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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: \(x \ne -\dfrac{1}{3};x \ne \dfrac{1}{3}\)
\( Pt \Leftrightarrow \dfrac{{12}}{{1 - 9{x^2}}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12 - {{\left( {1 - 3x} \right)}^2} - {{\left( {1 + 3x} \right)}^2}}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{12 + 12x}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow 12 + 12x = 0\\ \Leftrightarrow 12x = - 12\\ \Leftrightarrow x = - 1\left( {tm} \right) \)
8 tháng 6 2019
Tìm x:
1. 3x (2x + 3) - (2x + 5).(3x - 2) = 8
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=0 \)
\(\Leftrightarrow-2x+10=0\Leftrightarrow x=5\)
Vậy x = 5
2. 4x (x -1) - 3(x2 - 5) -x2 = (x - 3) - (x + 4)
\(\Leftrightarrow4x^2-4x-3x^2+15-x^2=x-3-x-4\)
\(\Leftrightarrow-4x+15=-7\)
\(\Leftrightarrow-4x=-22\Leftrightarrow x=\frac{11}{2}\)
Vậy x = \(\frac{11}{2}\)
3. 2 (3x -1) (2x +5) - 6 (2x - 1) (x + 2) = -6
\(\Leftrightarrow2\left(6x^2+15x-2x-5\right)-6\left(2x^2+4x-x-2\right)=-6\)
\(\Leftrightarrow12x^2+30x-4x-10-12x^2-24x+6x+12=-6\)
\(\Leftrightarrow8x=-8\Leftrightarrow x=-1\)
Vậy x = -1
4. 3 ( 2x - 1) (3x - 1) - (2x - 3) (9x - 1) - 3 = -3
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-18x^2+2x+27x-3-3=-3\)
\(\Leftrightarrow18x^2-6x-9x+3-18x^2+2x+27x-6=-3\)
\(\Leftrightarrow14x=0\Leftrightarrow x=0\)
Vậy x = 0
5. (3x - 1) (2x + 7) - ( x + 1) (6x - 5) = (x + 2) - (x - 5)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5=7\)
\(\Leftrightarrow18x=9\Leftrightarrow x=\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
6. 3xy (x + y) - (x + y) (x2 + y2 + 2xy) + y3 = 27
\(\Leftrightarrow3x^2y+3xy^2-\left(x+y\right)^3+y^3=27\)
\(\Leftrightarrow3x^2y+3xy^2-x^3-y^3-3x^2y-3xy^2+y^3=27\)
\(\Leftrightarrow-x^3=27\)
\(\Leftrightarrow x=-3\)
Vậy x = -3
7. 3x (8x - 4) - 6x (4x - 3) = 30
\(\Leftrightarrow24x^2-12x-24x^2+12x=30\)
\(\Leftrightarrow0=30\) ( vô lý)
Vậy pt vô nghiệm
8. 3x (5 - 2x) + 2x (3x - 5) = 20
\(\Leftrightarrow15x-6x^2+6x^2-10x=20\)
\(\Leftrightarrow5x=20\Leftrightarrow x=4\)
Vậy x = 4
25 tháng 6 2018
\(a.\dfrac{x+1}{2x+6}+2x=\dfrac{x+1+4x^2+12x}{2x+6}=\dfrac{4x^2+13x+1}{2x+6}\) ( x # -3)
\(b.\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\) ( x # - 3)
Các câu còn lại tương tự .
25 tháng 6 2018
\(a,\dfrac{x+1}{2x+6}+2x\)
\(=\dfrac{x+1}{2x+6}+\dfrac{2x\left(2x+6\right)}{2x+6}\)
\(=\dfrac{x+1+4x^2+12x}{2x+6}\)
\(=\dfrac{4x^2+13x+1}{2x+6}\)
\(b,\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3x}{2x^2+6x}-\dfrac{x-6}{2x^2-6x}\)
\(=\dfrac{2x-6}{2x^2+6x}=\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}=\dfrac{x-3}{x^2+3x}\)
\(c,\dfrac{x}{x-2y}+\dfrac{x}{x+2y}+\dfrac{4xy}{4y^2-x^2}\)
\(=\dfrac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\dfrac{x\left(x-2y\right)}{\left(x+2y\right)\left(x-2y\right)}-\dfrac{4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\dfrac{2x}{x+2y}\)
\(d,\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{4-9x^2}\)
\(=\dfrac{3x+2}{\left(3x+2\right)\left(3x-2\right)}-\dfrac{3x-2}{\left(3x+2\right)\left(3x-2\right)}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x+2}\)
TT
1 tháng 5 2018
a \(2x+2>4\\ \Leftrightarrow2\left(x+1\right)>4\\ \Leftrightarrow x+1>2\\ \Leftrightarrow x>1\)
b \(3x+2>-5\\ \Leftrightarrow3x>-7\\ \Leftrightarrow x>\dfrac{-7}{3}\)
c \(10-2x>2\\ \Leftrightarrow2\left(5-x\right)>2\\ \Leftrightarrow5-x>1\\ \Leftrightarrow-x>-4\\ \Leftrightarrow x< 4\)
d \(1-2x< 3\\ \Leftrightarrow-2x< 2\\ \Leftrightarrow2x>2\\ \Leftrightarrow x>1\)
KT
4 tháng 5 2018
a)2x+2>4
<=> 2x>4-2
<=>2x>2
<=>x>1
Vậy...
b)3x+2>-5
<=>3x>-5-2
<=>3x>-7
<=>x>\(\dfrac{-7}{3}\)
Vậy...
c)10-2x>2
<=>-2x>-10+2
<=>-2x>-8
<=>x<4
Vậy...
d)1-2x<3
<=>-2x<3-1
<=>-2x<2
<=>x>-1
Vậy...
e)10x+3-5\(\le\)14x+12
<=>10x-2\(\le\)14x+12
<=>10x-14x\(\le\)2+12
<=>-4x\(\le\)14
<=>x\(\ge\)\(\dfrac{-7}{2}\)
Vậy...
f)(3x-1)<2x+4
<=> 3x-2x<1+4
<=>x<5
Vậy...
11 tháng 8 2021
g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
cho tam giác ABC. các tia phân giác BD và CE. Tính góc A biết BE+CD+BC
giúp mình vs