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a) \(\dfrac{2x-9}{2x-5}+\dfrac{3x}{3x-2}=2\)
\(\Rightarrow\dfrac{2x-5-4}{2x-5}+\dfrac{3x-2+2}{3x-2}=2\)
\(\Rightarrow1-\dfrac{4}{2x-5}+1+\dfrac{2}{3x-2}=2\)
\(\Rightarrow\dfrac{4}{2x-5}+\dfrac{2}{3x-2}=0\)
\(\Rightarrow\dfrac{12x-8}{\left(2x-5\right)\left(3x-2\right)}+\dfrac{4x-10}{\left(2x-5\right)\left(3x-2\right)}=0\)
\(\Rightarrow\dfrac{16x-18}{\left(2x-5\right)\left(3x-2\right)}=0\)
\(\Rightarrow16x-18=0\)
\(\Rightarrow x=\dfrac{18}{16}=\dfrac{9}{8}\)
b) \(\dfrac{x+2}{2002}+\dfrac{x+5}{1999}+\dfrac{x+201}{1803}=-3\)
\(\Rightarrow\dfrac{x+2}{2002}+1+\dfrac{x+5}{1999}+1+\dfrac{x+201}{1803}+1=0\)
\(\Rightarrow\dfrac{x+2004}{2002}+\dfrac{x+2004}{1999}+\dfrac{x+2004}{1809}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2002}+\dfrac{1}{1999}+\dfrac{1}{1809}\right)=0\)
Vì \(\left(\dfrac{1}{2002}+\dfrac{1}{1999}+\dfrac{1}{1809}\right)\ne0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=-2004\)
c) \(\left(2x^2+3x+1\right)\left(2x^2+5x+3\right)=18\)
\(\Rightarrow\left(2x+1\right)\left(x+1\right)\left(x+1\right)\left(2x+3\right)=18\)
\(\Rightarrow\left(x+1\right)^2\left(2x+1\right)=18\)
Thôi, xử tiếp đi nhé :)))))
a: =>5-x+6=12-8x
=>-x+11=12-8x
=>7x=1
hay x=1/7
b: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
=>12x+10=6x+5
=>6x=-5
hay x=-5/6
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
a) \(\left(2x-1\right)^2-\left(3x+5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-1-3x-5\right)=0\\ \text{}\Leftrightarrow\left(2x-1\right)\left(-x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\-x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{2};-6\right\}\)
b) \(\dfrac{x+5}{4}-\dfrac{2x-3}{3}=\dfrac{2x-1}{12}\)
\(\Leftrightarrow3\left(x+5\right)-4\left(2x-3\right)=2x-1\\ \Leftrightarrow3x+15-8x+12=2x-1\\ \Leftrightarrow-5x+27=2x-1\\ \Leftrightarrow-5x-2x=-1-27\\ \Leftrightarrow-7x=-28\\ \Leftrightarrow x=4\)
Vậy \(S=\left\{4\right\}\)
\(c)\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{-4}{1-x^2}\)(ĐKXĐ: \(x\ne\pm1\))
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}-\dfrac{4}{\left(x-1\right)\left(x+1\right)}\\ \dfrac{\left(x+1\right)^2-\left(x-1\right)^2-4}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{4x-4}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{4}{x+1}=0\)
\(\Leftrightarrow4=0\)(vô lý)
Vậy .....
\(d)\dfrac{1}{x+1}+\dfrac{2x-1}{x^3+1}=\dfrac{2}{x^2-x+1}\)(ĐKXĐ: \(x\ne-1\))
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{2x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2}{x^2-x+1}=0\\ \Leftrightarrow\dfrac{x^2-x+1+2x-1-2\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-x+2x-2x-2}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-x-2}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow x^2+x-2x-2=0\\ \Leftrightarrow x\left(x+1\right)-2\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(KTM\right)\\x=2\left(TM\right)\end{matrix}\right.\)
Vậy ....
