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Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
3x2 + 2x - 1 = 0
=> 3x2 + 3x - x - 1 = 0
=> 3x(x + 1) - (x + 1) = 0
=> (3x - 1)(x + 1) = 0
=> \(\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)
x2 - 5x + 6 = 0
=> x2 - 2x - 3x + 6 = 0
=> x(x - 2) - 3(x - 2) = 0
=> (x - 3)(x - 2) = 0
=> \(\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
3x2 + 7x + 2 = 0
=> 3x2 + 6x + x + 2 = 0
=> 3x(x + 2) + (x + 2) = 0
=> (3x + 1)(x + 2) = 0
=> \(\orbr{\begin{cases}3x+1=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)
1, \(3x^2+2x-1=0\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\3x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}}\)
2, \(x^2-5x+6=0\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)
3, \(3x^2+7x+2=0\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{3}\end{cases}}}\)
a) Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy: S={-5;2}
b) Ta có: \(3x^2-7x+1=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)
c) Ta có: \(3x^2-7x+8=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)
Vậy: \(x\in\varnothing\)