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Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Sorry mình nhầm câu a
a) (2x - 1)2 + (x + 3)2 - 5(x + 7)(x - 7) = 0
b) (x + 2)(x2 - 2x + 4) - x(x2 + 2) = 15
c) (x + 3)3 - x(3x + 1)2 + (2x - 1)(4x2 - 2x + 1) = 28
d) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
Giải:
a) (2x - 1)2 + (x + 3)2 - 5(x + 7)(x - 7) = 0
\(\Leftrightarrow\) 4x2 - 4x + 1 + x2 + 6x + 9 - 5(x2 - 49) = 0
\(\Leftrightarrow\) 4x2 - 4x + 1 + x2 + 6x + 9 - 5x2 + 245 = 0
\(\Leftrightarrow\) 2x + 255 = 0
\(\Leftrightarrow\) 2x = - 255
\(\Leftrightarrow\) x = - 255 : 2
\(\Leftrightarrow\) x = \(-\frac{255}{2}\)
Vậy x = \(-\frac{255}{2}\)
b) (x + 2)(x2 - 2x + 4) - x(x2 + 2) = 15
\(\Leftrightarrow\) x3 + 8 - x3 - 2x = 15
\(\Leftrightarrow\) 8 - 2x = 15
\(\Leftrightarrow\) 2x = 8 - 1
\(\Leftrightarrow\) 2x = - 7
\(\Leftrightarrow\) x = - 7 : 2
\(\Leftrightarrow\) x = \(-\frac{7}{2}\)
Vậy x = \(-\frac{7}{2}\)
c) (x + 3)3 - x(3x + 1)2 + (2x - 1)(4x2 - 2x + 1) = 28
\(\Leftrightarrow\) x3 + 6x2 + 27x + 27 - x(9x2 + 6x + 1) + 8x3 - 1 = 28
\(\Leftrightarrow\) x3 + 6x2 + 27x + 27 - 9x3 - 6x2 - x + 8x3 - 1 = 28
\(\Leftrightarrow\) 26x + 26 = 28
\(\Leftrightarrow\) 26x = 28 - 26
\(\Leftrightarrow\) 26x = 2
\(\Leftrightarrow\) x = 2 : 26
\(\Leftrightarrow\) x = \(\frac{1}{13}\)
Vậy x = \(\frac{1}{13}\)
d) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
\(\Leftrightarrow\) x6 - 2x2 + 1 - (x6 - 1) = 0
\(\Leftrightarrow\) x6 - 2x2 + 1 - x6 + 1 = 0
\(\Leftrightarrow\) -2x2 + 2 = 0
\(\Leftrightarrow\) -2x2 = - 2
\(\Leftrightarrow\) x2 = - 2 : (- 2)
\(\Leftrightarrow\) x2 = 1
\(\Leftrightarrow\) x = 1 hoặc x = - 1
Vậy x \(\in\) {1; - 1}
a: \(\dfrac{4x^4+3x^3}{-x^3}+\dfrac{15x^3+6x}{3x}=0\)
\(\Leftrightarrow-4x-3+5x^2+2=0\)
\(\Leftrightarrow5x^2-4x-1=0\)
\(\Leftrightarrow5x^2-5x+x-1=0\)
=>(x-1)(5x+1)=0
=>x=1 hoặc x=-1/5
b: \(\dfrac{x^2-\dfrac{1}{2}x}{2x}-\dfrac{\left(3x-1\right)^2}{3x-1}=0\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{4}-3x+1=0\)
\(\Leftrightarrow\dfrac{-5}{2}x+\dfrac{3}{4}=0\)
\(\Leftrightarrow-\dfrac{5}{2}x=-\dfrac{3}{4}\)
hay \(x=\dfrac{3}{4}:\dfrac{5}{2}=\dfrac{3}{4}\cdot\dfrac{2}{5}=\dfrac{6}{20}=\dfrac{3}{10}\)
\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\\ \left(3x+2\right)\left(x^2-1\right)-\left(9x^2-4\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left[\left(x-1\right)-\left(3x-2\right)\right]=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(1-2x\right)=0\\ \left[{}\begin{matrix}3x+2=0\\x+1=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.\)
\(b.x\left(x+3\right)\left(x-3\right)-\left(x+2\right)\left(x^2-2x+4\right)=0\\ x\left(x^2-9\right)-\left(x^3+8\right)=0\\ x^3-9x-x^3-8=0\\ -9x-8=0\\ -9x=8\\ x=\frac{-8}{9}\)
