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Y
26 tháng 8 2018
a) \(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
\(=\dfrac{5x-1-5x+7}{3x+2-3x+1}\)
\(=\dfrac{-1+7}{2+1}\)
\(=\dfrac{6}{3}\)
\(=2\)
Với \(\dfrac{5x-1}{3x+2}=2\)
\(\Rightarrow5x-1=2\left(3x+2\right)\)
\(\Rightarrow5x-1-2\left(3x+2\right)=0\)
\(\Rightarrow5x-1-6x-4=0\)
\(\Rightarrow-x-5=0\)
\(\Rightarrow x=-5\)
27 tháng 7 2016
1. \(\left(4x+7\right)\left(3x+4\right)=\left(12x-5\right)\left(x-1\right)\)
\(12x^2+16x+21x+28=12x^2-12x-5x+5\)
\(12x^2+37x+28-12x^2+17x-5=0\)
54x+23=0
54x=-23
x=-23/54
27 tháng 7 2016
2. \(\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(15x^2-5x-3x+1=15x^2+10x-21x-14\)
\(15x^2-8x+1-15x^2+11x+14=0\)
3x+15=0
3x=-15
x=-5
3 tháng 7 2017
a, \(\left(x^2-y^2\right)-\left(5x+5y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
b, \(5x^3-5x^2y-10x^2+10xy\)
\(=5x^2\left(x-y\right)-10x\left(x-y\right)\)
\(=\left(5x-10x\right)\left(x-y\right)=5x\left(x-2\right)\left(x-y\right)\)
c, \(2x^2-5x=x\left(2x-5\right)\)
f, \(3x^2-7x-10=3x^2+3x^2-10x-10\)
\(=3x^2\left(x+1\right)-10\left(x+1\right)=\left(3x^2-10\right)\left(x+1\right)\)
d, \(x^3-3x^2+1-3x=x^3-3x^2-3x+1\)
\(=x^3+x^2-4x^2-4x+x+1\)
\(=x^2\left(x+1\right)-4x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x^2-4x+1\right)\left(x+1\right)\)
e, \(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left[\left(x-y\right)^2-4z^2\right]\)
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
g, \(x^4+1-2x^2=\left(x^2-1\right)^2\)
h, \(3x^2-3y^2-12x+12y=3\left(x^2-y^2\right)-12\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\)
\(=\left(x-y\right)\left(3x+3y-12\right)\)
\(=3\left(x-y\right)\left(x+y-4\right)\)
j, \(x^2-3x+2=x^2-2x-x+2=x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-1\right)\left(x-2\right)\)
3 tháng 7 2017
a. \(\left(x^2-y^2\right)-5\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-5\right)\)
b. \(5x^3-5x^2y-10x^2+10xy\)
\(=5\left[\left(x^3-x^2y\right)-\left(2x^2-2xy\right)\right]\)
\(=5\left[x^2\left(x-y\right)-2x\left(x-y\right)\right]\)
\(=5x\left(x-y\right)\left(x-2\right)\)
c. \(2x^2-5x=x\left(2x-5\right)\)
d. \(x^3-3x^2+1-3x\)
\(=\left(x^3+1\right)-\left(3x^2+3x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left[x^2-x+1-3x\right]\)
\(=\left(x+1\right)\left[x^2-4x+1\right]\)
\(=\left(x+1\right)\left[x^2-2.x.2+2^2-2^2+1\right]\)
\(=\left(x+1\right)\left[\left(x-2\right)^2-3\right]\)
\(=\left(x+1\right)\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)\)
e. \(3x^2-6xy+3y^2-12z^2\)
\(=3\left[x^2-2xy+y^2-4z^2\right]\)
\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=3\left(x-y+2z\right)\left(x-y-2z\right)\)
f. \(3x^2-7x-10\)
\(=3x^2-7x-7-3\)
\(=\left(3x^2-3\right)-\left(7x+7\right)\)
\(=3\left(x^2-1\right)-7\left(x+1\right)\)
\(=3\left(x+1\right)\left(x-1\right)-7\left(x+1\right)\)
\(=\left(x+1\right)\left[3\left(x-1\right)-7\right]\)
\(=\left(x+1\right)\left(3x-8\right)\)
g. \(x^4+1-2x^2=\left(x^2\right)^2-2.x^2+1=\left(x^2-1\right)^2\)
\(=\left(x+1\right)^2\left(x-1\right)^2\)
h. \(3x^2-3y^2-12x+12y\)
\(=3\left(x^2-y^2\right)-12\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\)
\(=\left(x-y\right)\left[3\left(x+y\right)-12\right]\)
