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19 tháng 6 2019
Giải pt :
a) \(2x\left(x+5\right)-\left(x-3\right)^2=x^2+6\)
\(\Leftrightarrow2x^2+10x-x^2+6x-9-x^2-6=0\)
\(\Leftrightarrow16x-15=0\)
\(\Leftrightarrow x=\frac{15}{16}\)
b) \(6\left(x-3\right)+\left(x-1\right)^2-\left(x+1\right)^2=2x\)
\(\Leftrightarrow2x-18=2x\)
\(\Leftrightarrow-18=0\)( vô lí )
=> x thuộc rỗng
c)d) tương tự
e) \(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
\(\Leftrightarrow\frac{5x-2}{6}+\frac{9-12x}{6}=\frac{12}{6}-\frac{2x+14}{6}\)
\(\Leftrightarrow5x-2+9-12x=12-2x-14\)
\(\Leftrightarrow-5x+9=0\)
\(\Leftrightarrow x=\frac{9}{5}\)
f) \(\frac{2x-1}{2}=\frac{2x+1}{4}-\frac{1-2x}{8}\)
\(\Leftrightarrow\frac{4\left(2x-1\right)}{8}=\frac{2\left(2x+1\right)}{8}-\frac{1-2x}{8}\)
\(\Leftrightarrow8x-4=4x+2-1+2x\)
\(\Leftrightarrow2x-5=0\)
\(\Leftrightarrow x=\frac{5}{2}\)
19 tháng 6 2019
Tìm x :
a) \(3x^3-27x=0\)
\(\Leftrightarrow3x\left(x^2-9\right)=0\)
\(\Leftrightarrow3x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b) \(2x^3-12x^2+18x=0\)
\(\Leftrightarrow2x\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
CB
1
16 tháng 8 2022
b: \(\Leftrightarrow x^3-4x-3\left(4x^2-4x+1\right)-2x-5=-6x^2-6x\)
\(\Leftrightarrow x^3-4x-12x^2+12x-3-2x-5=-6x^2-6x\)
\(\Leftrightarrow x^3-12x^2+6x-8+6x^2+6x=0\)
\(\Leftrightarrow x^3-6x^2+12x-8=0\)
=>x-2=0
hay x=2
c: \(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow6x^2+2-6x^2+12x-6=-10\)
=>12x-4=-10
=>12x=-6
hay x=-1/2
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
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...
TN
23 tháng 3 2020
\(a.\left(2x+5\right)\frac{6}{2}=75\\ \Leftrightarrow\left(2x+5\right)3=75\\ \Leftrightarrow6x+15=75\\\Leftrightarrow 6x=75-15\\\Leftrightarrow 6x=60\\ \Leftrightarrow x=10\)
\(b.\frac{x-3}{5}=6-\frac{1-2x}{3}\\ \Leftrightarrow\frac{3\left(x-3\right)}{15}=\frac{6.15}{15}-\frac{5\left(1-2x\right)}{15}\\ \Leftrightarrow3\left(x-3\right)=6.15-5\left(1-2x\right)\\ \Leftrightarrow3x-9=90-5+10x\\ \Leftrightarrow3x-9-90+5-10x=0\\ \Leftrightarrow-7x-94=0\\ \Leftrightarrow-7x=94\\ \Leftrightarrow x=\frac{-94}{7}\)
\(c.\frac{2x}{3}+\frac{2x-1}{6}=\frac{4-x}{3}\\ \Leftrightarrow\frac{2x.2}{6}+\frac{2x-1}{6}=\frac{2\left(4-x\right)}{6}\\ \Leftrightarrow2x.2+2x-1=2\left(4-x\right)\\ \Leftrightarrow4x+2x-1=8-2x\\ \Leftrightarrow4x+2x-1-8+2x=0\\ \Leftrightarrow8x-9=0\\ \Leftrightarrow8x=9\\ \Leftrightarrow x=\frac{9}{8}\)
\(d.\frac{x-1}{2}+\frac{x-1}{4}=\frac{1-x}{3}\\ \Leftrightarrow\frac{6\left(x-1\right)}{12}+\frac{3\left(x-1\right)}{12}=\frac{4\left(1-x\right)}{12}\\ \Leftrightarrow6\left(x-1\right)+3\left(x-1\right)=4\left(1-x\right)\\ \Leftrightarrow6x-6+3x-3=4-4x\\ \Leftrightarrow6x-6+3x-3-4+4x=0\\ \Leftrightarrow13x-13=0\\ \Leftrightarrow13x=13\\ \Leftrightarrow x=1\)
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
19 tháng 7 2017
6. \(-2x^2+2x-4=-2\left(x^2-x+2\right)\)
\(=-2\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+2\right)\)
\(=-2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right]\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{7}{2}\le\dfrac{-7}{2}< 0\)
-> ĐPCM.
7. 8. 9 Tương tự
10. \(6x^2+15x-21\)
\(=6\left(x^2+\dfrac{15}{6}x-\dfrac{7}{2}\right)\)
\(=6\left(x^2+2.x.\dfrac{5}{4}+\dfrac{25}{16}-\dfrac{25}{16}-\dfrac{7}{2}\right)\)
\(=6\left[\left(x+\dfrac{5}{4}\right)^2-\dfrac{81}{16}\right]\)
\(=6\left(x+\dfrac{5}{4}\right)^2-\dfrac{243}{8}\)
\(=\dfrac{243}{8}-6\left(x+\dfrac{5}{4}\right)^2\)
....
