7x+1=0\) \(\Leftrightarrow x=\dfrac{7\pm\sqrt{41}}{4}\)
Vậy phương trình có nghiệm \(x=\dfrac{7\pm\sqrt{41}}{4}\)
Giải các phương trình sau:
a, 3( x-1) + 2 = 2x - 1
b, ( x+1)( x-3) = 0
c, \(\dfrac{x}{x+1}-\dfrac{2x-3}{x-1}=\dfrac{x+3}{x^2-1}\)
a: 3(x-1)+2=2x-1
=>3x-3+2=2x-1
=>3x-1=2x-1
hay x=0
b: (x+1)(x-3)=0
=>x+1=0 hoặc x-3=0
=>x=-1 hoặc x=3
c: \(\Leftrightarrow x\left(x-1\right)-\left(2x-3\right)\left(x+1\right)=x+3\)
\(\Leftrightarrow x^2-x-2x^2-2x+3x+3=x+3\)
\(\Leftrightarrow-x^2-x=0\)
=>x=0(nhận) hoặc x=-1(loại)
Giaỉ các bất phương trình sau rồi biểu diễn tập nghiệm trên trục số
a) 2x+3>1-x b) 15-2(x-3) < -2x+5
c) (x+1)(x-3) ≤ (x+4) (x-1)
GIÚP mik nha mn
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 \(x\ge\dfrac{1}{5}\)
Bài1:(1,5 điểm)Giải các phương trình sau
a)3(2x-3)=5x+1
b)x+1/2021+x+2/2020+x+3/2019+x+2028/2=0
a) \(3\left(2x-x\right)=5x+1\)
\(\Leftrightarrow6x-3x=5x+1\)
\(\Leftrightarrow6x-3x-5x=1\)
\(\Leftrightarrow-2x=1\)
\(\Leftrightarrow x=\dfrac{1}{-2}=-\dfrac{1}{2}\)
b) \(\dfrac{x+1}{2021}+\dfrac{x+2}{2020}+\dfrac{x+3}{2019}+\dfrac{x+4}{2018}=0\)
\(\Leftrightarrow\dfrac{x+1}{2021}+1+\dfrac{x+2}{2020}+1=\dfrac{x+3}{2019}+1+\dfrac{x+4}{2018}+1\)
\(\Leftrightarrow\dfrac{x+2022}{2021}+\dfrac{x+2022}{2020}=\dfrac{x+2022}{2019}+\dfrac{x+2022}{2018}\)
\(\Leftrightarrow\left(x+2022\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}\right)\)
\(\Leftrightarrow x+2022=0\)
\(\Leftrightarrow x=-2022\)
a)3(2x-3)=5x+1
⇔6x-9=5x+1
⇔6x-5x=1+9
⇔x=10
vậy phương trình có nghiệm là S={10}
b)\(\dfrac{x+1}{2021}\)+\(\dfrac{x+2}{2020}\)+\(\dfrac{x+3}{2019}\)+\(\dfrac{x+2028}{2}\)=0
⇔2020(x+1)+2021(x+2)+2041210(x+2028)=0
⇔2045251x+4139579942=0
⇔2045251x=-4139579942=0
⇔x=-\(\dfrac{4139579942}{2045251}\)
vậy phương trình có tập nghiệm là S={\(-\dfrac{4139579942}{2045251}\)}
Giaỉ phương trình sau ;
4/x^2+2x-3 = 2x-5/x+3 - 2x/x-1
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 \(x=-\dfrac{1}{13}\)
bài 1 Giaỉ phương trình :
a ) \(\sqrt{2x+1}-\sqrt{x-2}=x+3\)
b ) \(\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{x^2+4x+3}\)
c )\(2\sqrt{x+3}=9x^2-x-4\)
ai giúp em với ạ
a, ĐK: \(x\ge2\)
\(\sqrt{2x+1}-\sqrt{x-2}=x+3\)
\(\Leftrightarrow\dfrac{x+3}{\sqrt{2x+1}+\sqrt{x-2}}=x+3\)
\(\Leftrightarrow\left(x+3\right)\left(\dfrac{1}{\sqrt{2x+1}+\sqrt{x-2}}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\sqrt{2x+1}+\sqrt{x-2}=1\left(vn\right)\end{matrix}\right.\)
Phương trình vô nghiệm.
b, ĐK: \(x\ge-1\)
\(\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{x^2+4x+3}\)
\(\Leftrightarrow\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{\left(x+3\right)\left(x+1\right)}\)
\(\Leftrightarrow-\sqrt{x+3}\left(\sqrt{x+1}-1\right)+2x\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=2x\\\sqrt{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x+3=4x^2\end{matrix}\right.\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
c, ĐK: \(x\ge-3\)
\(2\sqrt{x+3}=9x^2-x-4\)
\(\Leftrightarrow x+3+2\sqrt{x+3}+1=9x^2\)
\(\Leftrightarrow\left(\sqrt{x+3}+1\right)^2=9x^2\)
\(\Leftrightarrow\left(\sqrt{x+3}+1-3x\right)\left(\sqrt{x+3}+1+3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=3x-1\\\sqrt{x+3}=-3x-1\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}3x-1\ge0\\x+3=9x^2-6x+1\end{matrix}\right.\Leftrightarrow...\)
TH2: \(\left\{{}\begin{matrix}-3x-1\ge0\\x+3=9x^2+6x+1\end{matrix}\right.\Leftrightarrow...\)
Tự giải nha, t kh có máy tính ở đây.
Giaỉ các bất phương trình sau rồi biểu diễn tập nghiệm trên trục số
d)\(\dfrac{2x+1}{3}-\dfrac{1-x}{2}\) ≥\(1-\dfrac{x}{4}\)
e) \(\dfrac{x+1}{2}-\dfrac{2-x}{3}< \dfrac{2x-3}{4}\)
GIÚP MIK NHA MN
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 \(x< -\dfrac{13}{4}\)