\(c.2x\left(x-3\right)+5\left(x-3\right)=0\\ \left(x-3\right)\left(2x+5\right)=0\\ \left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-5}{2}\end{matrix}\right.\)
\(d.\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\\ \left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\\ \left(3x-1\right)\left[\left(x^2+2\right)-\left(7x-10\right)\right]=0\\ \left(3x-1\right)\left(x^2+2-7x+10\right)=0\\ \left(3x-1\right)\left(x^2-7x+12\right)=0\\ \left(3x-1\right)\left(x^2-4x-3x+12\right)=0\\ \left(3x-1\right)\left[\left(x^2-4x\right)+\left(-3x+12\right)\right]=0\\ \left(3x-1\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]=0\\ \left(3x-1\right)\left(x-4\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}3x-1=0\\x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\\x=3\end{matrix}\right.\)
\(e.\left(x+2\right)\left(3-4x\right)=x^2+4x+4\\ \left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\\ \left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\\ \left(x+2\right)\left[\left(3-4x\right)-\left(x+2\right)\right]=0\\ \left(x+2\right)\left(3-4x-x-2\right)=0\\ \left(x+2\right)\left(1-5x\right)=0\left[{}\begin{matrix}x+2=0\\1-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{1}{5}\end{matrix}\right.\)
\(f.x\left(2x-7\right)-4x+14=0\\ x\left(2x-7\right)-2\left(2x-7\right)=0\\ \left(2x-7\right)\left(x-2\right)=0\\ \left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=2\end{matrix}\right.\)
\(g.3x-15=2x\left(x-5\right)\\ 3\left(x-5\right)=2x\left(x-5\right)\\ 3\left(x-5\right)-2x\left(x-5\right)=0\\ \left(x-5\right)\left(3-2x\right)=0\\ \left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\end{matrix}\right.\)
\(h.\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \left(2x+1\right)\left[\left(3x-2\right)-\left(5x-8\right)\right]=0\\ \left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \left(2x+1\right)\left(6-2x\right)=0\\ \left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=3\end{matrix}\right.\)
1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)
\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)
\(\Leftrightarrow x=2\)
3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)
\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)
\(\Leftrightarrow6x=6\)
hay x=1
quá chuẩn luôn !!!!!!!!
NHỚ L.I.K.E cho mk nha
a) (x+2)(x^2-2x+4)-x(x^2+2)=15
<=> x^3 + 8 - x^3 - 2x = 15
<=> -2x = 7
<=> x = -7/2
b) (x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=28
<=> x^3 + 9x² + 27x + 27 - x(9x² + 6x + 1) + 8x^3 + 1 = 28
<=> x^3 + 9x² + 27x + 27 - 9x^3 - 6x² - x + 8x^3 + 1 - 28 = 0
<=> 3x² + 26x = 0
<=> x(3x + 26) = 0
Vậy x = 0 và x = -26/3
c) (x^2-1)^3-(x^4+x^2+1)(x^2-1)=0
<=> (x² - 1)[(x² -1)² - x^4 - x² - 1] = 0
<=> (x-1)(x+1)(x^4 - 2x² + 1 - x^4 - x² - 1 ) = 0
<=> -(x-1)(x+1)3x² = 0
Vậy nghiệm là x = 1 ; -1 ; 0