3.
a) \(2x+5=20-3x\)
\(\Leftrightarrow2x+3x=20-5\)
\(\Leftrightarrow5x=15\)
\(\Leftrightarrow x=3\)
Vậy \(S=\left\{3\right\}\)
b) \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[\left(2x-1\right)+\left(x+3\right)\right]\left[\left(2x-1\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(2x-1+x+3\right)\left(2x-1-x-3\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=4\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{2}{3};4\right\}\)
c) \(\dfrac{5x-4}{2}=\dfrac{16x+1}{7}\)
\(\Leftrightarrow\left(5x-4\right)7=\left(16x+1\right)2\)
\(\Leftrightarrow35x-28=32x+2\)
\(\Leftrightarrow35x-32x=2+28\)
\(\Leftrightarrow2x=30\)
\(\Leftrightarrow x=15\)
Vậy \(S=\left\{15\right\}\)
d) \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Rightarrow\left(2x+1\right)12-\left(x-2\right)18=\left(3-2x\right)24-72x\)
\(\Leftrightarrow24x+12-18x+36=72-48x-72x\)
\(\Leftrightarrow6x+48=72-120x\)
\(\Leftrightarrow6x+120x=72-48\)
\(\Leftrightarrow126x=24\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
Vậy \(S=\left\{\dfrac{4}{21}\right\}\)
bài 1:
b,\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)(ĐKXĐ:x ≠0,x≠-2)
<=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)
=>\(x^2+4x+4=x^2+5x+4+x^2\)
<=>\(x^2-x^2-x^2+4x-5x+4-4=0\)
<=>\(-x^2-x=0< =>-x\left(x+1\right)=0< =>\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(nhận\right)\end{matrix}\right.\)
vậy...............
d,\(\left(x+3\right)^2-25=0< =>\left(x+3-5\right)\left(x+3+5\right)=0< =>\left(x-2\right)\left(x+8\right)=0< =>\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
vậy............
bài 3:
g,\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}\)(ĐKXĐ:x khác -1,x khác 2)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-2x+x-2}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x\left(x-2\right)+\left(x-2\right)}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
<=>\(\dfrac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
=>\(4x-8-2x-2=x+3\)
<=>\(x=13\)
vậy..............
mấy ý khác bạn làm tương tụ nhé
chúc bạn học tốt ^ ^
c) \(8x^3-1=8x^2+4x+2\)
<=> \(\left(2x-3\right)\left(4x^2+2x+1\right)=0\)
<=> \(2x-3=0\) hoặc \(4x^2+2x+1=0\)
Th1: x=\(\dfrac{3}{2}\)
Th2: Vô nghiệm
Vậy x=\(\dfrac{3}{2}\)
\(\text{a) }\dfrac{2x^2-x-1}{2}-3x^2+x+4=\left(5-x\right)\left(2x+4\right)\\ \Leftrightarrow\left(\dfrac{2x^2-x-1}{2}-3x^2+x+4\right)2=\left(5-x\right)\left(2x+4\right)2\\ \Leftrightarrow2x^2-x-1-6x^2+2x+8=\left(5-x\right)\left(4x+8\right)\\ \Leftrightarrow-4x^2+x+7=20x+40-4x^2-8x\\ \Leftrightarrow-4x^2+x+4x^2-12x=40-7\\ \Leftrightarrow-11x=33\\ \Leftrightarrow x=-3\\ \text{Vậy }S=\left\{-3\right\}\)
