Tìm x: \(\left(2x+1\right)^2-3\left(2-x\right)^2+2\left(x-1\right)\left(x+2\right)=4-2\left(1-x\right)\)
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1 tháng 2 2022
\(\Leftrightarrow\left(2x-1\right)^3-\left(2x+3\right)^3-3\left(3x+1\right)^2-2\left(x-2\right)^2+\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3-36x^2-54x-27-3\left(9x^2+6x+1\right)-2\left(x^2-4x+4\right)+x^2+x-2=0\)
\(\Leftrightarrow-48x^2-48x-28-27x^2-18x-3-2x^2+8x-8+x^2+x-2=0\)
\(\Leftrightarrow-76x^2-57x-41=0\)
\(\Leftrightarrow76x^2+57x+41=0\)
\(\text{Δ}=57^2-4\cdot76\cdot41=-9215< 0\)
Vậy: Phương trình vô nghiệm
2 tháng 8 2017
1. (3x - 5)2 - (3x + 1)2 = 8
=> (3x - 5 - 3x - 1)(3x - 5 + 3x + 1) = 8
=> -6(6x - 4) = 8
=> 6x - 4 = \(\dfrac{-4}{3}\)
\(\Rightarrow x=\dfrac{4}{9}\)
2) 2x(8x - 3) - (4x - 3)2 = 27
=> 16x2 - 6x - 16x2 + 24x - 9 = 27
=> 18x - 9 = 27
=> x = 2
3) (2x - 3)2 - (2x + 1)2 = 3
=> (2x - 3 - 2x - 1)(2x - 3 + 2x +1) = 3
=> -4(4x - 2) = 3
=> 4x - 2 = \(\dfrac{-3}{4}\)
\(\Rightarrow x=\dfrac{5}{16}\)
4) (x + 5)2 - x2 = 45
=> (x + 5 - x)(x + 5 + x) = 45
=> 5(2x + 5) = 45
=> 2x + 5 = 9
=> x = 2
5) (x - 3)3 - (x - 3)(x2 + 3x + 9) + 9(x + 1)2 = 18
=> x3 - 9x2 + 27x - 27 - x3 + 27 + 9(x2 + 2x + 1) = 18
=> -9x2 + 27x + 9x2 + 18x + 9 = 18
=> 45x + 9 = 18
=> 45x = 9
=> x = \(\dfrac{1}{5}\)
6) x(x - 4)(x + 4) - (x - 5)(x2 + 5x + 25) = 13
=> x (x2 - 16) - (x3 - 125) = 13
=> x3 - 16x - x3 + 125 = 13
=> -16x = -112
=> x = 7.
NN
13 tháng 7 2016
\(\frac{\left(2-3x\right)^2}{3}-\frac{\left(1+2x\right)^2}{2}=\frac{3}{4}-2\left(x-1\right)\left(x+2\right)+x\left(1+x\right)\)
\(\frac{2^2-12x-3x^2}{3}-\frac{1^2+4x+2x^2}{2}=\frac{3}{4}-\left(x^2+x-2\right)+3x\)
\(\frac{2.\left(4-12x-3x^2\right)}{6}-\frac{3.\left(1+4x+2x^2\right)}{6}=\frac{11}{4}-x^2+2x\)
\(\frac{8-24x-6x^2}{6}-\frac{3+12x+2x^2}{6}=\frac{11}{4}-x^2+2x\)
\(\frac{8-24x-6x^2-3-12x-2x^2}{6}=\frac{11}{4}-x^2+2x\)
\(\frac{5-36x-8x^2}{6}=\frac{11}{4}-x^2+2x\)
Chỗ đây thì mk chịu
15 tháng 6 2016
(x-2)3+2.(1+2x)2=(1+x)3-3(x-2)2-(x-1)
<=>x3-6x2+12x-8+2.(1+4x+4x2)=1+3x2+3x+x3-3.(x2-4x+4)-x+1
