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1)
\(x-\dfrac{x-1}{3}+\dfrac{x+2}{6}>\dfrac{2x}{5}+5\\ \Leftrightarrow x-\dfrac{x-1}{3}+\dfrac{x+2}{6}-\dfrac{2x}{5}-5>0\\ \Leftrightarrow\dfrac{30x-10\left(x-1\right)+5\left(x+2\right)-2x\cdot6-5\cdot30}{30}>0\\ \Leftrightarrow30x-10x+10+5x+10-12x-150>0\\ \Leftrightarrow30x-10x=5x-12x>-10-10+150\\ \Leftrightarrow13x>130\\ \Leftrightarrow13x\cdot\dfrac{1}{13}>130\cdot\dfrac{1}{13}\\ \Leftrightarrow x>10\)
Vậy tập ngiệm của bât hương trình là {x/x>10}
mình mới học đến đây nên cách giải còn dài, thông cảm nha
2)
\(\dfrac{2x+6}{6}-\dfrac{x-2}{9}< 1\\ \Leftrightarrow\dfrac{2\left(x+3\right)}{6}-\dfrac{x-2}{9}< 1\\ \Leftrightarrow\dfrac{x+3}{3}-\dfrac{x-2}{9}-1< 0\\ \Leftrightarrow\dfrac{3\left(x+3\right)-x+2-9}{9}< 0\\ \Leftrightarrow3x+9-x+2-9< 0\\ \Leftrightarrow3x-x< -9+9-2\\ \Leftrightarrow2x< -2\\ \Leftrightarrow2x\cdot\dfrac{1}{2}< -2\cdot\dfrac{1}{2}\Leftrightarrow x< -1\)
Vậy tập nghiệm của bất phương trình là {x/x<-1}
a.
\(\dfrac{x+3}{x-2}+\dfrac{4+x}{2-x}\\ =\dfrac{x+3}{x-2}-\dfrac{4+x}{x-2}\\ =\dfrac{x+3-4-x}{x-2}\\ =-\dfrac{1}{x-2}\)
b. \(\dfrac{x+1}{2x+6}+\dfrac{2x+3}{x^2+3x}\)
\(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x}{2x\left(x+3\right)}+\dfrac{4x+6}{2x\left(x+3\right)}=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x^2+3x+2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{x\left(x+3\right)+2\left(x+3\right)}{2x\left(x+3\right)}=\dfrac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{x+2}{2x}\)
c. \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x}{2x\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
d. \(\dfrac{2x+6}{3x^2-x}:\dfrac{x^2+3x}{1-3x}\)
\(=\dfrac{2\left(x+3\right)}{x\left(3x-1\right)}:\dfrac{-x\left(x+3\right)}{3x-1}\)
\(=\dfrac{2\left(x+3\right)}{x\left(3x-1\right)}.\dfrac{-\left(3x-1\right)}{x\left(x+3\right)}\)
\(=-\dfrac{2}{x^2}\)
a) \(\frac{6-x}{3}-\frac{x}{4}=\frac{3+2x}{2}-1\)
\(\frac{4\left(6-x\right)}{12}-\frac{3x}{12}=\frac{3+2x}{2}-\frac{2}{2}\)
\(\frac{24-4x-3x}{12}=\frac{3+2x-2}{2}\)
\(\frac{24-7x}{12}=\frac{2x+1}{2}\)
\(\Rightarrow2\left(24-7x\right)=12\left(2x+1\right)\)
\(\Rightarrow48-14x=24x+12\)
\(\Rightarrow24x+14x=48-12\)
\(\Rightarrow38x=36\)
\(\Rightarrow x=\frac{18}{19}\)
b) \(-7x-\frac{x-3}{5}-\frac{x}{2}=x+\frac{2x+1}{3}\)
\(\frac{-70x}{10}-\frac{2\left(x-3\right)}{10}-\frac{5x}{10}=\frac{3x}{3}+\frac{2x+1}{3}\)
\(\frac{-70x-2x+6-5x}{10}=\frac{3x+2x+1}{3}\)
\(\frac{-77x+6}{10}=\frac{5x+1}{3}\)
\(\Rightarrow3\left(-77x+6\right)=10\left(5x+1\right)\)
\(\Leftrightarrow-231x+18=50x+10\)
\(\Leftrightarrow50x+231x=18-10\)
\(\Leftrightarrow281x=8\)
\(\Leftrightarrow x=\frac{8}{281}\)
Mấy câu kia tương tự
a) \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\) MTC: \(2x\left(x+3\right)\)
\(=\dfrac{3x}{2x\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-\left(x-6\right)}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
b) \(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5x+6}{4-x^2}\)
\(=\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x+6}{x^2-4}\)
\(=\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x+6}{\left(x-2\right)\left(x+2\right)}\) MTC: \(\left(x-2\right)\left(x+2\right)\)
\(=\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5x+6}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)+2\left(x+2\right)-\left(5x+6\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x-8+2x+4-5x-6}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x-10}{\left(x-2\right)\left(x+2\right)}\)
c) \(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)
\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{4x^2-2x}\)
\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{2x\left(2x-1\right)}\) MTC: \(2x\left(2x-1\right)\)
\(=\dfrac{\left(1-3x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{2x\left(3x-2\right)}{2x\left(2x-1\right)}-\dfrac{3x-2}{2x\left(2x-1\right)}\)
\(=\dfrac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)-\left(3x-2\right)}{2x\left(2x-1\right)}\)
\(=\dfrac{2x-1-6x^2+3x+6x^2-4x-3x+2}{2x\left(2x-1\right)}\)
\(=\dfrac{-2x+1}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(2x-1\right)}{2x\left(2x-1\right)}\)
\(=\dfrac{-1}{2x}\)
d) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(=\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\) MTC: \(\left(x-1\right)\left(x^2+x+1\right)\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2\left(x-2\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x^2+2\right)+2\left(x-2\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+2+2x-4-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-3+x}{\left(x-1\right)\left(x^2+x+1\right)}\)
4.
