Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Leftrightarrow x+x+x+x+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=1\)
\(\Leftrightarrow4x+\frac{15}{16}=1\Leftrightarrow4x=\frac{1}{16}\Leftrightarrow x=\frac{1}{16}:4=\frac{1}{64}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{4}+x+\frac{1}{8}+x+\frac{1}{16}=1\)
\(\Rightarrow4x+\frac{15}{16}=1\)
\(\Rightarrow4x=1-\frac{15}{16}=\frac{1}{16}\)
\(\Rightarrow x=\frac{1}{16}:4=\frac{1}{16}.\frac{1}{4}=\frac{1}{64}\)
a, (x-1) . 0,5 = 7,5 : (x-1)
=> = ( x - 1 ) 0,5 = \(\frac{x-1}{2}\)
\(=\frac{7,5}{x-1}=\frac{15}{2\left(x-1\right)}\)
=> x = - 1 \(\sqrt{15}\)
x = \(\sqrt{15+1}\)
đề sao sao ý
a) \(\left\{\left[\left(2x+14\right)\div2^2-3\right]\div2\right\}-1=0\)
\(\left[\left(2x+14\right)\div4-3\right]\div2=0+1\)
\(\left[\left(2x+14\right)\div4-3\right]=\left(0+1\right).2\)
\(\left(2x+14\right)\div4=\left(0+1\right).2+3\)
\(\left(2x+14\right)\div4=5\)
\(2x+14=5.4\)
\(2x+14=20\)
\(2x=20-14\)
\(2x=6\)
\(x=6\div2\)
\(x=3\)
b) Làm tương tự phần a)
a){[(2x+14)/22-3]/2}-1=0
{[(2x+14)/4-3]/2}-1 =0
[(2x+14)/4-3]/2 =0+1
[(2x+14)/4-3]/2 =1
(2x+14)/4-3 =1*2
(2x+14)/4-3 =2
(2x+14)/4 =2+3
(2x+14)/4 =5
2x+14 =5*4
2x+14 =20
2x =20-14
2x =6
x =6/2
x =3
gio minh dang ban nen chi giai phan a thoi nhe, khi nao ranh minh se giai not phan con lai sau nhe
a) \(\left(x-9\right)^4=\left(x-9\right)^7\)
\(\Rightarrow\left[{}\begin{matrix}x-9=1\\x-9=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=10\\x=9\end{matrix}\right.\)
b) \(\left(3x-15\right)^{10}=\left(3x-15\right)^{15}\)
\(\Rightarrow\left[{}\begin{matrix}3x-15=0\\3x-15=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{3}\\x=\dfrac{16}{3}\end{matrix}\right.\)
c) \(\left(x-8\right)^3=\left(x-8\right)^6\)
\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x-8=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=8\\x=9\end{matrix}\right.\)
a) (2x+3)-(5x-17)=9
2x+3-5x+17=9
-3x+20=9
-3x=9-20=-11
=>x=11/3
b) \(\left(\frac{1}{5}+\frac{4}{5}x\right)-\left(\frac{2}{5}x+\frac{1}{9}\right)=\frac{3}{7}\)
=> \(\frac{1}{5}+\frac{4}{5}x-\frac{2}{5}x-\frac{1}{9}=\frac{3}{7}\)
=> \(\left(\frac{1}{5}-\frac{1}{9}\right)+\left(\frac{4}{5}x-\frac{2}{5}x\right)=\frac{3}{7}\)
\(\frac{4}{45}+\frac{2}{5}x=\frac{3}{7}\)
\(\frac{2}{5}x=\frac{3}{7}-\frac{4}{45}=\frac{107}{315}\)
x=107/315:2/5=107/126
