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13 tháng 8 2019
a. 3(2x - 1)(3x - 1) - (2x - 3)(9x - 1) = 0
<=> 3(6x2-5x+1)-(18x2-29x+3)=0
<=> 14x=0
<=> x=0
b. (x - 3)(x - 5) + 3 (x - 1) = (x - 1)(x - 3)
<=> (x-3)(x-5-x+1)+3(x-1)=0
<=> -4(x-3)+3(x-1)=0
<=> -x+9=0
<=> x=9
c. (x - 1)(x - 2) - (x + 2)(x + 1) = 8
<=> x2-3x+2-(x2+3x+2)=8
<=> -6x=8
<=> \(x=\frac{-4}{3}\)
25 tháng 8 2019
a) \(\left(x-3\right)^2-4=0\)
\(\left(x-3\right)^2=0+4\)
\(\left(x-3\right)^2=4\)
\(\left(x-3\right)^2=\pm4\)
\(\left(x-3\right)^2=\pm2^2\)
\(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
\(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
\(4x^2+12x+9-4x^2+1=22\)
\(12x+10=22\)
\(12x=22-10\)
\(12x=12\)
\(x=1\)
c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)
\(16x^2-9-16x^2+40x-25=16\)
\(-34+40x=16\)
\(40x=16+34\)
\(40x=50\)
\(x=\frac{50}{40}=\frac{5}{4}\)
d) \(x^3-9x^2+27x-27=-8\)
\(x^3-9x^2+27x-27+8=0\)
\(x^3-9x^2+27x-19=0\)
\(\left(x^2-8x+19\right)\left(x-1\right)=0\)
Vì \(\left(x^2-8x+19\right)>0\) nên:
\(x-1=0\)
\(x=1\)
e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
\(x^3+2x^2+x+x^2+2x+1-x^2-3x^2=2\)
\(3x+1=2\)
\(3x=2-1\)
\(3x=1\)
\(x=\frac{1}{3}\)
NH
11 tháng 11 2020
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
HN
0
24 tháng 9 2021
\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)
\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)
7 tháng 9 2016
a ) \(x^3-6x^2+12x-8=0\)
\(\Leftrightarrow x^3-3.x^2.2+3.x.2^2-2^3=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow\left(x-2\right)=0\)
\(\Leftrightarrow x=2\)
b ) \(x^3+9x^2+27x+27=0\)
\(\Leftrightarrow x^3+3.x^2.3+3.x.3^2+3^3=0\)
\(\Leftrightarrow\left(x-3\right)^3=0\)
\(\Leftrightarrow\left(x-3\right)=0\)
\(\Leftrightarrow x=3\)
7 tháng 9 2016
a) x3 - 6x2 + 12x - 8 = 0
( x - 2 ) 3 = 0
x - 2 = 0
x = 2
b) x3 + 9x2 + 27x + 27 = 0
( x + 3 )3 = 0
x + 3 = 0
x = -3
LL
1
G
30 tháng 12 2020
\(a)\)\(\left(x+1\right)\left(x+3\right)-x\left(x-1\right)=8\)
\(\Leftrightarrow x^2+4x+3-x^2+x=8\)
\(\Leftrightarrow5x=5\)
\(\Leftrightarrow x=1\)
Vậy x = 1.
\(b)\)\(9x^2=1-\left(3x+1\right)\left(2x-9\right)\)
\(\Leftrightarrow\left(1-9x^2\right)-\left(3x+1\right)\left(2x-9\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(1-3x\right)-\left(3x+1\right)\left(2x-9\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(1-3x+9-2x\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(10-5x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\10-5x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=-1\\5x=10\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{3}\\x=2\end{cases}}\)
Vậy\(x=-\frac{1}{3}\)hoặc\(x=2\)
Dumflinz
NT
0
a) \(8\left(x-2\right)=3\)
\(\Leftrightarrow x-2=\dfrac{3}{8}\)
\(\Leftrightarrow x=\dfrac{3}{8}+2\)
\(\Leftrightarrow x=\dfrac{19}{8}\)
Vậy \(x=\dfrac{19}{8}\)
b) \(9^{x+1}-5.3^{2x}=324\)
\(\Rightarrow9^x.9-\left(3^2\right)^x.5=324\)
\(\Rightarrow9^x.9-9^x.5=324\)
\(\Rightarrow9^x\left(9-5\right)=324\)
\(\Rightarrow9^x.4=324\)
\(\Rightarrow9^x=\dfrac{324}{4}\)
\(\Rightarrow9^x=81\)
\(\Rightarrow9^x=9^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
a, \(8\left(x-2\right)=3\)
\(\Rightarrow x-2=\dfrac{3}{8}\Rightarrow x=\dfrac{19}{8}\)
b, \(9^{x+1}-5.3^{2x}=324\)
\(\Rightarrow9^x.9-5.9^x=324\)
\(\Rightarrow4.9^x=324\Rightarrow9^x=81=9^2\)
Vì \(9\ne\pm1;9\ne0\) nên \(x=2\)
Chúc bạn học tốt!!!