19 tháng 7 2017
1)
-2x2+2x-4
= -2(x2 -x +2)
= -2(x2 - 2.\(\dfrac{1}{2}\).x + \(\dfrac{1}{4}\)+\(\dfrac{7}{4}\))
= -2.(x-\(\dfrac{1}{2}\))2 - \(\dfrac{7}{2}\) \(\le\) - \(\dfrac{7}{2}\) với \(\forall\) x
=> -2x2+2x-4 luôn âm
=>đpcm
2)
-2x2 +6x -8
= -2 (x2 -3x + 4)
= -2(x2 - 2.\(\dfrac{3}{2}\).x +\(\dfrac{9}{4}\)+\(\dfrac{7}{4}\))
= -2.(x-\(\dfrac{3}{2}\))2 - \(\dfrac{7}{2}\) \(\le\) - \(\dfrac{7}{2}\) với \(\forall\) x
=> -2x2 +6x -8 luôn âm
=>đpcm
3)
-x2 + 4x -1
= - (x2 - 4x +1)
= -(x2 - 2.2.x + 4 -3)
= -(x - 2)2 +3 \(\le\) 3 với \(\forall\) x
=> -x2 + 4x -1 có thể không âm
=> sai đề
4)
-2x2 +6x -12
= -2(x2- 3x + 6)
= -2(x2 - 2.\(\dfrac{3}{2}\).x + \(\dfrac{9}{4}+\dfrac{15}{4}\))
= -2(x-\(\dfrac{3}{2}\))2 - \(\dfrac{15}{2}\) \(\le\) - \(\dfrac{15}{2}\) với \(\forall\) x
=> -2x2 +6x -12 luôn âm
=>đpcm
5)
6x2 +15x - 21
= 6(x2 + 2.\(\dfrac{15}{2}\)x + \(\dfrac{225}{4}\)- \(\dfrac{309}{4}\))
= 6.(x-\(\dfrac{15}{2}\))2 - \(\dfrac{927}{2}\) \(\ge\) - \(\dfrac{927}{2}\) với \(\forall\) x
=> 6x2 +15x - 21 có thể không âm
=> đề sai
1 tháng 6 2018
Tìm x:
1. \(25x^2-20x+4=0\)
⇔ \(\left(5x-2\right)^2=0\)
⇔ \(5x-2=0\)
⇔ \(5x=2\)
⇔ \(x=\dfrac{2}{5}\)
⇒ S = \(\left\{\dfrac{2}{5}\right\}\)
2. \(\left(2x-3\right)^2-\left(2x+1\right).\left(2x-1\right)=0\)
⇔ \(4x^2-12x+9-\left(4x^2-1\right)=0\)
⇔ \(4x^2-12x+9-4x^2+1=0\)
⇔ \(-12x+10=0\)
⇔ \(-12x=-10\)
⇔ \(x=\dfrac{5}{6}\)
⇒ S \(=\left\{\dfrac{5}{6}\right\}\)
3. \(\left(\dfrac{1}{2}x-1\right)\left(\dfrac{1}{2}x+1\right)-\left(\dfrac{1}{2}x-1\right)^2=0\)
⇔ \(\dfrac{1}{4}x^2-1-\left(\dfrac{1}{4}x^2-x+1\right)=0\)
⇔ \(\dfrac{1}{4}x^2-1-\dfrac{1}{4}x^2+x-1=0\)
⇔ \(-2+x=0\)
⇔ \(x=2\)
⇒ S \(=\left\{2\right\}\)
4. \(\left(2x-3\right)^2+\left(2x+5\right)^2=8\left(x+1\right)^2\)
⇔ \(4x^2-12x+9+4x^2+20x+25=8\left(x^2+2x+1\right)\)
⇔ \(8x^2+8x+34=8x^2+16x+8\)
⇔ \(8x+34=16x+8\)
⇔ \(8x-16x=8-34\)
⇔ \(-8x=-26\)
⇔ \(x=\dfrac{13}{4}\)
⇒ S \(=\left\{\dfrac{13}{4}\right\}\)
5.\(4x^2+12x-7=0\)
⇔ \(4x^2+14x-2x-7=0\)
⇔ \(2x\left(2x+7\right)-\left(2x+7\right)=0\)
⇔ \(\left(2x+7\right)\left(2x-1\right)=0\)
⇔ \(\left[{}\begin{matrix}2x+7=0\\2x-1=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{-7}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
⇒ S \(=\left\{\dfrac{-7}{2};\dfrac{1}{2}\right\}\)
6. \(\dfrac{1}{4}x^2+\dfrac{2}{3}x-\dfrac{5}{9}=0\)
⇔ \(9x^2+24x-20=0\)
⇔ \(9x^2+30x-6x-20=0\)
⇔ \(3x\left(3x+10\right)-2\left(3x+10\right)=0\)
⇔ \(\left(3x+10\right)\left(3x-2\right)=0\)
⇔ \(\left[{}\begin{matrix}3x+10=0\\3x-2=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
⇒ S \(=\left\{\dfrac{-10}{3};\dfrac{2}{3}\right\}\)
1 tháng 6 2018
7. \(24\dfrac{8}{9}-\dfrac{1}{4}x^2-\dfrac{1}{3}x=0\)
⇔ \(\dfrac{224}{9}-\dfrac{1}{4}x^2-\dfrac{1}{3}x=0\)
⇔ \(896-9x^2-12x=0\)
⇔ \(-896+9x^2+12x=0\)
⇔ \(9x^2+12x-896=0\)
⇔ \(9x^2-84x+96x-896=0\)
⇔ \(3x\left(3x-28\right)+32\left(3x-28\right)=0\)
⇔ \(\left(3x-28\right)\left(3x+32\right)=0\)
⇔ \(\left[{}\begin{matrix}3x-28=0\\3x+32=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=\dfrac{-32}{3}\end{matrix}\right.\)
⇒ S \(=\left\{\dfrac{-32}{3};\dfrac{28}{3}\right\}\)