\(64:4^{x}=256\)
giải chi tiết vs ạ
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Mình nghĩ đây là nâng cao tiểu học
Đặt S = 1/2 + 1/4 + 1/8 + 1/16 + ...
==> 2S = 1 + 1/2 + 1/4 + 1/8 + 1/16 + ...
2S = 1 + S
==> S = 1
Đây cũng là kết quả khi tính theo cấp số nhân khi n --> vô cùng
2A=1+1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512
2A-A=1-1/1024
A=1-1/1024
A=1023/1024

a
\(x+x^2-x^3-x^4=0\\ \Leftrightarrow x\left(1+x\right)-x^3\left(1+x\right)=0\\ \Leftrightarrow\left(1+x\right)\left(x-x^3\right)=0\\ \Leftrightarrow\left(1+x\right).x.\left(1-x^2\right)=0\\ \Leftrightarrow\left(1+x\right).x.\left(1-x\right)\left(1+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
b
x^3 chứ: )
\(x^3+27+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow x^3+3^3+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow\left(x+3\right).x.\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)

a: \(\Leftrightarrow x+2016=0\)
hay x=-2016
b: \(\Leftrightarrow x-100=0\)
hay x=100

`48/[x+4]+48/[x-4]=5` `ĐK: x \ne +-4`
`<=>[48(x-4)+48(x+4)]/[(x-4)(x+4)]=[5(x+4)(x-4)]/[(x-4)(x+4)]`
`=>48x-192+48x+192=5x^2-80`
`<=>5x^2-96x-80=0`
`<=>5x^2-100+4x-80=0`
`<=>5x(x-20)+4(x-20)=0`
`<=>(x-20)(5x+4)=0`
`<=>` $\left[\begin{matrix} x=20\\ x=\dfrac{-4}{5}\end{matrix}\right.$ (t/m)
Vậy `S={-4/5;20}`
ĐK : \(x\ne\pm4\)
\(\Leftrightarrow\cdot\dfrac{48\left(x+4\right)+48\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}=\dfrac{5\left(x+4\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\)
\(\Leftrightarrow48x+192+48x-192==5x^2-80\)
\(\Leftrightarrow96x=5x^2-80\)
\(\Leftrightarrow5x^2-96x-80=0\)
\(\Leftrightarrow5x^2+4x-100-80=0\)
\(\Leftrightarrow4\left(x-20\right)+5x\left(x-20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-20=0\\5x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-\dfrac{4}{5}\end{matrix}\right.\)

\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
=1/2-1/4+1/4-1/8+1/8-....+1/156-1/152
=1/2-1/152
=255/512
A=255/512

\(\dfrac{1}{3}+\dfrac{5}{6}\cdot\left(x-\dfrac{11}{5}\right)=\dfrac{3}{4}\)
\(\dfrac{5}{6}\cdot\left(x-\dfrac{11}{5}\right)=\dfrac{3}{4}-\dfrac{1}{3}\)
\(\dfrac{5}{6}\cdot\left(x-\dfrac{11}{5}\right)=\dfrac{5}{12}\)
\(x-\dfrac{11}{5}=\dfrac{5}{12}\cdot\dfrac{6}{5}\)
\(x-\dfrac{11}{5}=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+\dfrac{11}{5}\)
\(x=\dfrac{27}{10}\)
\(\dfrac{5}{6}\left(x-\dfrac{11}{5}\right)=\dfrac{3}{4}-\dfrac{1}{3}\)
\(\dfrac{5}{6}\left(x-\dfrac{11}{5}\right)=\dfrac{5}{12}\)
\(x-\dfrac{11}{5}=\dfrac{5}{12}:\dfrac{5}{6}\)
\(x-\dfrac{11}{5}=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+\dfrac{11}{5}=\dfrac{27}{10}\)

\(ĐK:x\ge0;x\ne4\\ P=\dfrac{5x+10\sqrt{x}-\left(3-\sqrt{x}\right)\left(\sqrt{x}-2\right)-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{5x+10\sqrt{x}-5\sqrt{x}+6+x-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{5\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{5\sqrt{x}}{\sqrt{x}-2}-\dfrac{3-\sqrt{x}}{\sqrt{x}+2}+\dfrac{6x}{4-x}\left(đk:x\ge0,x\ne4\right)\)
\(=\dfrac{5\sqrt{x}\left(\sqrt{x}+2\right)-\left(3-\sqrt{x}\right)\left(\sqrt{x}-2\right)-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{5x+10\sqrt{x}+x-5\sqrt{x}+6-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{5\sqrt{x}+6}{x-4}\)

\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
A=14 +18 +116 +132 +164 +1128 +1256 +1512
=12 −14 +14 −18 +....+1256 −1512
=12 −1512
=255512
Vậy A=255512
Phạm Long Khánh
\(64:4^4=256\)
\(\Rightarrow4^3:4^{x}=4^4\)
\(4^{x}=4^3:4^4\)
\(4^{x}=4^{-1}\)
⇒ x = -1
Vậy x = -1
\(daidaochungsinh\)
\(64-4^{x}=256\)
\(=>4^3:4^{x}=4^4\)
\(=>4^{x}=4^3:4^4\)
\(=>4^{x}=4^{-1}\)
\(=>x=-1\)
\(Vậy\) \(x=-1\)