2/3x-1/2x=3/4x^2
help me pl
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3 tháng 2 2017
a. \(3-4x\left(25-2x\right)-8x^2+x-300=0\)
\(\Leftrightarrow3-100x+8x^2-8x^2+x-300=0\)
\(\Leftrightarrow-297-99x=0\)
\(\Leftrightarrow x=3\)
Vậy \(n_0\) của PT là: x=3
b. \(\Leftrightarrow\frac{\left(2-6x\right)}{5}-2+\frac{3x}{10}=7-\frac{3x+3}{4}\)
\(\Leftrightarrow\frac{\left(4-12x\right)}{5}-\frac{20}{10}+\frac{3x}{10}=\frac{\left(28-3x-3\right)}{4}\)
\(\Leftrightarrow\frac{\left(-16-9x\right)}{10}=\frac{\left(25-3x\right)}{4}\)
\(\Leftrightarrow-64-36x=250-30x\)
\(\Leftrightarrow-6x=314\)
\(\Leftrightarrow x=-\frac{157}{3}\)
Vậy -\(n_0\) của PT là: \(x=\frac{-157}{3}\)
c. \(5x+\frac{2}{6}-8x-\frac{1}{3}=4x+\frac{2}{5}-5\)
\(\Leftrightarrow-3x=4x-\frac{23}{5}\)
\(\Leftrightarrow7x=\frac{23}{5}\)
\(\Leftrightarrow x=\frac{23}{35}\)
Vậy \(n_0\) của PT là: \(x=\frac{23}{35}\)
d. \(3x+\frac{2}{3}-3x+\frac{1}{6}=2x+\frac{5}{3}\)
\(\Leftrightarrow\frac{5}{6}=2x+\frac{5}{3}\)
\(\Leftrightarrow x=-\frac{5}{12}\)
Vậy \(n_0\) của Pt là: \(x=-\frac{5}{12}\)
26 tháng 7 2018
Thôi giúp luôn =.=
\(\left(x+3\right)^2+\left(2x+1\right)\left(3x-5\right)-2x\left(3-x\right)+4x+25\)
\(=x^2+6x+9+6x^2-10x+3x-5-6x+2x^2+4x+25\)
\(=9x^2-3x+29\)
ND
3 tháng 6 2018
1.
a) \(\left\{4x-2\left(x-3\right)-3\left[x-3\left(4-2x\right)+8\right]\right\}.\left(-3x\right)\)
= \(\left[4x-2x+6-3\left(x-12+6x\right)+8\right].\left(-3x\right)\)
\(=\left(4x-2x+6-3x+36-18x+8\right).\left(-3x\right)\)
= \(\left(-19x+50\right).\left(-3x\right)\)
\(=57x^2-150x\)
b) \(5\left(3x^2+4y^3\right)+\left[9\left(2x^2-y^3\right)-2\left(x^2-5y^3\right)\right]\)
\(=15x^2+20y^3+\left(18x^2-9y^3-2x^2+10y^3\right)\)
\(=15x^2+20y^3+16x^2+y^3\)
\(=31x^2+21y^3\)
2.
a) \(5x\left(1-2x\right)-3x\left(x+18\right)=0\)
\(\Rightarrow5x-10x^2-3x^2-54x=0\)
\(\Rightarrow-49x-13x^2=0\)
\(\Rightarrow x\left(-49-13x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-49}{13}\end{matrix}\right.\)
b)
\(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)
\(\Rightarrow5x-3\left[4x-2\left(4x-15x+6\right)\right]=182\)
\(\Rightarrow5x-3\left(4x-8x+30x-12\right)=182\)
\(\Rightarrow5x-12x+24x-90x+36=182\)
\(\Rightarrow-73x-146=0\)
\(\Rightarrow x=-2\)
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
S
14 tháng 12 2018
\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x^2+x}\)
b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep
c, tt
d, cx r
14 tháng 12 2018
a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)
\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)
\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)
\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)
30 tháng 9 2016
Bài 2
a) 4x(x-3)-3x+9
=4x(x-3)-3(x-3)
= (x-3)(4x-3)
b) x3+2x2-2x-4
=(x3+2x2)-(2x+4)
=x2(x+2)-2(x+2)
=(x+2)(x2-2)
c) 4x2-4y+4y-1
=4x2-1
=(2x-1)(2x+1)
d) x5-x
=x(x4-1)
=x(x2-1)(x2+1)
31 tháng 10 2018
a) 4x(x-3)-3x+9
= 4x(x-3) - 3(x-3)
= (x-3)(4x-3)
b)x3 + 2x2 - 2x - 4
= x2(x + 2) - 2(x + 2)
= (x+2)(x2-2)
c) 4x2 - 4y +4y -1
= [(2x)2-12] + (-4y+4y)
= (2x+1)(2x-1)
d) x5-x
= x(x4 - 1)
1 tháng 7 2017
Ta có : x2 + 3x
= x2 + \(2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\)
\(=\left(x+\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\)
AL
0
\(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{3}{4}x^2\\ \Leftrightarrow x\left(\dfrac{2}{3}-\dfrac{1}{2}\right)=\dfrac{3}{4}x^2\\\Leftrightarrow\dfrac{1}{6}x=\dfrac{3}{4}x^2\\ \Leftrightarrow\dfrac{3}{4}x^2-\dfrac{1}{6}x=0\\ \Leftrightarrow x\left(\dfrac{3}{4}x-\dfrac{1}{6}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{3}{4}x-\dfrac{1}{6}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{3}{4}x=\dfrac{1}{6}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{2}{9}\end{matrix}\right.\)
\(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{3}{4}x^2\\ \Leftrightarrow\dfrac{1}{6}x=\dfrac{3}{4}x^2\\ \Leftrightarrow\dfrac{3}{4}x\left(x-\dfrac{2}{9}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{2}{9}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{9}\end{matrix}\right.\)