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
Tìm x biết
a) (x+2).(x+3) - (x-2).(x+5)=10
b) (3x+2). (2x+9) - (x+2). (8x+11)=(x+1).(3-2x)
c) 3.(2x-1).(3x-1)-(2x-3).(9x-1)=0
d) (5x-8).(4x-5)-(3x-4).(2x+12)=12
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
a,4x-8/2x2+1=0 b,x2-x-6/x-3=0 c,x+5/3x-6-1/2=2x-3/2x-4 d,12/1-9x2=1-3x/1+3x-1+3x/1-3x
giúp mình với ;-;
ghi này chả hiểu j bn ak
ghi rõ ra coi
1.Giải phương trình:
a) 4x-8/2x^2+1 = 0
b)x^2-x-6/x-3 = 0
c)x+5/3x-6 - 1/2 = 2x-3/2x-4
d)12/1-9x^2 = 1-3x/1+3x - 1+3x/1-3x
2.Giải các phương trình:
a)5 + 96/x^2-16 = 2x-1/x+4 - 3x-1/4-x
b)3x+2/3x-2 - 6/2+3x = 9x^2/9x^2-4
c)x+1/x^2+x+1 - x-1/x^2-x+1 = 3/x(x^4+x^2+1)
Bài 1.
\( a)\dfrac{{4x - 8}}{{2{x^2} + 1}} = 0 (x \in \mathbb{R})\\ \Leftrightarrow 4x - 8 = 0\\ \Leftrightarrow 4x = 8\\ \Leftrightarrow x = 2\left( {tm} \right)\\ b)\dfrac{{{x^2} - x - 6}}{{x - 3}} = 0\left( {x \ne 3} \right)\\ \Leftrightarrow \dfrac{{{x^2} + 2x - 3x - 6}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{x\left( {x + 2} \right) - 3\left( {x + 2} \right)}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{\left( {x + 2} \right)\left( {x - 3} \right)}}{{x - 3}} = 0\\ \Leftrightarrow x - 2 = 0\\ \Leftrightarrow x = 2\left( {tm} \right) \)
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: \(x \ne -\dfrac{1}{3};x \ne \dfrac{1}{3}\)
\( Pt \Leftrightarrow \dfrac{{12}}{{1 - 9{x^2}}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12 - {{\left( {1 - 3x} \right)}^2} - {{\left( {1 + 3x} \right)}^2}}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{12 + 12x}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow 12 + 12x = 0\\ \Leftrightarrow 12x = - 12\\ \Leftrightarrow x = - 1\left( {tm} \right) \)
Bài 2.
\(a)5 + \dfrac{{96}}{{{x^2} - 16}} = \dfrac{{2x - 1}}{{x + 4}} - \dfrac{{3x - 1}}{{4 - x}}\)
ĐK: \(x\ne\pm4\)
\( Pt \Leftrightarrow \dfrac{{96}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} - \dfrac{{2x - 1}}{{x + 4}} - \dfrac{{3x - 1}}{{x - 4}} = - 5\\ \Leftrightarrow \dfrac{{96 - \left( {2x - 1} \right)\left( {x - 4} \right) - \left( {3x - 1} \right)\left( {x + 4} \right)}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = - 5\\ \Leftrightarrow \dfrac{{ - 5{x^2} - 2x + 96}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = - 5\\ \Leftrightarrow - 5{x^2} - 2x + 96 = - 5\left( {{x^2} - 16} \right)\\ \Leftrightarrow 96 - 2x = 80\\ \Leftrightarrow - 2x = - 16\\ \Leftrightarrow x = 8\left( {tm} \right)\\ b)\dfrac{{3x + 2}}{{3x - 2}} - \dfrac{6}{{2 + 3x}} = \dfrac{{9{x^2}}}{{9{x^2} - 4}} \)
ĐK: \(x \ne \dfrac{2}{3};x \ne -\dfrac{2}{3}\)
