Tìm x: x + 2phần-4 = -9phầnx + 2
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\(\left(x-4\right)^4=\left(x-4\right)^2\\ \Rightarrow\left(x-4\right)^2\left[\left(x-4\right)^2-1\right]=0\\ \Rightarrow\left(x-4\right)\left(x-4-1\right)\left(x-4+1\right)=0\\ \Rightarrow\left(x-4\right)\left(x-5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\\x=5\end{matrix}\right.\)
Tìm x biết :
45 - [ ( 72 - 8 . x ) : 4 + 7 ] . 3 = 0
[ ( 72 - 8 . x ) :4 + 7 ] . 3 = 45 - 0
[ ( 72 - 8 . x) : 4 + 7 ] . 3 = 45
( 72 - 8 . x ) : 4 + 7 = 45 : 3
( 72 - 8 . x ) : 4 + 7 = 15
( 72 - 8 . x ) : 4 = 15 - 7
( 72 - 8 . x ) : 4 = 8
72 - 8 . x = 8 × 4
72 - 8 . x = 32
8 . x = 72 - 32
8 . x = 40
x = 40 : 8
x = 5
Vậy x = 5
\([\left(72-8\cdot x\right):4+7]\cdot3=45-0\)
\([\left(72-8\cdot x\right):4+7]\cdot3=45\)
\(\left(72-8\cdot x\right):4+7=45:3\)
\(\left(72-8\cdot x\right):4+7=15\)
\(\left(72-8\cdot x\right):4=15-7\)
\(\left(72-8\cdot x\right):4=8\)
\(72-8\cdot x=8\cdot4\)
\(72-8\cdot x=32\)
\(8x=72-32\)
\(8x=40\)
\(x=40:8\)
\(x=5\)
\(\left(x-\dfrac{1}{2}\right).\dfrac{5}{2}=\dfrac{7}{4}-\dfrac{1}{2}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right).\dfrac{5}{2}=\dfrac{5}{4}\)
\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{2}\)
=> x = 1
Ta có: \(\left(x-\dfrac{1}{2}\right)\cdot\dfrac{5}{2}=\dfrac{7}{4}-\dfrac{1}{2}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)\cdot\dfrac{5}{2}=\dfrac{5}{4}\)
\(\Leftrightarrow x-\dfrac{1}{2}=\dfrac{1}{2}\)
hay x=1
\(\dfrac{x+2}{-4}=-\dfrac{9}{x+2}\\ \Rightarrow\left(x+2\right)^2=\left(-4\right).\left(-9\right)\\ \Rightarrow\left(x+2\right)^2=36\\ \Rightarrow\left(x+2\right)^2=\pm6^2\\ \Rightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
=>(x+2)^2=36
=>x+2=6 hoặc x+2=-6
=>x=4 hoặc x=-8