a.Tìm x biết :(x+1/2)+(x+1/4)+(x+1/8)+(x+1/16)=1
b.Tính:56,24+75,06x4,8-6,15x13,6=.....
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Trả lời
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)
\(\Leftrightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=\frac{23}{16}\)
\(\Leftrightarrow4x+\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right)=\frac{23}{16}\)
\(\Leftrightarrow4x+\frac{8+4+2+1}{16}=\frac{23}{16}\)
\(\Leftrightarrow4x+\frac{15}{16}=\frac{23}{16}\)
\(\Leftrightarrow4x=\frac{23}{16}-\frac{15}{16}\)
\(\Leftrightarrow4x=\frac{8}{16}\)
\(\Leftrightarrow4x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}:4\)
\(\Leftrightarrow x=\frac{1}{8}\)
Vậy x=\(\frac{1}{8}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)
\(\Leftrightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=\frac{23}{16}\)
\(\Leftrightarrow4x+\left(\frac{8+4+2+1}{16}\right)=\frac{23}{16}\)
\(\Leftrightarrow4x+\frac{15}{16}=\frac{23}{16}\)
\(\Leftrightarrow4x=\frac{23}{16}-\frac{15}{16}\)
\(\Leftrightarrow4x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}:4\)
\(\Leftrightarrow x=\frac{1}{8}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Leftrightarrow4\text{x}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=1\)
\(\Leftrightarrow4\text{x}+\frac{15}{16}=1\)
\(\Leftrightarrow4\text{x}=1-\frac{15}{16}\)
\(\Leftrightarrow4\text{x}=\frac{1}{16}\)
\(\Leftrightarrow x=\frac{1}{16}:4=\frac{1}{64}\)
(x+1/2)+(x+1/4)+(x+1/8)+(x+16)=1
(x.x.x.x)+(1/2+1/4+1/8+1/16=1
4x+(1/2+1/4+1/8+1/16)=1
4x+(8/16+4/16+2/16+1/16)=1
4x+15/16=1
4x=1-15/16
4x=1/16
x=1/16:4
=>x=1/64
tick cho mk nha bạn
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(4x+\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(4x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right)=1\)
\(4x+\left(1-\frac{1}{16}\right)=1\)
\(4x+\frac{15}{16}=1\)
\(4x=\frac{1}{16}\)
\(\Rightarrow x=\frac{1}{64}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(4x+\frac{15}{16}=1\)
\(4x=\frac{1}{16}\)
\(x=\frac{1}{16}\div4\)
\(x=\frac{1}{64}\)
Vậy ...
ta có:(x+x+x+x)+(1/2+1/4+1/8+1/16)=1 4x+15/16=1 4x=1-15/16 4x=1/16 x=1/16:4=1/64 vậy x=1/64 Mình trả lời trước nhé
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(x+\frac{1}{2}+x+\frac{1}{4}+x+\frac{1}{8}+x+\frac{1}{16}=1\)
\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(4x+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}\)
\(A=1-\frac{1}{2^4}\)
\(A=\frac{15}{16}\)
Thay A vào đẳng thức ta có
\(4x+\frac{15}{16}=1\)
\(4x=1-\frac{15}{16}\)
\(4x=\frac{1}{16}\)
\(x=\frac{1}{16}\div4\)
\(x=\frac{1}{64}\)
\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{4}+x+\frac{1}{8}+x+\frac{1}{16}=1\)
\(\Rightarrow4\times x+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(\Rightarrow4\times x+\frac{8+4+2+1}{16}=1\)
\(\Rightarrow4\times x+\frac{15}{16}=1\)
\(\Rightarrow4\times x=1-\frac{15}{16}=\frac{1}{16}\)
\(\Rightarrow x=\frac{1}{16}:4=\frac{1}{16}\times\frac{1}{4}\)
\(\Rightarrow x=\frac{1}{64}\).
1) Áp dụng HTL:
\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}=\dfrac{1}{4^2}+\dfrac{1}{\left(4\sqrt{2}\right)^2}=\dfrac{3}{32}\Rightarrow AH=\dfrac{4\sqrt{6}}{3}\left(cm\right)\)
Áp dụng đ/lý Pytago:
\(BC^2=AB^2+AC^2\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{4^2+\left(4\sqrt{2}\right)^2}=4\sqrt{3}\left(cm\right)\)
Bài 2:
a) \(pt\Leftrightarrow\sqrt{\left(2x+1\right)^2}=3\Leftrightarrow\left|2x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3\\2x+1=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
b) \(A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=2\sqrt{x}.\dfrac{\sqrt{x}+1}{\sqrt{x}}=2\sqrt{x}+2\)
(x+1/2)+(x+1/4)+(x+1/8)+(x+1/16)=1
4x+(1/2+1/4+1/8+1/16)=1
4x+15/16=1
4x=1-15/16
4x=1/16
x=1/64