2x^3-2x^2+18x
Bài:Chia 1 biến đã sắp xếp 1)(2x^3+11x^2+18x-3):(2x+3) 2)(2x^3+11x^2+18x-3):(3x+3) 3)(2x^3+9x^2+5x+41):(2x^2-x+9) 4)(13x+41x^2+35x^3-14):(5x-2) 5)(5x^2-3x^3+15-9x):(5-3x) 6)(-4x^2+x^3-20+5x):(x-4)
1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)
\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)
\(=x^2+4x+3-\dfrac{12}{2x+3}\)
tìm x biết ;
(x-3)^3 - (2x+1).(4x^2+1^2) = (x+2)^3-(2x-3)^3-18x (2x-3)
\(\left(x-3\right)^3-\left(2x+1\right)\left(4x^2+1^2\right)=\left(x+2\right)^3-\left(2x-3\right)^3-18x\left(2x-3\right)\)\(\Leftrightarrow x^3-9x^2+27x-27-\left(8x^3+2x+4x^2+1\right)=x^3+6x^2+12x+8-\left(8x^3-36x^2+54x-27\right)-36x^2+54x\)\(\Leftrightarrow x^3-9x^2+27x-27-8x^3-2x-4x^2-1-x^3-6x^2-12x-8+8x^3-36x^2+54x-27+36x^2-54x=0\)\(\Leftrightarrow-19x^2+13x-9=0\)
\(\Leftrightarrow-19\left(x^2-\dfrac{13}{19}+\dfrac{169}{1444}\right)-\dfrac{515}{76}=0\)
\(\Leftrightarrow-19\left(x-\dfrac{13}{38}\right)^2=\dfrac{515}{76}\Rightarrow\left[{}\begin{matrix}x-\dfrac{13}{38}=\sqrt{\dfrac{515}{76}}\\x-\dfrac{13}{38}=-\sqrt{\dfrac{515}{76}}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{515}{76}}+\dfrac{13}{38}\\x=-\sqrt{\dfrac{515}{76}}+\dfrac{13}{38}\end{matrix}\right.\)
tìm x
(x-3)3-(2x+1)( 4x2-2x+1)=(x+2)3-(2x-3)3-18x(2x-3)
(x3 - 9x2 + 27x - 27) - (8x3 + 1) - (x3 + 6x2 + 12x + 8) + (2x - 3)3 + 3.2x.3.(2x - 3) = 0
x3 - 9x2 + 27x - 27 - 8x3 - 1 - x3 - 6x2 - 12x - 8 + (2x)3 - 33 = 0
-15x2 + 15x - 63 = 0
Tìm x
( 2x + 3 )3 - ( x + 2 )2 + ( x + 5 )( x - 4 ) = ( 2x - 1)( 4x2 + 2x + 1 ) + 18x( 2x + 3 )
\(\left(2x+3\right)^3-\left(x+2\right)^2+\left(x+5\right)\left(x-4\right)=\left(2x-1\right)\left(4x^2+2x+1\right)+18x\left(2x+3\right)\)
\(\Leftrightarrow8x^3+36x^2+54x+27-x^2-4x-4+x^2+x-20=8x^3-1+36x^2+54x\)
\(\Leftrightarrow8x^3+36x^2+51x+3=8x^3+36x^2+54x-1\)
\(\Leftrightarrow-3x=-4\Leftrightarrow x=\dfrac{4}{3}\)
Chúc bạn học tốt !!!!!
giải phương trình 27x^3+18x^2-9x+(27x^2+2x-1)√2x-1-125=0
số nghiệm của phương trình 9x^4+18x^3-2x^2-2x-1=0
Tìm x
a) (12x-5)(3x-1)-(18x-1)(2x+3)=5
b) (x+2)(x-3)-(x-2)(x+5)=2(x+3)
c) (2x+3)(2x-1)-(2x+5)-(2x-3)=12
a, 2x²-18x+28=0. b, x-2/x²-9+3x-1/x+3=2x+1/x-3+1
\(a,2x^2-18x+28=0\)
\(\Leftrightarrow2\left(x^2-9x+14\right)=0\)
\(\Leftrightarrow x^2-9x+14=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=2\end{matrix}\right.\)
\(b,\dfrac{x-2}{x^2-9}+\dfrac{3x-1}{x+3}=\dfrac{2x+1}{x-3}+1\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{x-2}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(3x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(2x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-1=0\)
\(\Leftrightarrow\dfrac{x-2}{\left(x-3\right)\left(x+3\right)}+\dfrac{3x^2-10x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{2x^2+7x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=0\)\(\Rightarrow x-2+3x^2-10x+3-2x^2-7x-3-x^2+9=0\)
\(\Leftrightarrow-16x+7=0\)
\(\Leftrightarrow-16x=-7\)
\(\Leftrightarrow x=\dfrac{7}{16}\left(tm\right)\)
\(VậyS=\left\{\dfrac{7}{16}\right\}\)
a: =>x^2-9x+14=0
=>(x-2)(x-7)=0
=>x=2 hoặc x=7
b: =>x-2+(3x-1)(x-3)=(2x+1)(x+3)+x^2-9
=>x-2+3x^2-9x-x+3=2x^2+7x+3+x^2-9
=>3x^2-9x+1=3x^2+7x-6
=>-16x=-7
=>x=7/16
2x^3-12^2+18x