Tìn x, biết:
(x4)2 = \(\frac{x^{12}}{x^{15}}\)
Giúp mik nha =o=
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a) 2 x 12 x 43 + 3 x 32 x 8 + 25 x 6 x4
=(2 x12) x43+(3 x8) x32+(6 x4) x25
=24 x43+24 x32+ 24 x25
=24 x(43+32+25)
=24 x100
=2400
b) 78 x 31 + 78 x 24 + 17 x 78 + 28 x 78
=78x(31+24+17+28)
=78x100
=7800
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
X x X : 2 = 15
x2 : 2 = 15
x2 = 15 x 2
x2 = 30
=> x = Căn bậc 2 của 30
a, 10+15+20+....+295+x.300+x=67
10+15+20+...+295+x(300+1)=67
10+15+20+...+295+x.301=67
8845+x.301=67
67-8845=x.301
-8878=x.301
x=-29/149/301
b,
\(\frac{1}{7.6}+\frac{1}{6.5}+\frac{1}{5.4}+\frac{1}{4.3}+\frac{1}{3.2}+\frac{1}{2.1}-\frac{1}{x+1}=\frac{59}{77}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}-\frac{1}{x+1}=\frac{59}{77}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)\(1-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)
\(\frac{6}{7}-\frac{1}{x+1}=\frac{59}{77}\)
\(\frac{1}{x+1}=\frac{6}{7}-\frac{59}{77}\)
\(\frac{1}{x+1}=\frac{1}{11}\)
suy ra x+1=11
suy ra x=10
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(\Leftrightarrow-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
=> x \(\in\) {-1;0;1;2;3;4;5;6}
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(\Leftrightarrow\)\(\frac{9-10}{12}\le\frac{x}{12}< 1-\left(\frac{8-3}{12}\right)\)
\(\Leftrightarrow\)\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
\(\Leftrightarrow-1\le x< 7\)
Mà x nguyên
=>x={-1;0;1;2;3;4;5;6}
\(2x-\frac{5}{12}=x-\frac{1}{2}\)
\(\Leftrightarrow2x-x=-\frac{1}{2}-\frac{5}{12}\)
\(\Leftrightarrow x=-\frac{11}{12}\)
Vậy ...
sr bạn .-. ở trên sai r, mk lm lại :333
\(2x-\frac{5}{12}=x-\frac{1}{2}\)
\(\Leftrightarrow2x-x=-\frac{1}{2}+\frac{5}{12}\)
\(\Leftrightarrow x=-\frac{1}{12}\)
Vậy ..
=>x8=x3
=>x=0;x=1
\(\left(x^4\right)^2=\frac{x^{12}}{x^{15}}\)
\(x^8=x^{12}:x^{15}=x^{12}\cdot\frac{1}{x^{15}}\)
\(\Rightarrow x^8-x^{12}\cdot\frac{1}{x^{15}}=0\)
\(x^8.\left(1-x^4\cdot\frac{1}{x^{15}}\right)=0\)
=> x8 = 0 => x = 0 (Loại, vì x12/x15, x15 khác 0)
\(1-x^4:x^{15}=0\)
\(1-\frac{1}{x^{11}}=0\)
\(\frac{1}{x^{11}}=1=\frac{1}{1}\)
=> x11 =1
=> x = 1 (TM)
KL: x = 1