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ĐKXĐ: \(x\ne\left\{-3;-2;-1;0\right\}\)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}=\dfrac{x}{x\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}=\dfrac{x}{x\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{x}{x\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{3}{x\left(x+3\right)}=\dfrac{x}{x\left(x+3\right)}\)
\(\Leftrightarrow x=3\)
a) \(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+38\)
\(\Leftrightarrow\left[\left(3x+2\right)-\left(3x-2\right)\right]\left[\left(3x+2\right)+\left(3x-2\right)\right]=5x+38\)
\(\Leftrightarrow\left(3x+2-3x+2\right)\left(3x+2+3x-2\right)=5x+38\)
\(\Leftrightarrow4\cdot6x=5x+38\)
\(\Leftrightarrow24x-5x=38\)
\(\Leftrightarrow19x=38\Leftrightarrow x=\dfrac{38}{19}=2\)
Vậy \(S=\left\{2\right\}\)
b) \(\left(x+1\right)\left(x^2-2x+1\right)-2x=2\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow x^3-2x^2+x+x^2-2x+1-2x=2\left(x^2-1\right)\)
\(\Leftrightarrow x^3-2x^2+x+x^2-2x+1-2x=2x^2-2\)
\(\Leftrightarrow x^3-2x^2+x+x^2-2x+1-2x-2x^2+2=0\)
\(\Leftrightarrow x^3-3x^2-3x+3=0\)
PT vô nghiệm , không tìm được x
Vậy \(S=\varnothing\)
c) \(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)
\(\Leftrightarrow3\left(x^2-2x+4\right)+9\left(x-1\right)=3\left(x^2+x-3\right)\)
\(\Leftrightarrow3x^2-6x+12+9x-9=3x^2+3x-9\)
\(\Leftrightarrow3x^2-6x+12+9x-9-3x^2-3x+9=0\)
\(\Leftrightarrow0x+12=0\)
PT vô nghiệm
Vậy \(S=\varnothing\)
Câu cuối tương tự
a) (3x + 2)2 - (3x - 2)2 = 5x + 38
<=> 6x.4 = 5x + 38 <=> 19x = 38 <=> x = 2
b) 3(x - 2)2 + 9(x - 1) = 3(x2 + x - 3)
<=> 3x2 - 12x + 12 + 9x - 9 = 3x2 + 3x - 9
<=> -6x = -12 <=> x = 2
c) (x + 3)2 - (x - 3)2 = 6x + 8
<=> 2x.6 = 6x + 8 <=> 6x = 8 <=> x = 4/3
d) (x - 1)3 - x(x + 1)2 = 5x(2 - x) - 11(x + 2)
<=> x3 - 3x2 + 3x - 1 - x3 - 2x2 - x = 10x - 5x2 - 11x - 22
<=> 3x = -21 <=> x = -7
e) (x + 1)(x2 - x + 1) - 2x = x(x - 1)(x + 1)
<=> x3 - 1 - 2x = x3 - x
<=> x = -1
a, \(12-2\left(1-x\right)^2=\left(3x-2\right)\left(2x-3\right)\)
\(< =>12-2\left(1-2x+x^2\right)=6x^2-9x-4x+6\)
\(< =>12-2+4x-2x^2=6x^2-13x+6\)
\(< =>10+4x-2x^2-6x^2+13x-6=0\)
\(< =>-8x^2+17x+4=0< =>\orbr{\begin{cases}x=\frac{17-\sqrt{417}}{16}\\x=\frac{17+\sqrt{417}}{16}\end{cases}}\)
b, \(10x+3-5x=4x+12< =>5x+3-4x-12=0\)
\(< =>x-9=0< =>x=9\)
c, \(11x+42-2x=100-9x-22< =>9x+42-100+9x+22=0\)
\(< =>18x+64-100=0< =>18x-36=0< =>x=\frac{36}{18}=2\)
d, \(2x-\left(3-5x\right)=4\left(x+3\right)< =>2x-3+5x=4x+12\)
\(< =>7x-3-4x-12=0< =>3x-15=0< =>x=\frac{15}{3}=5\)
e, \(2\left(x-3\right)+5x\left(x-1\right)=5x^2< =>2x-6+5x^2-5=5x^2\)
\(< =>2x-11+5x^2-5x^2=0< =>2x-11=0< =>x=\frac{11}{2}\)
f, \(-6\left(1,5-2x\right)=3\left(-15+2x\right)< =>-6\left(\frac{3}{2}-2x\right)=3\left(2x-15\right)\)
\(< =>-9+12x-6x+45=0< =>6x+36=0< =>x=-6\)
g, \(14x-\left(2x+7\right)=3x+12x-13< =>14x-2x-7=15x-13\)
\(< =>12x-7-15x+13=0< =>-3x+6=0< =>x=-2\)
h, \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)
\(< =>x^2-16-6x+4=x^2-8x+16\)
\(< =>x^2-6x-12-x^2+8x-16=0\)
\(< =>2x-28=0< =>x=\frac{28}{2}=14\)
q, \(4\left(x-2\right)-\left(x-3\right)\left(2x-5\right)=?\)thiếu đề
a: 3x-5>15-x
=>4x>20
hay x>5
b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)
=>3x2+x>3x2-12
=>x>-12
