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a) (x-2)\(^4\) = 256
(x-2)\(^4\) = \(4^4\)
x-2=4
x=4+2
x=6
b) \(\dfrac{x^4}{256}=81\)
\(\dfrac{x^4}{4^4}=81\)
\(\dfrac{x}{4}=81\)
x =81.4
x = 324
c) \(125\left(x+\dfrac{4}{5}\right)^3=729\)
\(\left(x+\dfrac{4}{5}\right)^3=729:125=\dfrac{729}{125}\)
\(\left(x+\dfrac{4}{5}\right)^3=\left(\dfrac{9}{5}\right)^3\)
\(x+\dfrac{4}{5}=\dfrac{9}{5}\)
x = \(\dfrac{9}{5}-\dfrac{4}{5}\)
x \(=1\)
a)(x-2)^4=4^4
=(x-2)=4 sr x=4+2
sr x=6
xin lỗi nha mình chỉ biết làm bài a thôi mong cậu thông cảm và kb với mình nhé
Mình biết bài nào làm bài đó thôi nhé
a) (x-2)4 = 256
=> x-2 = 4
x = 4+2
x = 6
b) \(\frac{x^4}{256}=81\)
\(\Rightarrow\frac{x^4}{4^4}=3^4\)
Từ đây có thể làm theo 2 cách khác nhau
C1 : \(\frac{x^4}{4^4}=81\)
\(\Rightarrow\left(\frac{x}{4}\right)^4=81\)
\(\Rightarrow\frac{x}{4}=3\)
x = 3.4
x = 12
C2 : \(\frac{x^4}{4^4}=3^4\)
=> x4 = 34.44
x4 = (3.4)4
x4 = 124
<=> x = 4
c) \(125.\left(x+\frac{4}{5}\right)^3=729\)
\(\left(x+\frac{4}{5}\right)^3=729:125\)
\(\left(x+\frac{4}{5}\right)^3=5,832\)
\(\Rightarrow x+\frac{4}{5}=1,8=\frac{9}{5}\)
\(x=\frac{9}{5}-\frac{4}{5}\)
\(x=\frac{5}{5}=1\)
ta có : \(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}\Leftrightarrow\dfrac{x^2}{36}=\dfrac{y^2}{16}=\dfrac{z^2}{9}\)
áp dụng tính chất dãy tỉ số bằng nhau
ta có : \(\dfrac{x^2-y^2-z^2}{36-16-9}=\dfrac{1331}{11}=121\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x^2}{36}=121\\\dfrac{y^2}{16}=121\\\dfrac{z^2}{9}=121\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2=4356\\y^2=1936\\z^2=1089\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\pm66\\y=\pm44\\z=\pm33\end{matrix}\right.\)
vậy \(x=\pm66;y=\pm44;z=\pm33\)
Ta có : \(\dfrac{x}{6}=\dfrac{x^2}{36};\dfrac{y}{4}=\dfrac{y^2}{16};\dfrac{z}{3}=\dfrac{z^2}{9}\)
Áp dụng tính chất của dãy tỉ số bằng nhau có :
\(\dfrac{x^2}{36}=\dfrac{y^2}{16}=\dfrac{z^2}{9}=\dfrac{x^2-y^2-z^2}{36-16-9}=\dfrac{1331}{11}=121\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x^2}{36}=121\\\dfrac{y^2}{16}=121\\\dfrac{z^2}{9}=121\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2=4356\\y^2=1936\\z^2=1089\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=66\\y=44\\z=33\end{matrix}\right.\)
(\(\dfrac{1}{9}\))2 = (\(\dfrac{1^2}{3^2}\))2= ((\(\dfrac{1}{3}\))2)2= (\(\dfrac{1}{3}\))4
a
=> \(x=\dfrac{2}{7}:\dfrac{2}{3}=\dfrac{2.3}{2.7}=\dfrac{3}{7}\)
b
=> \(x=\dfrac{2}{5}:\dfrac{3}{5}=\dfrac{2.5}{3.5}=\dfrac{2}{3}\)
c
=> \(x=\dfrac{13}{7}.\dfrac{8}{13}=\dfrac{13.8}{7.13}=\dfrac{8}{7}\)
d
=> \(x=\dfrac{3}{2}:\dfrac{7}{4}=\dfrac{3.2.2}{2.7}=\dfrac{6}{7}\)
a: 2/3*x=2/7
=>x=2/7:2/3=3/7
b: x*3/5=2/5
=>x=2/5:3/5=2/5*5/3=10/15=2/3
c: x:8/13=13/7
=>x=13/7*8/13=8/7
d: 3/2:x=7/4
=>x=3/2:7/4=3/2*4/7=12/14=6/7
Bài 1 :
a) \(\frac{12}{21}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{4}{7}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{1}{7}-\frac{2}{3}=-\frac{11}{21}\)
b) \(\left(-\frac{25}{13}\right)+\left(-\frac{9}{17}\right)+\frac{12}{13}+\left(-\frac{25}{17}\right)\)
\(=\left[\left(-\frac{25}{13}\right)+\frac{12}{13}\right]+\left[\left(-\frac{9}{17}\right)+\left(-\frac{25}{17}\right)\right]\)
\(=-1+\left(-2\right)=-1-2=-3\)
c) \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\cdot1=\frac{5}{9}\)
Bài 2 :
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}=-\frac{29}{70}\)
=> \(x=\left(-\frac{29}{70}\right):\frac{2}{3}=\left(-\frac{29}{70}\right)\cdot\frac{3}{2}=-\frac{87}{140}\)
b) \(x:\frac{5}{2}-\frac{1}{2}=-\frac{2}{3}\)
=> \(x:\frac{5}{2}=-\frac{2}{3}+\frac{1}{2}=-\frac{1}{6}\)
=> \(x=\left(-\frac{1}{16}\right)\cdot\frac{5}{2}=-\frac{5}{32}\)
c) Bạn chỉ cần xét hai trường hợp âm và dương thôi :>
11/13-(5/42-x)=(15/28-11/13)
11/13-(5/42-x)=-37/182
(5/42-x)=11/13+37/182
(5/42-x)=191/182
x=5/42-191/182
x=-254/273
vậy x=-254/273
\(\left(2-\frac{x}{13}\right)^3=\frac{1331}{2197}\)
=> \(\left(2-\frac{x}{13}\right)^3=\left(\frac{11}{13}\right)^3\)
=> \(2-\frac{x}{13}=\frac{11}{13}\)
=> \(\frac{x}{13}=2-\frac{11}{13}=\frac{2\cdot13-11}{13}=\frac{15}{13}\)
=> x = 15