\(=\left(x-y\right).3.\left(x+y-4\right)\)
j. \(x^2-3x+2=x^2-x-2x+2\)
\(=x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(x-2\right)\)
P/s: ( Có j sai ns nha nhiều số quá tui rối đầu )
KT
9 tháng 8 2018
a) \(2x^3-5x^2+8x-3\)
\(=\left(2x^3-4x^2+6x\right)-\left(x^2-2x+3\right)\)
\(=2x\left(x^2-2x+3\right)-\left(x^2-2x+3\right)\)
\(=\left(2x-1\right)\left(x^2-2x+3\right)\)
b) bn ktra lại đề
c) \(12x^2+5x-12y^2+12y-10xy-3\)
\(=\left(12x^2-18xy+9x\right)+\left(8xy-12y^2+6y\right)-\left(4x-6y+3\right)\)
\(=3x\left(4x-6y+3\right)+2y\left(4x-6y+3\right)-\left(4x-6y+3\right)\)
\(=\left(3x+2y-1\right)\left(4x-6y+3\right)\)
BD
9 tháng 8 2018
a﴿ 2x − 5x + 8x − 3 = 2x − 4x + 6x − x − 2x + 3 = 2x x − 2x + 3 − x − 2x + 3 = 2x − 1 x − 2x + 3
c﴿ 12x + 5x − 12y + 12y − 10xy − 3
= 12x − 18xy + 9x + 8xy − 12y + 6y − 4x − 6y + 3
= 3x 4x − 6y + 3 + 2y 4x − 6y + 3 − 4x − 6y + 3
= 3x + 2y − 1 4x − 6y + 3 3 2
31 tháng 7 2018
a) ta có : \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x=30\Leftrightarrow15x=30\Leftrightarrow x=2\)
b) điều kiện : \(x\ne\dfrac{1}{5};x\ne1;x\ne\dfrac{3}{5}\)
ta có : \(\dfrac{3}{5x-1}+\dfrac{2}{3-3x}=\dfrac{4}{\left(1-5x\right)\left(5x-3\right)}\)
\(\Leftrightarrow\dfrac{3\left(3-3x\right)+2\left(5x-1\right)}{\left(5x-1\right)\left(3-3x\right)}=\dfrac{4}{\left(1-5x\right)\left(5x-3\right)}\)
\(\Leftrightarrow\dfrac{x+7}{3-3x}=\dfrac{4}{3-5x}\Leftrightarrow\left(x+7\right)\left(3-5x\right)=4\left(3-3x\right)\)
\(\Leftrightarrow-5x^2-20+9=0\)
ta có : \(\Delta'=\left(10\right)^2+5\left(9\right)=145>0\) \(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(x=\dfrac{10+\sqrt{145}}{-5};x=\dfrac{10-\sqrt{145}}{-5}\)
9 tháng 12 2018
a, 15x3y5z : 5x2y3 = 3xy2z.
b, 12x4y2 : ( - 9xy2 ) = \(\frac{3}{4}x^3\).
c, ( 30x4y3 - 25x2y3 - 3x4y4 ) : 5x2y3 = \(6x^2-5-\frac{3}{5}x^2y.\)
d, ( 4x4 - 8x2y2 + 12x5y ) : ( - 4x2 ) = -x2 + 2y2 - 3x3y.
18 tháng 4 2020
a) ĐKXĐ: \(x\ne1\)
Ta có: \(\frac{7x-3}{x-1}=\frac{2}{3}\)
\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow21x-9-2x+2=0\)
\(\Leftrightarrow19x-7=0\)
\(\Leftrightarrow19x=7\)
hay \(x=\frac{7}{19}\)
Vậy: \(x=\frac{7}{19}\)
b) ĐKXĐ: \(x\ne-1\)
Ta có: \(\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}\)
\(\Leftrightarrow4\left(3-7x\right)=1+x\)
\(\Leftrightarrow12-28x-1-x=0\)
\(\Leftrightarrow11-29x=0\)
\(\Leftrightarrow29x=11\)
hay \(x=\frac{11}{29}\)
Vậy: \(x=\frac{11}{29}\)
c) ĐKXĐ: \(x\notin\left\{\frac{-2}{3};\frac{1}{3}\right\}\)
Ta có: \(\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\)
\(\Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(\Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\)
\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)
\(\Leftrightarrow15x^2-8x+1-15x^2+11x+14=0\)
\(\Leftrightarrow3x+15=0\)
\(\Leftrightarrow3x=-15\)
hay x=-5
Vậy: x=-5
d) ĐKXĐ: \(x\notin\left\{1;\frac{-4}{3}\right\}\)
Ta có: \(\frac{4x+7}{x-1}=\frac{12x+5}{3x+4}\)
\(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\)
\(\Leftrightarrow12x^2+16x+21x+28=12x^2-12x+5x-5\)
\(\Leftrightarrow12x^2+37x+28=12x^2-7x-5\)
\(\Leftrightarrow12x^2+37x+28-12x^2+7x+5=0\)
\(\Leftrightarrow44x+33=0\)
\(\Leftrightarrow44x=-33\)
hay \(x=\frac{-3}{4}\)
