(x2-9) \(\times\)(3-5x)=0
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\(a,5x\left(x^2-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,3\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow3\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ c,x^2-9x-10=0\\ \Leftrightarrow x^2+x-10x-10=0\\ \Leftrightarrow x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)
a, 5\(x\)(\(x^2\) - 9) = 0
\(\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy \(x\) \(\in\) { -3; 0; 3}
b, 3.(\(x+3\)) - \(x^2\) - 3\(x\) = 0
3.(\(x+3\)) - \(x\).( \(x\) + 3) = 0
(\(x+3\))( 3 - \(x\)) = 0
\(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -3; 3}
c, \(x^2\) - 9\(x\) - 10 = 0
\(x^2\) + \(x\) - 10\(x\) - 10 = 0
\(x.\left(x+1\right)\) - 10.( \(x-1\)) = 0
(\(x+1\))(\(x-10\)) = 0
\(\left[{}\begin{matrix}x+1=0\\x-10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -1; 10}
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1: Tìm x
a) 2x - 15 = -27
b) 2 (x + 1) – 3 = 7
c) 14 – (40 – x) = -27
d) 96 – 2(4 – 5x) = -12
e) – 40 – (– 3 – 33) + (40 – x) = – (– 47)
f) x(3x – 9). (121 – x2) = 0
g) – 62 – (38 + x) + 2x = – 100
h) (x + 1)2.(x2 + 1) = 0
i) (x – 12) – (2x + 31) = 6
k) 17/ (x + 3)3 : 3 – 1 = – 10
Bài 1: a) 2x - 15 = 27 => 2x = 27 + 15 = 42 => x = 42 : 2 = 21. b) 2(x + 1) - 3 = 7 => 2(x + 1) = 7 + 3 = 10 => x + 1 = 10 : 2 = 5 => x = 5 - 1 = 4. c) 14 - (40 - x) = -27 => 40 - x = 14 - (-27) = 41 => x = 40 - 41 = -1. d) 96 - 2(4 - 5x) = -12 => 2(4 - 5x) = 96 - (-12) = 108 => 4 - 5x = 108 : 2 = 54 => 5x = 4 - 54 = -50 => x = (-50) : 5 = -10.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\)
\(f\left(x\right)=2x.\left(x^2-3\right)-4.\left(1-2x\right)+x^2.\left(x-2\right)+\left(5x+3\right)\)\(=2x^3-6x-4+8x+x^3-2x^2+5x+3=3x^3+7x-1-2x^2=3x^3-2x^2+7x-1\)\(g\left(x\right)=-3.\left(1-x^2\right)-2.\left(x^2-2x-1\right)=-3+3x^2-2x^2+4x+2=-1+x^2+4x=x^2+4x-1\)
\(b)\)
\(h\left(x\right)=f\left(x\right)-g\left(x\right)=\left(3x^3-2x^2+7x-1\right)-\left(-1+x^2+4x\right)=x^2+4x-1=3x^3-2x^2+7x-1+1-x^2-4x=3x^3-3x^2+3x\)
\(\text{Xét}:\)
\(3x^3-3x^2+3x=0\)
\(\rightarrow3x.\left(x^2-x+1\right)=0\)
\(\rightarrow x.\left(x^2-x+1\right)=0\)
\(\rightarrow\orbr{\begin{cases}3x.\left(x^2-x+1\right)=0\\x.\left(x^2-x+1\right)=0\end{cases}}\) \(\rightarrow\orbr{\begin{cases}x=0\\x^2-x+1=0\end{cases}}\)
\(\rightarrow\orbr{\begin{cases}x=0\\x\notinℝ\end{cases}}\) \(\rightarrow x=0\)
\(\text{Vậy nghiệm của}\)\(h\left(x\right)\)\(\text{là}:\)\(0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,25x^2-1=15\)\(< =>x^2=\dfrac{16}{25}< =>x=\pm\dfrac{4}{5}\)
\(b,\left(x-4\right)^2-\left(5x+2\right)^2=0\)\(< =>\left(-4x-6\right)\left(6x-2\right)=0\)
\(< =>\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(c,\left(x-1\right)\left(x-9\right)=0< =>\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,x^2+4x=-3\Leftrightarrow x^2+4x+3=0\Leftrightarrow\left(x+1\right)\left(x+3\right)=0\)
\(\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
\(b,3x^2+4x-4=0\Leftrightarrow3x^2+6x-2x-4=0\Leftrightarrow3x\left(x+2\right)-2\left(x+2\right)=0\Leftrightarrow\left(3x-2\right)\left(x+2\right)=0\)
\(\left[{}\begin{matrix}x=-2\\3x=2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-2\\x=\frac{2}{3}\end{matrix}\right.\)
\(c,x^2+5x-6=0\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
\(\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\)
\(d,x^2-6x=-9\Leftrightarrow x^2+6x+9=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)16x^2-24x+9=25`
`<=>(4x-3)^2=25`
`+)4x-3=5`
`<=>4x=8<=>x=2`
`+)4x-3=-5`
`<=>4x=-2`
`<=>x=-1/2`
`b)x^2+10x+9=0`
`<=>x^2+x+9x+9=0`
`<=>x(x+1)+9(x+1)=0`
`<=>(x+1)(x+9)=0`
`<=>` \(\left[ \begin{array}{l}x=-9\\x=-1\end{array} \right.\)
`c)x^2-4x-12=0`
`<=>x^2+2x-6x-12=0`
`<=>x(x+2)-6(x+2)=0`
`<=>(x+2)(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\)
`d)x^2-5x-6=0`
`<=>x^2+x-6x-6=0`
`<=>x(x+1)-6(x+1)=0`
`<=>(x+1)(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\)
`e)4x^2-3x-1=0`
`<=>4x^2-4x+x-1=0`
`<=>4x(x-1)+(x-1)=0`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-\dfrac14\end{array} \right.\)
`f)x^4+4x^2-5=0`
`<=>x^4-x^2+5x^2-5=0`
`<=>x^2(x^2-1)+5(x^2-1)=0`
`<=>(x^2-1)(x^2+5)=0`
Vì `x^2+5>=5>0`
`=>x^2-1=0<=>x^2=1`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x^2-9\right).\left(3-5x\right)=0\)
\(\Rightarrow\hept{\begin{cases}x^2-9=0\\3-5x=0\end{cases}\Rightarrow}\hept{\begin{cases}x^2=0+9\\5x=3-0\end{cases}\Rightarrow\hept{\begin{cases}x^2=9\\5x=3\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x^2=3^2\\x=3:5\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=\frac{3}{5}\end{cases}}}\)
Vậy .....................
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\(\orbr{\begin{cases}x^2-9=0\\3-5x=0\end{cases}}\) chớ không phải \(\hept{\begin{cases}x^2-9=0\\3-5x=0\end{cases}}\)
Cần phân biệt "hoặc" và "và"
( x2 - 9 ) . ( 3 - 5x ) = 0
=> \(\orbr{\begin{cases}x^2-9=0\\3-5x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x^2=9\\5x=3\end{cases}}\)=> \(\orbr{\begin{cases}x=\pm3\\x=\frac{3}{5}\end{cases}}\)
Vậy ....
\(\Leftrightarrow\orbr{\begin{cases}x^2-9=0\\3-5x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\pm3\\x=\frac{3}{5}\end{cases}}}\)
Vaayj
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