Tìm x, biết
b) (x - 7)2 - x(x-9)=14
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b) \(\Leftrightarrow3x^3+12x-2x^2-8=0\\ \Leftrightarrow\left(3x^3-2x^2\right)+\left(12x-8\right)=0\\ \Leftrightarrow x^2\left(3x-2\right)+4\left(3x-2\right)=0\\ \Leftrightarrow\left(x^2+4\right)\left(3x-2\right)=0\)
Vì \(x^2+4>0\Rightarrow3x-2=0\Rightarrow x=\dfrac{2}{3}\)
c) \(x^2+5x=0\\ \Leftrightarrow x\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
d) \(\Leftrightarrow x^3-27+x\left(4-x^2\right)=36\\ \Leftrightarrow x^3+4x-x^3=63\\ \Leftrightarrow4x=63\\ \Leftrightarrow x=\dfrac{63}{4}\)
b) 3x(x\(^3\) +12x-2x\(^2\)-8=0
3x(x\(^2\)+4)-2(x\(^2\)+4)=0
(x\(^2\)+4)(3x-2)=0
\(\Leftrightarrow\left[{}\begin{matrix}X^2+4=0\\3X-2=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x\in Z\\X=\dfrac{2}{3}\end{matrix}\right.\)
a) x\(^2\)+5x=0
x(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
c)(x-3)(x\(^2\)+3x+9)+x(x+2)(2-x)=36
x\(^3\)-27+x(x+2)(2-x)=36
4x-27=36
4x=36+27
4x=63
x=\(\dfrac{63}{4}\)
B-(\(3x^6-4xy^5+\dfrac{1}{3}xy^2\))=
B= \(\left(7x^6-\dfrac{1}{2}xy^5-xy^2-\dfrac{1}{3}\right)+\left(3x^6-4xy^5+\dfrac{1}{3}xy^2-\dfrac{3}{2}\right)\)
B= \(7x^6-\dfrac{1}{2}xy^5-xy^2-\dfrac{1}{3}+3x^6-4xy^5+\dfrac{1}{3}xy^2-\dfrac{3}{2}\)
B= \(7x^6+3x^6-\dfrac{1}{2}xy^5-4xy^5-xy^2+\dfrac{1}{3}xy^2-\dfrac{1}{3}+\dfrac{2}{3}\)
B= \(10x^6-\dfrac{9}{2}xy^5-\dfrac{2}{3}xy^2+\dfrac{1}{3}\)
Áp dụng tc dtsbn:
\(3x=7y\Rightarrow\dfrac{x}{7}=\dfrac{y}{3}=\dfrac{x-y}{7-3}=\dfrac{-16}{4}=-4\\ \Rightarrow\left\{{}\begin{matrix}x=-28\\y=-12\end{matrix}\right.\)
a) x – 9 = -14
x = -14 + 9
x = -5
b) 2( x + 7 ) = -16
2( x + 7 ) = 2 . ( -8 )
x + 7 = -8
x = -8 – 7 = -15
c) | x – 9 | = 7
x – 9 = 7 hoặc x – 9 = -7
x = 7 + 9 hoặc x = -7 + 9
x = 16 hoặc x = 2
d) ( x – 5 )( x + 7 ) = 0
x – 5 = 0 hoặc x + 7 = 0
x = 5 hoặc x = -7
a) x + 4 = – 14 – 9
x + 4 = – 23
x = – 23 – 4
x = – 27
b) 3x = – 14 + 2
3x = – 12
x = – 12 : 3
x = – 4
c) 2| x | = 4 – (– 8)
2| x | = 12
| x | = 6
x = 6 hoặc x = – 6
d) |x – 2| = 7
x – 2 = 7 hoặc x – 2 = – 7
x = 9 hoặc x = – 5
a.\(\left(-12\right)x-14=-2\)
\(\left(-12\right)x=-2+14\)
\(\left(-12\right)x=12\)
\(x=12:\left(-12\right)\)
\(x=-1\)
\(b,\left(-8\right)x=\left(-5\right)\left(-7\right)-3\)
\(\left(-8\right)x=35-3\)
\(\left(-8\right)x=32\)
\(x=32:\left(-8\right)\)
\(x=-4\)
\(c,\left(-9\right)x+3=\left(-2\right)\left(-7\right)+16\)
\(\left(-9\right)x+3=14+16\)
\(\left(-9\right)x+3=30\)
\(\left(-9\right)x=30-3\)
\(\left(-9\right)x=27\)
\(x=27:\left(-9\right)\)
\(x=-3\)
\(x:\dfrac{3}{4}\) = \(\dfrac{9}{14}-\dfrac{1}{7}\)
\(x:\dfrac{3}{4}\) = \(\dfrac{1}{2}\)
\(x=\dfrac{1}{2}\) x \(\dfrac{3}{4}\)
\(x=\dfrac{3}{8}\)
\(x\)x \(\dfrac{1}{2}-\dfrac{1}{4}\)=\(\dfrac{2}{3}\)
\(x\) x \(\dfrac{1}{4}\) =\(\dfrac{2}{3}\)
\(x\) = \(\dfrac{2}{3}:\dfrac{1}{4}\)
\(x\) \(\dfrac{8}{3}\)
\(1\)/
\(a\)) \(=\left(\dfrac{7}{5}-\dfrac{8}{7}\right)+\dfrac{17}{5}:0,6\)
\(=\dfrac{9}{35}+\dfrac{17}{3}\)
\(=\dfrac{622}{105}\)
\(b\)) \(=\dfrac{11}{6}+\dfrac{-14}{15}\)
\(=\dfrac{9}{10}\)
\(c\)/ \(=\dfrac{7}{4}-\dfrac{2}{3}\)
\(=\dfrac{13}{12}\)
\(\left(x-7\right)^2-x\left(x-9\right)=14\)
\(\Leftrightarrow x^2-14x+49-x^2+9x=14\)
\(\Leftrightarrow-5x=-35\)
<=> x = 7
\(\left(x-7\right)^2-x\left(x-9\right)=14\)
\(\Leftrightarrow x^2-14x+49-x^2+9x=14\)
\(\Leftrightarrow49-5x=14\)
\(\Leftrightarrow5x=49-14\)
\(\Leftrightarrow5x=35\)
\(\Leftrightarrow x=35:5\)
\(\Leftrightarrow x=7\)