Tìm x
\(\dfrac{15x}{x^2+3x-4}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)\)
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Y
26 tháng 11 2023
\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\left(dkxd:x\ne0,x\ne5\right)\\ =\dfrac{3x-x-1}{x\left(x-5\right)}=\dfrac{2x-1}{x^2-5x}\)
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\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\left(dkxd:x\ne0,y\ne-2\right)\\ =\dfrac{8}{4}.\dfrac{15x^2.x^3}{3x^2}=10x^3\)
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\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\left(dkxd:x\ne1,x\ne-1\right)\\ =\dfrac{8\left(y-1\right)}{3\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)^2}{4\left(y-1\right)^3}\\ =\dfrac{2\left(x-1\right)}{3\left(x+1\right)\left(y-1\right)^2}\)
OY
29 tháng 1 2018
\(\dfrac{15x}{x^2+3x-4}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-4\\x\ne1\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{15x-x^2-3x+4}{\left(x-1\right)\left(x+4\right)}=12\left(\dfrac{3x-3+x+4}{3\left(x+4\right)\left(x-1\right)}\right)\)
\(\Leftrightarrow\dfrac{3(12x-x^2+4)}{3\left(x-1\right)\left(x+4\right)}=12\left(\dfrac{4x+1}{3\left(x+4\right)\left(x-1\right)}\right)\)
\(\Leftrightarrow-x^2+12x+4=16x+4\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=-4\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
HT
18 tháng 8 2017
a) \(\Leftrightarrow\dfrac{15x}{x^2+3x-4}-1=\dfrac{12}{x+4}+\dfrac{4}{x-1}\)
\(\Leftrightarrow\dfrac{15x}{x^2+4x-x-4}-\dfrac{12}{x+4}-\dfrac{4}{x-1}=1\)
\(\Leftrightarrow\dfrac{15x}{\left(x-1\right)\left(x+4\right)}-\dfrac{12}{x+4}-\dfrac{4}{x-1}=1\)
\(\Leftrightarrow\dfrac{15x-12x+12-4x-16}{\left(x-1\right)\left(x+4\right)}=1\)
\(\Leftrightarrow\dfrac{-1}{x-1}=1\)
\(\Leftrightarrow x-1=-1\)
\(\Rightarrow x=0\)
tick cho t vs hik
HT
18 tháng 8 2017
b) \(\Leftrightarrow\left|x-2\right|+3=5\)
\(\Leftrightarrow\left|x-2\right|=5-3\)
\(\Leftrightarrow\left|x-2\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
KK
13 tháng 3 2017
b)
\(\dfrac{1}{x-1}+\dfrac{1}{x-2}=\dfrac{1}{x+2}+\dfrac{1}{x+1}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x+1}=\dfrac{1}{x+2}-\dfrac{1}{x-2}\)
\(\Leftrightarrow\dfrac{2}{\left(x-1\right)\left(x+1\right)}=\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{2}{x^2-1}=\dfrac{-4}{x^2-4}\)
\(\Leftrightarrow2x^2-8=-4x^2+4\) ( điều kiện \(x\ne\pm1,x\ne\pm2\) )
\(\Leftrightarrow6x^2=12\)
\(\Rightarrow x=\pm\sqrt{2}\)
KK
13 tháng 3 2017
a )
\(\dfrac{15x}{x^2+3x-4}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)\)
\(\Leftrightarrow\dfrac{15x-\left(x^2+3x-4\right)}{x^2+3x-4}=\dfrac{12}{x+4}+\dfrac{12}{3x-3}\)