<=>x3-6x2+12x-8+2+8x+8x2=1+3x2+3x+x3-3x2+12x-12-x+1
<=>x3+2x2+20x-6=x3+14x+2
<=>2x2+6x-8=0
<=>2x2-2x+8x-8=0
<=>2x.(x-1)+8.(x-1)=0
<=>2(x-1)(x+4)=0
<=>x-1=0 hoặc x+4=0
<=>x=1 hoặc x=-4
2 tháng 2 2022
\(\Leftrightarrow20\left(x^2-4x+3\right)-24\left(4x^2-4x+1\right)=15\left(9x^2+6x+1\right)+90x\left(x-1\right)\)
\(\Leftrightarrow20x^2-80x+60-96x^2+96x-24=135x^2+90x+15+90x^2-90x\)
\(\Leftrightarrow-301x^2+16x+21=0\)
\(\text{Δ}=16^2-4\cdot\left(-301\right)\cdot21=25540\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là
\(\left\{{}\begin{matrix}x_1=\dfrac{-16-\sqrt{25540}}{-602}=\dfrac{16+\sqrt{25540}}{602}\\x_2=\dfrac{16-\sqrt{25540}}{602}\end{matrix}\right.\)
31 tháng 8 2016
(x -2)\(^3\) +(3x-2)\(^2\) -5x (x+1) = (1+x)\(^3\) - 2(2x+1)\(^2\)
<=> (x\(^3\) -3.x\(^2\).2+3.x.2\(^2\) -2\(^3\)) + [(3x)\(^2\) - 2.3x.2 +2\(^2\)] - (5x.x+ 5x .1) = (1\(^3\) + 3.1\(^2\).x+ 3.1.x\(^2\) + x\(^3\) )- [2((2x)\(^2\) +2.2x.1+ 1\(^2\))]
<=> (x\(^3\) - 6x\(^2\) + 12x - 8) + (9x\(^2\) -12x+ 4)- (5x\(^2\) + 5x) = (1+3x + 3x\(^2\) + x\(^3\)) - [ 2.(4x\(^2\) + 4x +1]= (1+3x + 3x\(^2\) + x\(^3\)) - ( 8x\(^2\)+ 8x +2)
<=> x\(^3\) - 6x\(^2\) + 12x - 8 + 9x\(^2\) -12x+ 4 - 5x\(^2\) - 5x = 1+3x + 3x\(^2\) + x\(^3\) - 8x\(^2\) -8x - 2
<=> x\(^3\) +(- 6x\(^2\) + 9x\(^2\) - 5x\(^2\) ) +(12x- 12x - 5x) + (-8 +4) = (1-2) + ( 3x-8x) +( 3x\(^2\) - 8x\(^2\) ) + x\(^3\)
<=> x\(^3\) +( -2x\(^2\)) + (-5x) + (-4) = -1 + (-5x) +( -5x\(^2\))+ x\(^3\)<=> x\(^3\) -2x\(^2\) -5x-4= -1 - 5x - 5x\(^2\) +x\(^3\)<=> -2x\(^2\) -4 = -1 -5x\(^2\)<=> -2x\(^2\) + 5x\(^2\) = -1 +4 ( chuyển vế )<=> 3x\(^2\) = 3<=> x\(^2\) = 3:3<=> x\(^2\) = 1<=> x = \(\sqrt{1}\)<=> x= 1 CHÚC BẠN HỌC TỐT
\(\Leftrightarrow4x^2+4x+1-3\left(x^2-4x+4\right)+2\left(x^2+x-2\right)=4-2+2x\)
\(\Leftrightarrow4x^2+4x+1-3x^2+12x-12+2x^2+2x-4=2x+2\)
\(\Leftrightarrow3x^2+18x-15-2x-2=0\)
\(\Leftrightarrow3x^2+16x-17=0\)
\(\text{Δ}=16^2-4\cdot3\cdot\left(-17\right)=460>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-16-2\sqrt{115}}{6}=\dfrac{-8-\sqrt{115}}{3}\\x_2=\dfrac{-8+\sqrt{115}}{3}\end{matrix}\right.\)