tìm x
a,x^3+3x^2=4x+12 b,49x^2=(3x+2)^2 c,3x^2(x-5)+12(5-x)=0 d,x^2(x-5)+45-9x=0
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NH
11 tháng 11 2020
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
NQ
10
20 tháng 3 2020
Bài 1.
\( a)\dfrac{{4x - 8}}{{2{x^2} + 1}} = 0 (x \in \mathbb{R})\\ \Leftrightarrow 4x - 8 = 0\\ \Leftrightarrow 4x = 8\\ \Leftrightarrow x = 2\left( {tm} \right)\\ b)\dfrac{{{x^2} - x - 6}}{{x - 3}} = 0\left( {x \ne 3} \right)\\ \Leftrightarrow \dfrac{{{x^2} + 2x - 3x - 6}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{x\left( {x + 2} \right) - 3\left( {x + 2} \right)}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{\left( {x + 2} \right)\left( {x - 3} \right)}}{{x - 3}} = 0\\ \Leftrightarrow x - 2 = 0\\ \Leftrightarrow x = 2\left( {tm} \right) \)
20 tháng 3 2020
Bài 2.
\(c)\dfrac{{x + 5}}{{3x - 6}} - \dfrac{1}{2} = \dfrac{{2x - 3}}{{2x - 4}}\)
ĐK: \(x\ne2\)
\( Pt \Leftrightarrow \dfrac{{x + 5}}{{3x - 6}} - \dfrac{{2x - 3}}{{2x - 4}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{x + 5}}{{3\left( {x - 2} \right)}} - \dfrac{{2x - 3}}{{2\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{2\left( {x + 5} \right) - 3\left( {2x - 3} \right)}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{ - 4x + 19}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow 2\left( { - 4x + 19} \right) = 6\left( {x - 2} \right)\\ \Leftrightarrow - 8x + 38 = 6x - 12\\ \Leftrightarrow - 14x = - 50\\ \Leftrightarrow x = \dfrac{{27}}{5}\left( {tm} \right)\\ d)\dfrac{{12}}{{1 - 9{x^2}}} = \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} \)
ĐK: \(x \ne -\dfrac{1}{3};x \ne \dfrac{1}{3}\)
\( Pt \Leftrightarrow \dfrac{{12}}{{1 - 9{x^2}}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12 - {{\left( {1 - 3x} \right)}^2} - {{\left( {1 + 3x} \right)}^2}}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{12 + 12x}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow 12 + 12x = 0\\ \Leftrightarrow 12x = - 12\\ \Leftrightarrow x = - 1\left( {tm} \right) \)
VN
16 tháng 1 2017
a)
<=> 3x - 3 + x - 2 = 2x - 2 - x + 1
<=> 3x + x - 2x + x = -2 + 1 + 3 + 2
<=> 3x = 4
<=> x = 4/3
Các câu sau làm tương tự
29 tháng 9 2018
\(\left(3x-3\right)+\left(x-2\right)=\left(2x-2\right)-\left(x-1\right)\)
<=> \(3x-3+x-2=2x-2-x+1\)
<=> \(4x-5=x-1\)
<=> \(3x=4\)
<=> \(x=\frac{4}{3}\)
Vậy....
3 tháng 10 2022
a: \(\Leftrightarrow\left(x+12-3x\right)\left(x+12+3x\right)=0\)
=>(-2x+12)(4x+12)=0
=>x=-3 hoặc x=6
b: \(\Leftrightarrow20x^3-15x^2+45x-45=0\)
=>\(x\simeq0.93\)
d: =>-4x+28+11x=-x+3x+15
=>7x+28=2x+15
=>5x=-13
=>x=-13/5
e: \(\Leftrightarrow4x^3-12x+x=4x^3-3x+5\)
=>-9x=-3x+5
=>-6x=5
=>x=-5/6
KN
2 tháng 8 2019
a) \(x^2-36=0\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow x=\pm\sqrt{36}=\pm6\)
KN
2 tháng 8 2019
b) \(\left(3x-5\right)^2-\left(x+6\right)^2=0\)
\(\Leftrightarrow\left(3x-5-x-6\right)\left(3x-5+x+6\right)=0\)
\(\Leftrightarrow\left(2x-11\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=\frac{-1}{4}\end{cases}}\)
22 tháng 3 2021
a) ĐKXĐ: \(x\ne3\)
Ta có: \(\dfrac{x^2-x-6}{x-3}=0\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)
Suy ra: x+2=0
hay x=-2(thỏa ĐK)
Vậy: S={-2}
22 tháng 3 2021
d)
ĐKXĐ: \(x\notin\left\{1;3\right\}\)
Ta có: \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)
\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\)
Suy ra: \(x^2-3x+5x-15=x^2-1-8\)
\(\Leftrightarrow2x-15+9=0\)
\(\Leftrightarrow2x-6=0\)
hay x=3(loại)
Vậy: \(S=\varnothing\)
\(a,x^3+3x^2=4x+12\)
\(x^2\left(x+3\right)=4\left(x+3\right)\)
\(\Rightarrow\left(x+3\right)\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm2\end{cases}}\)
\(b,49x^2=\left(3x+2\right)^2\)
\(7x=3x+2\)
\(\Rightarrow7x-3x=2\)
\(\Rightarrow4x=2\)
\(\Rightarrow x=\frac{1}{2}\)
các câu còn lại tương tự nha
\(a,x^3+3x^2=4x+12\)
\(x^3+3x^2-4x-12=0\)
\(\Rightarrow x^2\left(x+3\right)-4\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\\left(x+2\right)\left(x-2\right)=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm2\end{cases}}\)
\(b,49x^2=\left(3x+2\right)^2\)
\(\Rightarrow\left(7x\right)^2=\left(3x+2\right)^2\)
\(\Rightarrow7x=3x+2\)
\(\Rightarrow7x-3x=2\)
\(\Rightarrow4x=2\)
\(\Rightarrow x=\frac{1}{2}\)
\(c,3x^2\left(x-5\right)+12\left(5-x\right)=0\)
\(3x^2\left(x-5\right)-12\left(x-5\right)=0\)
\(\left(x-5\right)\left(3x^2-12\right)=0\)
\(\Rightarrow3.\left(x-5\right)\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\pm2\end{cases}}}\)
\(d,x^2\left(x-5\right)+45-9x=0\)
\(x^2\left(x-5\right)+9\left(5-x\right)=0\)
\(x^2\left(x-5\right)-9\left(x-5\right)=0\)
\(\left(x-5\right)\left(x^2-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x^2-9=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=\pm3\end{cases}}\)