tính:
a)(5x-2)2=64
b)25x+1=64
giúp lẹ
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\(=\left(5x-1\right)^2+\left(5x+1\right)^2-2\left(5x-1\right)\left(5x+1\right)\)
\(=\left(5x-1-5x-1\right)^2\)
\(=\left(-2\right)^2=4\)
\(2^3-\left(5x\right)^3-5\times\left(5x\right)^2\)+ 5
\(=8-125x^3-125x^2+5\)
\(=-125x^2\left(x-1\right)+13\)
a, \(\dfrac{4x+13}{5x\left(x-7\right)}-\dfrac{x-48}{5x\left(7-x\right)}\)
\(=\dfrac{4x+13}{5x\left(x-7\right)}+\dfrac{x-48}{5x\left(x-7\right)}\)
\(=\dfrac{4x+13+x-48}{5x\left(x-7\right)}\)
\(=\dfrac{5x-35}{5x\left(x-7\right)}\)
\(=\dfrac{5\left(x-7\right)}{5x\left(x-7\right)}=\dfrac{1}{x}\)
b, \(\dfrac{1}{x-5x^2}-\dfrac{25x-15}{25x^2-1}\)
\(=\dfrac{1}{x\left(1-5x\right)}+\dfrac{25x-15}{\left(1-5x\right)\left(1+5x\right)}\)
\(=\dfrac{1+5x}{x\left(x-5x\right)\left(1+5x\right)}+\dfrac{x\left(25x-15\right)}{x\left(1-5x\right)\left(1+5x\right)}\)
\(=\dfrac{1+5x+25x^2-15x}{x\left(1-5x\right)\left(1+5x\right)}\)\(=\dfrac{25x^2-10x+1}{x\left(1-5x\right)\left(1+5x\right)}=\dfrac{\left(5x-1\right)^2}{x.\left(1-5x\right)\left(1+5x\right)}\)
\(=\dfrac{\left(5x-1\right)^2}{-x\left(5x-1\right)\left(1+5x\right)}\) \(=\dfrac{-\left(5x-1\right)}{x\left(1+5x\right)}\)
d: ta có: \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)
`a)1/[x-5x^2]-[25x-15]/[25x^2-1]`
`=[-(5x+1)-x(25x-15)]/[x(5x-1)(5x+1)]`
`=[-5x-1-25x^2+15x]/[x(5x-1)(5x+1)]`
`=[-25x^2+10x-1]/[x(5x-1)(5x+1)]`
`=[-(5x-1)^2]/[x(5x-1)(5x+1)]`
`=[1-5x]/[x(5x+1)]`
________________________________________________-
`b)(-1/[x^2-4x]+2/[16-x^2]-[-1]/[4x+16]):1/[4x]`
`=[-4(x+4)-8x+x(x-4)]/[4x(x-4)(x+4)].4x`
`=[-4x-16-8x+x^2-4x]/[(x-4)(x+4)]`
`=[x^2-16x-16]/[x^2-16]`
Đk: \(x\ne\dfrac{3}{5};x\ne\dfrac{1}{5}\)
Pt \(\Leftrightarrow\dfrac{4}{\left(5x-3\right)\left(1-5x\right)}=\dfrac{-3\left(5x-3\right)}{\left(1-5x\right)\left(5x-3\right)}-\dfrac{2x\left(1-5x\right)}{\left(1-5x\right)\left(5x-3\right)}\)
\(\Rightarrow4=-3\left(5x-3\right)-2x\left(1-5x\right)\)
\(\Leftrightarrow-10x^2+17x-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17+\sqrt{89}}{20}\left(tmpt\right)\\x=\dfrac{17-\sqrt{89}}{20}\left(ktmpt\right)\end{matrix}\right.\)
Vậy...
a) ( 5x - 2 )2 = 64
( 5x - 2 )2 = 82
\(\Rightarrow\)5x - 2 = 8
5x = 10
x = 2
b) 25x+1 = 64
25x + 1 = 26
\(\Rightarrow\)5x + 1 = 6
5x = 5
x = 1
Hk tốt
a) ( 5x - 2 )2 = 64
( 5x - 2)2 = 82
5x - 2 = 8
5x = 10
x = 10 : 5 = 2
b) 25x + 1 = 64
25x+1 = 26
5x + 1 = 6
5x = 5
x = 5 : 5 = 1