\(\Leftrightarrow\dfrac{12x-x^2+4}{x^2+4x-x-4}=\dfrac{48x+12}{\left(x+4\right)\left(3x-3\right)}\)
\(\Leftrightarrow\dfrac{12x-x^2+4}{x\left(x+4\right)-\left(x+4\right)}=\dfrac{48x+12}{3\left(x+4\right)\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{12x-x^2+4}{\left(x+4\right)\left(x-1\right)}=\dfrac{48x+12}{3\left(x+4\right)\left(x-1\right)}\)
\(\Leftrightarrow12x-x^2+4=\dfrac{48x+12}{3}\)
\(\Leftrightarrow12x-x^2+4=16x+4\)
\(\Leftrightarrow x^2+8x=0\)
\(\Delta=b^2-4ac\)
\(\Delta=64\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-8+\sqrt{64}}{2}=0\left(nhận\right)\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-8-\sqrt{64}}{2}=-8\left(loại\right)\end{matrix}\right.\)
Do \(x=-8\) không thỏa mãn phương trình
Vậy \(x=0\)
19 tháng 12 2021
c: \(=\dfrac{3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}=\dfrac{3x}{x^2+1}\)
20 tháng 1 2022
Câu 1:
=>15(2x+1)-8(3x-1)=100
=>30x+15-24x+8=100
=>6x+23=100
hay x=77/6
Câu 2:
=>2(5x-3)+12-3(7x-1)=x+2
=>10x-6+12-21x+3-x-2=0
=>-12x=-7
hay x=7/12
Câu 3:
\(\Leftrightarrow2\left(x^2-1\right)+3\left(x+1\right)=2\left(x^2-4x+4\right)\)
\(\Leftrightarrow2x^2-2+3x+3-2x^2+8x-8=0\)
=>11x-7=0
hay x=-7/11
20 tháng 1 2022
Câu 4:
(x - 4)^3/6 + 1 = x(x + 1)/2 - (x - 5)(x + 5)/3
<=> (x - 4)^3 + 6/6 = x^2 + x/2 - x^2 - 25/3
<=> (x - 4)^3 + 6/6 = 3x^2 + 3x - 2x^2 + 50/6
<=> (x - 4)^3 + 6 = 3x^2 + 3x - 2x^2 + 50
<=> x^3 - 12x^2 + 48x - 58 = x^2 + 3x + 50
<=> x^3 -13x^2 + 45x - 108 = 0
Đến đây bạn bấm máy nhẩm nghiệm là ra nhé
Câu 5:
3(x + 2)^3/5 - (x - 1)^2/10 = (x - 3)(x + 3)/2
<=> 6(x + 2)^3 - (x - 1)^2/10 = 5(x^2 - 9)/10
<=> 6(x + 2)^3 - (x - 1)^2 = 5(x^2 - 9)
<=> 6x^3 + 36x^2 + 72x + 48 - x^2 + 2x - 1 - 5x^2 + 45 = 0
<=> 6x^3 + 30x^2 + 74x + 92 = 0
Đến đây bạn bấm máy nhẩm nghiệm như câu 4 nhé
11 tháng 7 2023
a: =>2(2x-3)-9=5-3x-2
=>4x-6-9=-3x+3
=>4x-15=-3x+3
=>7x=18
=>x=18/7
b: =>\(\dfrac{2}{3x}-\dfrac{3}{12}=\dfrac{4}{5}-\dfrac{21}{3x}+2\)
=>\(\dfrac{23}{3x}=\dfrac{4}{5}+2+\dfrac{1}{4}=\dfrac{61}{20}\)
=>3x=460/61
=>x=460/183
Theo đề ta có:
\(\dfrac{15x}{x^2+3x-4}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)\)
\(\Leftrightarrow\dfrac{15x}{\left(x-1\right)\left(x+4\right)}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3\left(x-1\right)}\right)\)
ĐKXĐ : \(x\ne1;x\ne-4\)
\(45x-3\left(x-1\right)\left(x+4\right)=36\left(x-1\right)+12\left(x+4\right)\)
\(\Leftrightarrow45x+3x^2-3x+12=36x-36+12x+48\)
\(\Leftrightarrow-3x^2-6x=0\)
\(\Leftrightarrow-3x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐK\right)\\x=-2\left(TMĐK\right)\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)