Vậy: \(x=\frac{-3}{4}\)
18 tháng 4 2020
a)
\(\frac{7x-3}{x-1}=\frac{2}{3}\\ \Leftrightarrow\frac{21x-9}{3\cdot\left(x-1\right)}-\frac{2x-2}{3\cdot\left(x-1\right)}=0\\ \Leftrightarrow\frac{21x-9-2x+2}{3\cdot\left(x-1\right)}=0\\ \Leftrightarrow\frac{19x-7}{3\cdot\left(x-1\right)}=0\\ \Rightarrow19x-7=0\\ \Rightarrow x=\frac{7}{19}\)
b)
\(\frac{2\cdot\left(3-7x\right)}{1+x}=\frac{1}{2}\\ \Leftrightarrow\frac{12-28x}{2\cdot\left(1+x\right)}-\frac{1+x}{2\cdot\left(1+x\right)}=0\\ \Leftrightarrow\frac{12-28x-1-x}{2\cdot\left(1+x\right)}=0\\ \Leftrightarrow\frac{11-29x}{2\cdot\left(1+x\right)}=0\\\Rightarrow11-29x=0\\ \Rightarrow x=\frac{11}{29}\)
c)
\(\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\\ \Leftrightarrow\frac{15x^2-8x+1}{\left(3x+2\right)\cdot\left(3x-1\right)}-\frac{15x^2-11x-14}{\left(3x+2\right)\cdot\left(3x-1\right)}=0\\ \Leftrightarrow\frac{15x^2-8x+1-15x^2+11x+14}{\left(3x+2\right)\cdot\left(3x-1\right)}=0\\ \Leftrightarrow\frac{3x+15}{\left(3x+2\right)\cdot\left(3x-1\right)}=0\\ \Rightarrow3x+15=0\\ \Rightarrow x=-5\)
d)
\(\frac{4x+7}{x-1}=\frac{12x+5}{3x+4}\\ \Leftrightarrow\frac{12x^2+37x+28}{\left(x-1\right)\cdot\left(3x+4\right)}-\frac{12x^2-7x-5}{\left(x-1\right)\cdot\left(3x+4\right)}=0\\ \Leftrightarrow\frac{12x^2+37x+28-12x^2+7x+5}{\left(x-1\right)\cdot\left(3x+4\right)}=0\\ \Leftrightarrow\frac{44x+33}{\left(x-1\right)\cdot\left(3x+4\right)}=0\\ \Leftrightarrow44x+33=0\\ \Rightarrow x=-\frac{3}{4}\)
25 tháng 5 2022
4: \(3x^3-5x^2+5x-2\)
\(=3x^3-2x^2-3x^2+2x+3x-2\)
\(=x^2\left(3x-2\right)-x\left(3x-2\right)+\left(3x-2\right)\)
\(=\left(3x-2\right)\left(x^2-x+1\right)\)
5: \(5x^3-12x^2+14x-4\)
\(=5x^3-2x^2-10x^2+4x+10x-4\)
\(=\left(5x-2\right)\left(x^2-2x+2\right)\)
A
4 tháng 4 2020
a) \(\frac{9x-0,7}{4}\)\(-\)\(\frac{5x-1,5}{7}\)=\(\frac{12x-2,1}{3}\)
⇔\(\frac{21\left(9x-0,7\right)}{84}\)\(-\)\(\frac{12\left(5x-1,5\right)}{84}\)=\(\frac{28\left(12x-2,1\right)}{84}\)
⇒189x\(-\)14,7\(-\)60x+18=336x\(-\)58,8
⇔\(-\)207x=\(-\)62,1
⇔x=\(\frac{3}{10}\)
Vậy tập nghiệm của phương trình đã cho là:S={\(\frac{3}{10}\)}
3 tháng 9 2016
1) \(\frac{8xy\left(3x-1\right)^3}{12x^3\left(1-3x\right)}=-\frac{8xy\left(3x-1\right)^3}{12x^3\left(3x-1\right)}=-\frac{2y\left(3x-1\right)^2}{3x^2}\)
2) \(\frac{5x^3+5x}{x^4-1}=\frac{5x\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2-1\right)}=\frac{5x}{x^2-1}\)
3) \(\frac{9-\left(x+5\right)^2}{x^2+4x+4}=\frac{\left(3-x-5\right)\left(3+x+5\right)}{\left(x+2\right)^2}=\frac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}=-\frac{x+8}{x+2}\)
3) \(\frac{32x-8x^2+2x^3}{x^3+64}=\frac{2x\left(16-4x+x^2\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{2x}{x+4}\)
-ĐKXĐ: \(-3x\ge0\Leftrightarrow x\le0\)
\(\left|5x^2-12x\right|=-3x\)
\(\Leftrightarrow\left[{}\begin{matrix}5x^2-12x=-3x\\5x^2-12x=3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x^2-9x=0\\5x^2-15x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(5x-9\right)=0\\5x\left(x-5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5x-9=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=\dfrac{9}{5}\left(loại\right)\\x=5\left(loại\right)\end{matrix}\right.\)
-Vậy \(S=\left\{0\right\}\)
cảm ơn cậu nha!