\(\text{b) }\dfrac{\left(2x-5\right)\left(3x+7\right)}{4}+2x-1=\dfrac{\left(x-1\right)\left(2x+4\right)}{2}+1\\ \Leftrightarrow\dfrac{\left(2x-5\right)\left(3x+7\right)}{4}+2x-1=\left(x-1\right)\left(x+2\right)+1\\ \Leftrightarrow\left(\dfrac{\left(2x-5\right)\left(3x+7\right)}{4}+2x-1\right)4=\left(x^2-x+2x-2+1\right)4\\ \Leftrightarrow\left(2x-5\right)\left(3x+7\right)+8x-4=\left(x^2+x-1\right)4\\ \Leftrightarrow6x^2-15x+14x-35+8x-4=4x^2+4x-4\\ \Leftrightarrow6x^2+7x-39=4x^2+4x-4\\ \Leftrightarrow6x^2+7x-4x^2-4x-39+4=0\\ \Leftrightarrow2x^2+3x-35=0\\ \Leftrightarrow2x^2+10x-7x-35=0\\ \Leftrightarrow\left(2x^2+10x\right)-\left(7x+35\right)=0\\ \Leftrightarrow2x\left(x+5\right)-7\left(x+5\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-5\end{matrix}\right.\\ \\ \text{Vậy }S=\left\{\dfrac{7}{2};-5\right\}\)
\(\text{c) }8x^3-1=8x^2+4x+2\\ \Leftrightarrow\left(2x-1\right)\left(4x^2+2x+1\right)=2\left(4x^2+2x+1\right)\\ \Leftrightarrow2x-1=2\\ \Leftrightarrow2x=3\\ \Leftrightarrow x=\dfrac{3}{2}\\ \text{Vậy }S=\left\{\dfrac{3}{2}\right\}\)
\(\text{d) }\left(x^2+x+1\right)\left(x^2-x+1\right)=x^6-1\\ \Leftrightarrow\left(x^3+1\right)\left(x^3-1\right)=\left(x^2+x+1\right)\left(x^2-x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x^2-x+1\right)=\left(x^2+x+1\right)\left(x^2-x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x-1\right)=1\\ \Leftrightarrow x^2-1=1\\ \Leftrightarrow x^2=2\\ \Leftrightarrow x=\sqrt{2}\\ \text{Vậy }S=\left\{\sqrt{2}\right\}\)
\(\text{e) }\left(x^3+2x\right)\left(x^2+4\right)=\left(x^2+6x^2+8\right)\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left(x^2+2x^2+4x^2+8\right)\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left[\left(x^2+2x^2\right)+\left(4x^2+8\right)\right]\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left[x^2\left(x^2+2\right)+4\left(x^2+2\right)\right]\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left(x^2+4\right)\left(x^2+2\right)\left(3-2x\right)\\ \Leftrightarrow x=3-2x\\ \Leftrightarrow3x=3\\ \Leftrightarrow x=1\\ \text{Vậy }S=\left\{1\right\}\)
f) Kiểm tra lại hạng tử thứ 2 ở vế phải.
a.
\(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2x+10-x^2-5x=0\)
\(\Leftrightarrow-x^2-3x+10=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow\left(x^2+5x\right)-\left(2x+10\right)=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
b.
\(2x^2+3x-5=0\)
\(\Leftrightarrow2x^2-2x+5x-5=0\)
\(\Leftrightarrow\left(2x^2-2x\right)+\left(5x-5\right)=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}\\x=1\end{matrix}\right.\)
bài 2:
ĐKXĐ: x khác -1
\(\dfrac{1-x}{x+1}+3=\dfrac{2x+3}{x+1}\)
\(\Leftrightarrow\dfrac{1-x+3\left(x+1\right)}{x+1}=\dfrac{2x+3}{x+1}\)
\(\Leftrightarrow1-x+3x+3=2x+3\)
\(\Leftrightarrow0x=-1\)
\(\Leftrightarrow x\in\varnothing\)
Suy ra pt vô nghiệm
b.
ĐKXĐ: x khác \(\dfrac{3}{2}\)
\(\dfrac{\left(x+2\right)^2}{2x-3}-1=\dfrac{x^2+10}{2x-3}\)
\(\Leftrightarrow\dfrac{x^2+4x+4}{2x-3}-\dfrac{2x-3}{2x-3}=\dfrac{x^2+10}{2x-3}\)
\(\Leftrightarrow x^2+4x+4-2x+3=x^2+10\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\) ( loại)
a.
