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12 tháng 7 2015
\(1;x^2+7x+10=0\Rightarrow x^2+2x+5x+10=0\Rightarrow x\left(x+2\right)+5\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x+5\right)=0\)
=> x + 2 = 0 hoặc x + 5 = 0
=> x = -2 hoặc x = - 5
2, x^4 - 5x^2 + 4 = 0
x^4 - 4x^2 - x^2 + 4 = 0
x^2 ( x^2 - 4) - ( x^2 - 4) = 0
( x^2 - 1)( x^2 - 4) = 0
( x - 1 )( x + 1)( x - 2)( x + 2) = 0
=> x= 1 hoặc x= -1 hoặc x = 2 hoặc x = - 2
Đúng cho mi8nhf mình giải tiếp cho
26 tháng 5 2016
a) (2x + 1)(3x - 2) = (5x - 8)(2x + 1)
<=> 6x2 - x - 2 = 10x2 - 11x - 8
<=> 6x2 - 10x2 - x + 11x -2 + 8 = 0
<=> -4x2 + 10x + 6 = 0
<=> -2 (2x2 - 5x - 3) = 0
<=> 2x2 - 5x - 3 = 0
<=> 2x2 - 6x + x - 3 = 0
<=> x (2x + 1) - 3 (2x + 1) = 0
<=> (x - 3) (2x + 1) = 0
* x - 3 = 0 => x = 3
* 2x + 1 = 0 => x = -1/2
S = {-1/2; 3}
b) 4x2 – 1 = (2x +1)(3x -5)
<=> 4x2 – 1 - (2x +1)(3x -5) = 0
<=> (2x - 1) (2x + 1) - (2x + 1)(3x - 5) = 0
<=> (2x + 1) (2x - 1 - 3x + 5) = 0
<=> (2x + 1) (-x + 4) = 0
* 2x + 1 = 0 <=> x = -1/2
* -x + 4 = 0 <=> x = 4
S = {-1/2; 4}
c) (x + 1)2 = 4(x2 – 2x + 1)
<=> (x + 1)2 - 4(x2 – 2x + 1) = 0
<=> (x + 1)2 - 4(x2 – 1)2 = 0
* (x + 1)2 = 0 <=> x = -1
* 4(x2 - 1)2 = 0 <=> x = 1 và x = -1
S = {-1; 1}
d) 2x3 + 5x2 – 3x = 0
<=> x (2x2 + 5x - 3) = 0
<=> x (2x2 + 6x - x - 3) = 0
<=> x [x(2x - 1) + 3 (2x - 1)] = 0
<=> x (2x - 1) (x + 3) = 0
* x = 0
* 2x - 1 = 0 <=> x = 1/2
* x + 3 = 0 <=> x = -3
S = { -3; 0; 1/2}
NT
1
17 tháng 5 2018
1, bạn làm hai cái mũ 4 ra là làm đc
2) Ta có : x4 - x3 - x + 1 = 0
<=> x3(x - 1) - (x - 1) = 0
<=> (x - 1)(x3 - 1) = 0
<=> (x - 1)(x - 1)(x2 + x + 1) = 0
<=> (x - 1)2(x2 + x + 1) = 0
<=> x - 1 = 0 (vì x2 + x + 1 > 0 với mọi x)
<=> x = 1
23 tháng 2 2020
a) Ta có: \(\left(2x+3\right)^2-\left(5+x\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3+5+x\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-3\\3x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-3}{2}\\x=\frac{-8}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-3}{2};\frac{-8}{3}\right\}\)
b) Ta có: \(\left(2x+5\right)^2-\left(2x-5\right)^2=6x+8\)
\(\Leftrightarrow\left(2x+5+2x-5\right)\left(2x+5-2x+5\right)-6x-8=0\)
\(\Leftrightarrow40x-6x-8=0\)
\(\Leftrightarrow34x=8\)
\(\Leftrightarrow x=\frac{8}{34}=\frac{4}{17}\)
Vậy: \(x=\frac{4}{17}\)
c) Ta có: \(\left(4x+3\right)^2=4\left(x-1\right)^2\)
\(\Leftrightarrow16x^2+24x+9=4\left(x^2-2x+1\right)\)
\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)
\(\Leftrightarrow12x^2+32x+5=0\)
\(\Leftrightarrow12x^2+2x+30x+5=0\)
\(\Leftrightarrow2x\left(6x+1\right)+5\left(6x+1\right)=0\)
\(\Leftrightarrow\left(6x+1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6x+1=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=-1\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{6}\\x=\frac{-5}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-1}{6};\frac{-5}{2}\right\}\)
d) Ta có: \(\left(7x-1\right)\left(3x-2\right)-49x^2+14x=1\)
\(\Leftrightarrow\left(7x-1\right)\left(3x-2\right)-\left(49x^2-14x+1\right)=0\)
\(\Leftrightarrow\left(7x-1\right)\left(3x-2\right)-\left(7x-1\right)^2=0\)
\(\Leftrightarrow\left(7x-1\right)\left[3x-2-\left(7x-1\right)\right]=0\)
\(\Leftrightarrow\left(7x-1\right)\left(3x-2-7x+1\right)=0\)
\(\Leftrightarrow\left(7x-1\right)\left(-4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x-1=0\\-4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=1\\-4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{7}\\x=\frac{-1}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{7};\frac{-1}{4}\right\}\)
1. Với \(x^2-2\ge0\Rightarrow\orbr{\begin{cases}x\ge\sqrt{2}\\x\le-\sqrt{2}\end{cases}}\)
Pt\(\Leftrightarrow x^4-4x^2+5x^2-10+8=0\Rightarrow x^4+x^2-2=0\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2+1\right)=0\Rightarrow x^2=2\Rightarrow\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}\left(tm\right)}\)
Với \(x^2-2< 0\Rightarrow-\sqrt{2}< x< \sqrt{2}\)
Pt \(\Leftrightarrow x^4-4x^2+10-5x^2+8=0\Leftrightarrow x^4-9x^2+18=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-6=0\\x^2-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=6\\x^2=3\end{cases}\left(l\right)}\)vì \(x\notin\left(-\sqrt{2};\sqrt{2}\right)\)
2. \(2x^4-20x+18=0\Rightarrow x^4-10x+9=0\)
\(\Rightarrow\left(x^4-x^3\right)+\left(x^3-x^2\right)+\left(x^2-x\right)-\left(9x-9\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x^3+x^2+x-9\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x^3+x^2+x-9=0\end{cases}}\)
\(\Rightarrow x=1\)