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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: \(x\in\varnothing\)
a) Ta có: \(x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)hay x=1
Vậy: S={1}
c) Ta có: \(x+x^4=0\)
\(\Leftrightarrow x\left(x^3+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)=0\)
mà \(x^2-x+1>0\forall x\)
nên x(x+1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy: S={0;-1}
a) \(x\left(x+2\right)-3x-6=0\)
\(x\left(x+2\right)-3\left(x+2\right)=0\)
\(\left(x+2\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
b) \(\left(x^3+3x^2+3x+1\right)-3x^2-3x=0\)
\(x^3+1=0\)
\(\left(x+1\right)\left(x^2-x+1\right)=0\)
\(x=-1\)
c) \(4x^2-25=0\)
\(\left(2x-5\right)\left(2x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
\(a,x\left(x+2\right)-3x-6=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)\(b,\left(x^3+3x^2+3x+1\right)-3x^2-3x=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-3x^2-3x=0\)
\(\Leftrightarrow x^3+1=0\)
\(\Rightarrow x^3=1\Rightarrow x=1\)
\(c,4x^2-25=0\)
\(\Leftrightarrow\left(2x+5\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}2x+5=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
1.
<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
2.
<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
3.
<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)
4.
<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
5.
<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)
6,7. ko đủ điều kiện tìm
c(x-1)^2=4
x^2-2x+1=4
x^2-2x+1-4=0
x^2-2x-3=0
x^2-3x+x-3=0
x(x-3)+(x-3)=0
(x-3)(x+1)=0
\(\Rightarrow\hept{\begin{cases}x-3=0\\x+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=-1\end{cases}}}\)
d, x^3+2x^2-x-2=0
x^2(x+2)-(x+2)=0
(x+2)(x^2-1)=0
\(\Rightarrow\hept{\begin{cases}x=-2\\x=+-1\end{cases}}\)
a)
\(\Rightarrow x\left(x-5\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x-5=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)
b)
\(\Rightarrow3x\left(x-2\right)-2\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(3x-2\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-2=0\\3x-2=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{2}{3}\end{array}\right.\)
c)
\(\Rightarrow\left(3x-1\right)\left(5x+x-2\right)=0\)
\(\Rightarrow\left(3x-2\right)^2.2=0\)
\(\Rightarrow3x-2=0\)
\(\Rightarrow x=\frac{2}{3}\)
a)
\(3x^2+12x-66=0\)
\(\Leftrightarrow x^2+4x-22=0\)
\(\Leftrightarrow x^2+4x+4=26\Leftrightarrow (x+2)^2=26\)
\(\Rightarrow x+2=\pm \sqrt{26}\Rightarrow x=-2\pm \sqrt{26}\)
b)
\(9x^2-30x+225=0\)
\(\Leftrightarrow (3x)^2-2.3x.5+25+200=0\)
\(\Leftrightarrow (3x-5)^2=-200< 0\) (vô lý nên pt vô nghiệm)
c)
\(x^2+3x-10=0\)
\(\Leftrightarrow x^2-2x+5x-10=0\)
\(\Leftrightarrow x(x-2)+5(x-2)=0\Leftrightarrow (x+5)(x-2)=0\)
\(\Rightarrow x=-5\) hoặc $x=2$
d)
$3x^2-7x+1=0$
$\Leftrightarrow 3(x^2-\frac{7}{3}x)+1=0$
$\Leftrightarrow 3(x^2-\frac{7}{3}x+\frac{7^2}{6^2})=\frac{37}{12}$
$\Leftrightarrow 3(x-\frac{7}{6})^2=\frac{37}{12}$
$\Leftrightarrow (x-\frac{7}{6})^2=\frac{37}{36}$
$\Rightarrow x-\frac{7}{6}=\frac{\pm \sqrt{37}}{6}$
$\Rightarrow x=\frac{7\pm \sqrt{37}}{6}$
e)
$3x^2+7x+2=0$
$\Leftrightarrow 3(x^2+\frac{7}{3}x+\frac{7^2}{6^2})=\frac{25}{12}$
$\Leftrightarrow 3(x+\frac{7}{6})^2=\frac{25}{12}$
$\Leftrightarrow (x+\frac{7}{6})^2=\frac{25}{36}$
$\Rightarrow x+\frac{7}{6}=\pm \frac{5}{6}$
$\Rightarrow x=\frac{-1}{3}$ hoặc $x=-2$
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\int^{x=1}_{x=2}\)