\( Pt \Leftrightarrow \dfrac{{3x + 2}}{{3x - 2}} - \dfrac{6}{{2 + 3x}} - \dfrac{{9{x^2}}}{{9{x^2} - 4}} = 0\\ \Leftrightarrow \dfrac{{{{\left( {2 + 3x} \right)}^2} - 6\left( {3x - 2} \right) - 9{x^2}}}{{\left( {3x - 2} \right)\left( {2 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{16 - 6x}}{{\left( {3 - 2x} \right)\left( {2 + 3x} \right)}} = 0\\ \Leftrightarrow 16 - 6x = 0\\ \Leftrightarrow - 6x = - 16\\ \Leftrightarrow x = \dfrac{8}{3}\left( {tm} \right)\\ c)\dfrac{{x + 1}}{{{x^2} + x + 1}} - \dfrac{{x - 1}}{{{x^2} - x + 1}} = \dfrac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}} \)
Ta có: \(x(x^4+x^2+1)=x[(x^2+1)^2-x^2]=x(x^2+x+1)(x^2-x+1)\)
Do \(\left\{ \begin{array}{l} {x^2} + x + 1 = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\forall x\\ {x^2} - x + 1 = \left( {x - \dfrac{1}{2}} \right) + \dfrac{3}{4} > 0\forall x \end{array} \right.\) nên phương trình xác định với mọi $x \ne 0$
Quy đồng, rồi biến đổi phương trình về dạng \(2x=3 \Leftrightarrow x =\dfrac{3}{2} (tm)\)
Tìm x biết :
a)(3x-3)+(x-2)=(2x-2)-(x-1).
b)(4x-3)+(3x+5)=3x-2.
c)(6x-8)-5(x+2)=2x-12.
d)(9x-2)-4(2x+5)=-12.
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
\(\left(3x-3\right)+\left(x-2\right)=\left(2x-2\right)-\left(x-1\right)\)
<=> \(3x-3+x-2=2x-2-x+1\)
<=> \(4x-5=x-1\)
<=> \(3x=4\)
<=> \(x=\frac{4}{3}\)
Vậy....
Tìm x
a)(x+12)^2-9x^2=0
b)20x^3-15x^2+7x=45-38x
c)16x^4-40x^3+10x^2=80x^3-20x^2+196x
d)-4.(x-7)+11x=-x+3.(x+5)
e)4x.(x^2-3)+x=4x^3-3x+5
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
GIẢI PT
a) 4x-8/ 2x^2 +1=0
b) x^2 -x-6 / x-3=0
c) x+5 /3x-6 - 1/2 =2x-3 /2x -4
d) 12 / 1-9x^2 = 1-3x / 1+3x - 1+3x / 1-3x
<=>4x-8=0
<=>4x=8
=.x=2(nhan)
Tìm x,biết:
a) x^2-36=0
b) (3x-5)^2-(x+6)^2=0
c) (5x-4)^2-49x^2=0
d)4x^3-36x=0
e) 2/3x(x^2-4)=0
a) \(x^2-36=0\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow x=\pm\sqrt{36}=\pm6\)
b) \(\left(3x-5\right)^2-\left(x+6\right)^2=0\)
\(\Leftrightarrow\left(3x-5-x-6\right)\left(3x-5+x+6\right)=0\)
\(\Leftrightarrow\left(2x-11\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=\frac{-1}{4}\end{cases}}\)
d) \(4x^3-36x=0\)
\(\Leftrightarrow4x\left(x^2-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x=0\\x^2-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}}\)
Vậy...
a,4x-8/2x^2+1=0 b,x^2-x-6/x-3=0 c,x+5/3x-6-1/2=2x-3/2x-4 d,12/1-9x^2=1-3x/1+3x-1+3x/1-3x
ai giúp với huhu ;-;
Giải phương trình:
a) \(\dfrac{x^2-x-6}{x-3}=0\)
b) \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)
c) \(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\)
d) \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)
e) \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)Thể loại truyện
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}
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: \(S=\varnothing\)