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a, \(2^x-15=17\)
\(\Rightarrow2^x=17+15\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
b, \(\left(7x-11\right)^3=2^5.5^2+200\)
\(\Rightarrow\left(7x-11\right)^3=32.25+200\)
\(\Rightarrow\left(7x-11\right)^3=1000\)
\(\Rightarrow\left(7x-11\right)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(\Rightarrow7x=10+11\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=21:7\)
\(\Rightarrow x=3\)
c, \(x^{10}=1^x\)
\(\Rightarrow x\in\left\{1;0\right\}\)
\(2^x-15=17\)
\(\Rightarrow2^x=17+15\)
\(\Rightarrow2^x=32=2^4\)
\(\Rightarrow x=4\)
\(\left(7x-11\right)^3=2^5.5^2+200\)
Phần này mk ko bt làm đâu
\(x^{10}=1^x\)
\(\Rightarrow\)\(x^{10}=1\)
\(\Rightarrow x=1\)
a) 2x - 15 = 17
2x = 25
b) ( 7x - 11 )3 = 25. 52 + 200
(7x - 11)3 = 32 . 25 + 200
(7x -11)3 = 1000
(7x-11)3 = 103
7x - 11 = 10
7x = 10+11
7x = 21
x = 21 : 7
x = 3
like nha
e) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\cdot\left(2x-15\right)^2-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\cdot\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\left(2x-15\right)^3=0\) hoặc \(\left(2x-15\right)^2-1=0\)
+)TH1: \(\left(2x-15\right)^3=0\)
\(\Rightarrow2x-15=0\)
\(\Rightarrow2x=15\)
\(\Rightarrow x=\frac{15}{2}\)
+)TH2: \(\left(2x-15\right)^2-1=0\)
\(\Rightarrow\left(2x-15\right)^2=1\)
\(\Rightarrow\left[{}\begin{matrix}2x-15=1\\2x-15=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2x=16\\2x=14\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
Vậy \(x=\frac{15}{2}\) hoặc \(x=8\) hoặc \(x=7\)
a) \(2^x-17=15\Rightarrow2^x=32\)
Mà \(2^5=32\Rightarrow x=5\)
Vậy x = 5
b)\(\left(7x-11\right)^3=2^5\cdot5^2+200\)
\(\Rightarrow\left(7x-11\right)^3=1000\)
\(\Rightarrow\left(7x-11\right)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=3\)
Vậy x = 3
c)\(x^{10}=1^x\Rightarrow x^{10}=1\)(số 1 có luỹ thừa là bao nhiêu thì vẫn là 1 thui)\(\Rightarrow x=1\)
Vậy x = 1
d) \(x^{10}=x\Rightarrow x^{10}-x=0\)
\(\Rightarrow x\left(x^9-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x^9-1=0\)
+)TH1: \(x=0\)
+)TH2: \(x^9-1=0\Rightarrow x^9=1\Rightarrow x=1\)
Vậy x = 0 hoặc x = 1
a) (x + 1/2) . (2/3 − 2x) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
b) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}\)
\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}\)
\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}\)
\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}\)
\(\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}\)
\(\Rightarrow x.\frac{44}{7}=-\frac{8}{7}\)
\(\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}\)
\(\Rightarrow x=-\frac{2}{11}\)
c) \(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)
\(\Rightarrow x\left(3\frac{1}{4}-\frac{7}{6}\right)=\frac{5}{12}+\frac{5}{3}\)
\(\Rightarrow x\left(\frac{13}{4}-\frac{7}{6}\right)=\frac{25}{12}\)
\(\Rightarrow x.\frac{25}{12}=\frac{25}{12}\)
\(\Rightarrow x=\frac{25}{12}:\frac{25}{12}\)
\(\Rightarrow x=1\)
d) \(5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}\)
\(\Rightarrow\left(5\frac{8}{17}-\frac{4}{17}\right):x+\frac{22}{7}:\frac{52}{3}=\frac{4}{11}\)
\(\Rightarrow5\frac{4}{17}:x+\frac{33}{182}=\frac{4}{11}\)
\(\Rightarrow\frac{89}{17}:x=\frac{4}{11}-\frac{33}{182}\)
\(\Rightarrow\frac{89}{17}:x=\frac{365}{2002}\)
\(\Rightarrow x=\frac{89}{17}:\frac{365}{2002}\)
\(\Rightarrow x\approx28,7\) (số hơi lẻ)
e) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)