3/5x-1+3/3-5x=4/(1-5x)(5x-3)
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TG
0
LD
2
BA
7 tháng 6 2020
\(\frac{3}{\left(5x-1\right)}+\frac{2}{3-5x}=\frac{4}{\left(5x-1\right)\left(3-5x\right)}\)
\(\Rightarrow9-15x+10x-2=4\)
\(\Leftrightarrow3-5x=0\)
\(\Leftrightarrow x=\frac{3}{5}\)
TT
7 tháng 6 2020
\(\frac{3}{\left(5x-1\right)}+\frac{2}{3-5x}=\frac{4}{\left(5x-1\right)\left(3-5x\right)}\)
⇒9 − 15x + 10x − 2 = 4
⇔3 − 5x = 0
⇔x =\(\frac{3}{5}\)
DM
1
TT
21 tháng 2 2020
\(ĐKXĐ:x\ne\frac{1}{5},x\ne\frac{3}{5}\)
Ta có : \(\frac{3}{5x-1}=\frac{2}{3-5x}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\)
\(\Leftrightarrow\frac{3\left(3-5x\right)}{\left(5x-1\right)\left(3-5x\right)}-\frac{2\left(5x-1\right)}{\left(5x-1\right)\left(3-5x\right)}+\frac{4}{\left(5x-1\right)\left(5x-3\right)}=0\)
\(\Rightarrow9-15x-10x+2+4=0\)
\(\Leftrightarrow-25x=-15\)
\(\Leftrightarrow x=\frac{3}{5}\) ( không thỏa mãn \(ĐKXĐ\) )
Vậy pt đã cho vô nghiệm
DH
4 tháng 7 2017
a, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(\Rightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)
\(\Rightarrow-80x=-480\Rightarrow x=6\)
b, \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x+12\right)+1\)
\(\Rightarrow15x+25-8x+12=5x+6x+36+1\)
\(\Rightarrow15x-8x-5x-6x=36+1-25-12\)
\(\Rightarrow-4x=0\Rightarrow x=0\)
c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow10x-12x-12x=-16+11+16-15\)
\(\Rightarrow-14x=-4\Rightarrow x=\dfrac{2}{7}\)
d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)
\(\Rightarrow5x-3\left[4x-2\left(4x-15x+6\right)\right]=182\)
\(\Rightarrow5x-3\left(4x-8x+30x-12\right)=182\)
\(\Rightarrow5x-12x+24x-90x+36=182\)
\(\Rightarrow-73x=182-36\)
\(\Rightarrow-73x=146\Rightarrow x=-2\)
Chúc bạn học tốt!!!
14 tháng 4 2023
a: \(=2x^3:\dfrac{-3}{2}x+4x:\dfrac{3}{2}x-5:\dfrac{3}{2}\)
=-4/3x^2+8/3-10/3
=-4/3x^2-2/3
d: \(\dfrac{3x^3-5x+2}{x-3}=\dfrac{3x^3-9x^2+9x^2-27x+22x-66+68}{x-3}\)
\(=3x^2+9x+22+\dfrac{68}{x-3}\)
AH
Akai Haruma
Giáo viên
14 tháng 6 2021
Lời giải:
Tại $x=4$ thì:
\(A=5(x^5-x^4+x^3-x^2+x-1)-1\)
\(=(x+1)(x^5-x^4+x^3-x^2+x-1)-1=x^6+1-1=x^6\)
\(=4^6=4096\)
24 tháng 8 2019
\(a,-5x\left(x-3\right)\left(2x+4\right)-\left(x+3\right)\left(x-3\right)+\left(5x-2\right)\left(3x+4\right)\)
\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)
\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)
\(=-10x^3+19x^2+74x+1\)
\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)
\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)
\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)
\(=-5x^4-11x^3+24x^2+12x+7\)
\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)
\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)
\(=-2x^2-27x+57\)
24 tháng 8 2019
\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)
\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)
\(=-x^3+4x^2+22x+5\)
\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)
\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)
\(=-9x^3-55x^2+4x+35\)
\(g,\left(x-1\right)^2-\left(x+2\right)^2\)
\(=x^2-2x+1-x^2-4x-4\)
\(=-6x-3\)
\(\dfrac{3}{{5x - 1}} + \dfrac{2}{{3 - 5x}} = \dfrac{4}{{\left( {1 - 5x} \right)\left( {x - 3} \right)}}\)
ĐKXĐ: \(x \ne \dfrac{1}{5};x\ne \dfrac{3}{5};x \ne 3\)
\( \Leftrightarrow 3\left( {3 - 5x} \right)\left( {x - 3} \right) + 2\left( {5x - 1} \right)\left( {x - 3} \right) + 4\left( {3 - 5x} \right) = 0\\ \Leftrightarrow 9x - 27 - 15{x^2} + 45x + 10{x^2} - 30x - 2x + 6 + 12 - 20x = 0\\ \Leftrightarrow - 5{x^2} + 2x - 9 = 0 \)
\(\Rightarrow\) Phương trình vô nghiệm.
Cảm ơn bn