2\(^{x-1}\)=128
tim x
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=> (x - 1)2 = 64 => (x - 1)2 = 82 => x - 1 = 8 => x = 9
b/ => x2 - x = 0
=> x(x - 1) = 0
=> x = 0 hoặc x - 1 = 0 => x = 1
a) 2 . (x - 1)2 = 128
(x - 1)2 = 64 = + 8
=> x = 9 hoặc x = -7
b) x2 = x
<=> x \(\in\) {0; -1; 1}
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(2\times A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2\times A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\)
\(A=1-\frac{1}{128}\)
\(A=\frac{127}{128}\)
\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(2\times B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)
\(B=1-\frac{1}{16}=\frac{15}{16}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Leftrightarrow4\times x+\frac{15}{16}=1\)
\(\Leftrightarrow4\times x=\frac{1}{16}\)
\(\Leftrightarrow x=\frac{1}{64}\)
a) x-7 = - 5
<=> x= 2
b) 128 - 3.x.(x+4) = 23
<=> - 3x2 - 12x +105=0
<=> -x2 -4x + 35 =0
\(\Delta'=156\Rightarrow\sqrt{\Delta'}=2\sqrt{39}\)
=> \(\left[{}\begin{matrix}x=-2+\sqrt{39}\\x=-2-\sqrt{39}\end{matrix}\right.\)
vậy S= \(\left\{-2+\sqrt{39};-2-\sqrt{39}\right\}\)
\(\left(\frac{3}{20}+\frac{1}{20}-x\right):\frac{32}{9}=\frac{21}{128}\)
\(\Leftrightarrow\frac{4}{20}-x=\frac{21}{128}.\frac{32}{9}\)
\(\Leftrightarrow\frac{1}{5}-x=\frac{7}{12}\)
\(\Leftrightarrow x=\frac{1}{5}-\frac{7}{12}\)
\(\Leftrightarrow x=-\frac{23}{60}\)
Vậy \(x=\frac{-23}{60}\)
Bài làm
\(\left(\frac{3}{20}+\frac{1}{20}-x\right):\frac{32}{9}=\frac{21}{128}\)
\(\left(\frac{3}{20}+\frac{1}{20}-x\right)=\frac{21}{128}.\frac{32}{9}\)
\(\frac{4}{20}-x=\frac{7}{4}.\frac{1}{3}\)
\(\frac{4}{20}-x=\frac{7}{12}\)
\(x=\frac{4}{20}-\frac{7}{12}\)
\(x=\frac{12}{60}-\frac{35}{60}\)
\(x=-\frac{23}{60}\)
Vậy \(x=-\frac{23}{60}\)
\(x=-\frac{37}{15}\)
Khó quá thì nên hoc24 .vn
Nếu đúng thì ấnĐúng 2 không những thees sau khi ấn sẽ may mắn cả năm
\(2^{x-1}=128\)
\(2^x:2^1=128\)
\(2^x=128\cdot2\)
\(2^x=256\)
Mà \(256=2^8\)
Do đó \(x=8\)
\(2^{x-1}=128\)
\(\Leftrightarrow2^{x-1}=2^7\)
\(\Leftrightarrow x-1=7\)
\(\Leftrightarrow x=7+1=8\)
Tự kết luận nha ^^