Tìm x biết
a, x^23 = 64 . x^20
b, ( 4x-3 )^4 = ( 3-4x )
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\(a,A=\left(x^2-x\right)\left(x^2-x-12\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)+36-36\\ A=\left(x^2-x+6\right)^2-36\ge-36\\ A_{min}=-36\Leftrightarrow x^2-x+6=0\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ b,B=4x^4+4x^3+5x^2+4x+3\\ B=\left(4x^4+4x^3+x^2\right)+\left(x^2+4x+4\right)-1\\ B=x^2\left(2x+1\right)^2+\left(x+2\right)^2-1\ge-1\\ B_{min}=-1\Leftrightarrow\left\{{}\begin{matrix}x\left(2x+1\right)=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy dấu \("="\) không xảy ra
\(\Leftrightarrow-\dfrac{2}{5}\left(4x-3\right)^2=-\dfrac{5}{18}\)
\(\Leftrightarrow\left(4x-3\right)^2=\dfrac{25}{36}\)
\(\Leftrightarrow4x-3\in\left\{\dfrac{5}{6};-\dfrac{5}{6}\right\}\)
hay \(x\in\left\{\dfrac{23}{24};\dfrac{13}{24}\right\}\)
x2 + 4x + 3 = 0
\(\Leftrightarrow\)x2 + x + 3x + 3 = 0
\(\Leftrightarrow\)x(x + 1) + 3(x + 1) = 0
\(\Leftrightarrow\)(x + 1)(x + 3) = 0
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)
Vậy....
\(a,\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow4x-3-x-5=30-3x\\ \Leftrightarrow4x-x+3x=30+5+3\\ \Leftrightarrow6x=38\\ \Leftrightarrow x=\dfrac{19}{3}\)
câu a+b dùng quy tắc chuyển vế
c, 3.(1/2-x)-5.(x-1/10)=-7/4
=>(3.1/2-3x)-(5x-5.1/10)=-7/4
=>3/2-3x-5x+1/2=-7/4
=>(3/2+1/2)-(3x+5x)=-7/4
=> 2-8x=-7/4
=>8x=15/4
=>x=15/4:8
=>x=15/32
a) 2.(1/4 - 3x) = 1/5 - 4x
=> 1/2 - 6x = 1/5 -4x
=> -6x + 4x = 1/5 - 1/2
=> -2x = -3/10 = 3/20
b) 4.(1/3 - x) + 1/2 = 5/6 +x
=> 4/3 - 4x + 1/2 = 5/6 +x
=> -4x - x = 5/6 - 4/3 - 1/2
=> -5x = -1
=> x= 1/5
c) 3. (1/2 - x) -5. ( x - 1/10) = -7/4
=> 3/2 - 3x - 5x + 1/2 = -7/4
=> -3x - 5x = -7/4 - 3/2 - 1/2
=> -8x = -15/4
=> x = 15/32
4x + 5 : 3 - 121 : 11
=4 . 4 + 5 : 3 - 121 : 11
= 16 + \(\frac{5}{3}\) - 11
= \(\frac{20}{3}\)
a) x23= 64 . x20
x23: x20= 64
x3= 64
=> x3= 43
=> x =4
Vậy...
a)\(x^{23}=64.x^{20}\)
\(\Leftrightarrow\frac{x^{23}}{x^{20}}=64\)
\(\Leftrightarrow x^3=64\Rightarrow x=4\)
b)\(\left(4x-3\right)^4=3-4x\)
\(\Leftrightarrow\left(3-4x\right)^4=3-4x\)
\(\Leftrightarrow\left(3-4x\right)^3=1\)
\(\Leftrightarrow3-4x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)