\(\dfrac{x+1}{99}+\dfrac{x+3}{97}+\dfrac{x+5}{95}=\dfrac{x+7}{93}+\dfrac{x+9}{91}+\dfrac{x+11}{89}\\ \Rightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+5}{95}+1\right)=\left(\dfrac{x+7}{93}+1\right)+\left(\dfrac{x+9}{91}+1\right)+\left(\dfrac{x+11}{89}+1\right)\\ \Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{97}++\dfrac{x+100}{95}=\dfrac{x+100}{93}+\dfrac{x+100}{91}+\dfrac{x+100}{89}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{93}-\dfrac{1}{91}-\dfrac{1}{89}\right)=0\\ \Leftrightarrow x+100=0\Leftrightarrow x=-100\)
\(\text{1) }\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}=\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\\ \Leftrightarrow\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}\cdot24=\left[\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\right]24\\ \Leftrightarrow3\left(4x^2-9\right)=4\left(x^2-8x+16\right)+8\left(x^2-4x+4\right)\\ \Leftrightarrow12x^2-27=4x^2-32x+64+8x^2-32x+32\\ \Leftrightarrow12x^2-27=12x^2-64x+96\\ \Leftrightarrow12x^2-12x^2+64x=96+27\\ \Leftrightarrow64x=123\\ \Leftrightarrow x=\dfrac{123}{64}\\ \text{Vậy }S=\left\{\dfrac{123}{64}\right\}\\ \)
\(\text{2) }x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}=\dfrac{7x-\dfrac{x-3}{2}}{5}\\ \Leftrightarrow\left(x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}\right)15=\dfrac{7x-\dfrac{x-3}{2}}{5}\cdot15\\ \Leftrightarrow15x+30-2x-\dfrac{2x-5}{6}=21x-\dfrac{3x-9}{2}\\ \Leftrightarrow15x-2x-\dfrac{2x-5}{6}-21x+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow\left(-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}\right)6=-30\cdot6\\ \Leftrightarrow-48x-2x+5+9x-27=-180\\ \Leftrightarrow-41x==-158\\ \Leftrightarrow x=\dfrac{158}{41}\\ \text{Vậy }S=\left\{\dfrac{158}{41}\right\}\)
\(\text{3) }1-\dfrac{x-\dfrac{1+x}{3}}{3}=\dfrac{x}{2}-\dfrac{2x-\dfrac{10-7}{3}}{2}\\ \Leftrightarrow\left(1-\dfrac{x-1-x}{3}\right)6=\left(\dfrac{x}{2}-\dfrac{2x-1}{2}\right)6\\ \Leftrightarrow6+2=-3x+3\\ \Leftrightarrow-3x=8-3\\ \Leftrightarrow-3x=5\\ \Leftrightarrow x=-\dfrac{5}{3}\\ \\ \text{Vậy }S=\left\{-\dfrac{5}{3}\right\}\)
b: \(=\dfrac{x}{2\left(x-3\right)}+\dfrac{4}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2+3x+8}{2\left(x-3\right)\left(x+3\right)}\)
c: \(=\dfrac{\left(x+1\right)^2}{\left(x-1\right)^2}\cdot\dfrac{4\left(x-1\right)^2}{2\left(x+1\right)^2}=\dfrac{4}{2}=2\)
d: \(=\dfrac{2x+1}{x-2}\cdot\dfrac{-\left(x-2\right)}{2x+1}=-1\)
Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
đk: x khác -1; 3
\(\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=\dfrac{x}{2\left(3-x\right)}\)
<=> \(\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}+\dfrac{x}{2\left(x-3\right)}=0\)
<=> \(\dfrac{x\left(x-3\right)-4x+x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}=0\)
<=> \(x^2-3x-4x+x^2+x=0\)
<=> 2x2 - 6x = 0
<=> 2x(x-3) = 0
Mà x khác 3
<=> 2x = 0
<=> x = 0
Cảm ơn nha