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3) \(x^2-7x+6=0\)
\(\Leftrightarrow x^2-6x-x+6=0\)
\(\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
S=\(\left\{6;1\right\}\)
\(\)
a)\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6.\)
\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)
\(\Leftrightarrow-4x+7=0\)
\(\Leftrightarrow4x=7\Leftrightarrow x=1,75\)
\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10.\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-4x^2+1-10=0\)
\(\Leftrightarrow-24x+27=0\)
\(\Leftrightarrow24x=27\Leftrightarrow x=1,125\)
Rút gọn các phân thức:
a) \(\frac{\left(3x+2\right)^2-\left(x+2\right)^2}{x^3-x^2}=\frac{9x^2+12x+4-x^2-4x-4}{x^3-x^2}=\frac{8x^2+8x}{x^3-x^2}=\frac{8x\left(x+1\right)}{x^2\left(x-1\right)}=\frac{8\left(x+1\right)}{x-1}\)
b) \(\frac{x^4-1}{x^3+2x^2-x-2}=\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x^3-x\right)+\left(2x^2-2\right)}=\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x+2\right)\left(x^2-1\right)}=\frac{x^2+1}{x+2}\)
c) \(\frac{x^2+7x+12}{x^2+5x+6}=\frac{\left(x^2+3x\right)+\left(4x+12\right)}{\left(x^2+3x\right)+\left(2x+6\right)}=\frac{\left(x+3\right)\left(x+4\right)}{\left(x++3\right)\left(x+2\right)}=\frac{x+4}{x+2}\)
d) \(\frac{x^{10}-x^8+x^6-x^4+x^2-1}{x^4-1}=\frac{\left(x^{10}-x^8\right)+\left(x^6-x^4\right)+\left(x^2-1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\frac{\left(x^2-1\right)\left(x^8+x^4+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\frac{x^8+x^4+1}{x^2+1}\)
\(\left(x^2+5x\right)+10\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2+5x\right)-10\left(x^2+5x\right)+24=0\)
\(\Leftrightarrow\left(x^2+5x\right)\left(1-10\right)+14=0\)
\(\Leftrightarrow\left(-9\right)\left(x^2+5x\right)+14=0\)
\(\Leftrightarrow-9\left(x^2+5x\right)=-14\)
\(\Leftrightarrow x^2+5x=\frac{14}{9}\)
\(\Leftrightarrow x=0,2938.....\)
a) \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)
\(\Leftrightarrow-4x+7=0\Leftrightarrow x=\dfrac{7}{4}\)
b) \(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-1\right)=10\)
\(\Leftrightarrow4x^2-24x+36-4x^2+1-10=0\)
\(\Leftrightarrow-24x+27=0\Leftrightarrow x=\dfrac{9}{8}\)
c) đề có sai ko vậy bạn :
\(\left(x-4\right)^2-\left(x+2\right)\left(x-2\right)=6\)
\(\Leftrightarrow\left(x^2-8x+16\right)-\left(x^2-4\right)=6\)
\(\Leftrightarrow x^2-8x+16-x^2+4-6=0\)
\(\Leftrightarrow-8x+14=0\Leftrightarrow x=\dfrac{7}{4}\)
d) \(9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)
\(\Leftrightarrow9\left(x^2+2x+1\right)-\left(9x^2-4\right)=10\)
\(\Leftrightarrow9x^2+18x+9-9x^2+4-10=0\)
\(\Leftrightarrow18x+3=0\Leftrightarrow x=\dfrac{1}{6}\)
a) ( x - 2 )2 - ( x - 3 )( x+ 3) = 6
( x2 - 4x + 4 ) - ( x2 - 9 ) = 6
x2 - 4x + 4 - x2 + 9 = 6
-4x + 13 = 6
-4x = -7
x = 7/4
\(\frac{x-1}{x^2-x+1}-\frac{x+1}{x^2+x+1}=\frac{10}{x\left(x^4+x^2+1\right)}\)
<=> \(\frac{x\left(x-1\right)\left(x^2+x+1\right)}{x\left(x^2-x+1\right)\left(x^2+x+1\right)}-\frac{x\left(x+1\right)\left(x^2-x+1\right)}{x\left(x^2-x+1\right)\left(x^2+x+1\right)}-\frac{10}{x\left(x^2-x+1\right)\left(x^2+x+1\right)}=0\)
<=> \(x\left(x^3-1\right)-x\left(x^3+1\right)-10=0\)
<=> \(x^4-x-x^4-x=10\)
<=> - 2x = 10
<=> x = - 10 : 2
<=> x = - 5
Vậy S = {- 5}
a) \(6x^n:x^2-6x^n+2x.3x^{n-1}+2x\)
\(=6x^{n-2}-6x^n+6x^{n-1+1}+2x\)
\(=6x^{n-2}+2x\)
b) \(6^6-4^3.3^6+4^3\)
\(=6^6-\left(2^2\right)^3.3^6+4^3\)
\(=6^6-2^6.3^6+4^3\)
\(=6^6-\left(2.3\right)^6+4^3=4^3\)
c) \(10-x-x^3-x^2+x+x^2+x^3\)
\(=10\)
a ) \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Leftrightarrow x^2-4x+4-x^2+9=6\)
\(\Leftrightarrow-4x+13=6\)
\(\Leftrightarrow-4x=-7\)
\(\Leftrightarrow x=\frac{7}{4}\)
Vậy \(x=1\).
b ) \(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-1\right)=10\)
\(\Leftrightarrow4x^2-24x+36-4x^2+1=10\)
\(\Leftrightarrow-24x+37=10\)
\(\Leftrightarrow-24x=27\)
\(\Leftrightarrow x=\frac{9}{8}.\)
Mấy pài kia tương tự . :D
(x-6)2-x2=x-1
<=> (x-6-x)(x-6+x)=x-1
<=> -6(2x-6)=x-1
<=> -12x+36=x-1
<=> 13x=35
<=> x=\(\frac{35}{13}\)
Câu kia bn xem lại đề nhé
Chúc hok tốt
Thank you Trần Thùy Linh. Giúp mình câu này nhé:
( x - 3 ) nhân ( x - 2 ) = ( x - 4 )