\(\left(2x-1\right)^3+6\left(3x-1\right)^3=2\left(x+1\right)^3+6\left(x+2\right)^3\)
\(\Leftrightarrow\left(2x\right)^3-3.\left(2x\right)^2.1+3.2x.1+1^3+6.\left[\left(3x\right)^3-3.\left(3x\right)^2.1+3.3x.1+1^3\right]=2\left(x^3+3x^2+3x+1\right)+6\left(x^2+3.x^2.2+3.x.2^2+2^3\right)\)
a) \(\left(2x+1\right)^2-\left(x+2\right)^2>0\)
\(\Leftrightarrow\left(2x+1-x-2\right)\left(2x+1+x+2\right)>0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+3\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\3x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\3x+3< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
Vậy tập nghiệm của bất phương trình là x > 1 hoặc x < -1
b) Sửa lại rồi làm câu b nèk\(\dfrac{5x-3x}{5}+\dfrac{3x+1}{4}>\dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)
\(\Leftrightarrow4\left(5x-3x\right)+5\left(3x+1\right)>10\left(x+2x\right)-30\)\(\Leftrightarrow20x-12x+15x+5>10x+20x-30\)\(\Leftrightarrow20x-12x+15x-10x-20x>-30-5\)\(\Leftrightarrow-7x>-35\)
\(\Leftrightarrow x< 5\)
c) \(\dfrac{-1}{2x+3}< 0\)
dễ nhé mình học bài hóa mai kt 15 phút nên ko có time để giúp
\(\text{a) }\dfrac{5x^2-3x}{5}+\dfrac{3x+1}{4}< \dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\\ \Leftrightarrow4\left(5x^2-3x\right)+5\left(3x+1\right)< 10x\left(2x+1\right)-15\\ \Leftrightarrow20x^2-12x+15x+5< 20x^2+10x-15\\ \Leftrightarrow20x^2+3x-20x^2-10x< -15-5\\ \Leftrightarrow-7x< -20\\ \Leftrightarrow x>\dfrac{20}{7}\)
Vậy bất phương trình có nghiệm \(x>\dfrac{20}{7}\)
\(\text{b) }\dfrac{5x-20}{3}-\dfrac{2x^2+x}{2}\ge\dfrac{x\left(1-3x\right)}{3}-\dfrac{5x}{4}\\ \Leftrightarrow4\left(5x-20\right)-6\left(2x^2+x\right)\ge4x\left(1-3x\right)-15x\\ \Leftrightarrow20x-80-12x^2-6x\ge4x-12x^2-15x\\ \Leftrightarrow-12x^2+14x+12x^2+11x\ge80\\ \Leftrightarrow25x\ge80\\ \Leftrightarrow x\ge\dfrac{16}{5}\)
Vậy bất phương trình có nghiệm \(x\ge\dfrac{16}{5}\)
\(\text{c) }\left(x+3\right)^2\le x^2-7\\ \Leftrightarrow x^2+6x+9\le x^2-7\\ \Leftrightarrow x^2+6x-x^2\le-7-9\\ \Leftrightarrow6x\le-16\\ \Leftrightarrow x\le-\dfrac{8}{3}\)
Vậy bất phương trình có nghiệm \(x\le-\dfrac{8}{3}\)
a) \(\left(2x-1\right)^2=3x\left(2x-1\right)\)
\(\left(2x-1\right)^2-3x\left(2x-1\right)=0\)
\(\left(2x-1\right)\left(2x-1-3x\right)=0\)
\(\left(2x-1\right)\left(-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
\(a,\left(2x-1\right)^2=3x\left(2x-1\right)\\ \Leftrightarrow4x^2-4x+1=6x^2-3x\\ \Leftrightarrow6x^2-3x-4x^2+4x-1=0\\ \Leftrightarrow2x^2+x-1=0\\ \Leftrightarrow2x^2+2x-x-1=0\\ \Leftrightarrow2x\left(x+1\right)-\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(b,ĐKXĐ:x\ne0,2\\ \dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\\ \Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x^2+2x-x+2-2}{x\left(x-2\right)}=0\\ \Rightarrow x^2+x=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)