giúp mk đc k😔
(2/5)^3x=8/125
K
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NT
2
18 tháng 8 2018
\(\left(\frac{3}{5}-\frac{2}{3}x\right)^3=\left(\frac{-4}{5}\right)^3\)
\(\frac{3}{5}-\frac{2}{3}x=\frac{-4}{5}\)
\(\frac{2}{3}x=\frac{7}{5}\)
\(x=\frac{21}{10}\)
T
18 tháng 8 2018
\(\left(\frac{3}{5}-\frac{2}{3}.x\right)^3=\frac{-64}{125}\)
\(\Leftrightarrow\left(\frac{3}{5}-\frac{2}{3}.x\right)^3=\left(-\frac{4}{5}\right)^3\)
\(\Rightarrow\frac{3}{5}-\frac{2}{3}.x=-\frac{4}{5}\)
\(\Leftrightarrow\frac{2}{3}.x=\frac{3}{5}-\left(-\frac{4}{5}\right)\)
\(\Leftrightarrow\frac{2}{3}.x=\frac{3}{5}+\frac{4}{5}\)
\(\Leftrightarrow\frac{2}{3}.x=\frac{7}{5}\)
\(\Leftrightarrow x=\frac{7}{5}:\frac{2}{3}\)
\(\Leftrightarrow x=\frac{21}{10}\)
Vậy \(x=\frac{21}{10}\)
CC
1
18 tháng 8 2018
\(\left(\frac{3}{5}-\frac{2}{3}x\right)^3=\frac{-64}{125}\)
\(\left(\frac{3}{5}-\frac{2}{3}x\right)^3=\left(\frac{-4}{5}\right)^3\)
\(\frac{3}{5}-\frac{2}{3}x=\frac{-4}{5}\)
\(\frac{9}{15}-\frac{10x}{15}=\frac{-12}{15}\)
\(-\frac{10x}{15}=\frac{9}{15}-\frac{-12}{15}\)
\(\text{ }-10x=9+12\)
\(-10x=21\)
\(x=\frac{-21}{10}\)
k nha
AH
Akai Haruma
Giáo viên
4 tháng 7 2021
Bài 1.
a. $=a^2+2.a.12+12^2=a^2+24a+144$
b. $=(3a)^2+2.3a.\frac{1}{3}+(\frac{1}{3})^2=9a^2+2a+\frac{1}{9}$
c. $=(5a^2)^2+2.5a^2.6+6^2=25a^4+60a^2+36$
d. $=\frac{1}{4}+2.\frac{1}{2}.4b+(4b)^2$
$=\frac{1}{4}+4b+16b^2$
e.
$=(a^m)^2+2.a^m.b^n+(b^n)^2$
$=a^{2m}+2a^mb^n+b^{2n}$
AH
Akai Haruma
Giáo viên
4 tháng 7 2021
Bài 2.
$(x-0,3)^2=x^2-0,6x+0,09$
$(6x-3y)^2=36x^2-36xy+9y^2$
$(5-2xy)^2=25-20xy+4x^2y^2$
$(x^4-1)^2=x^8-2x^4+1$
$(x^5-y^3)^2=x^{10}-2x^5y^3+y^6$
Y
23 tháng 6 2023
\(1,\left(3x+2\right)\left(5-x^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\5-x^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\-x^2=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\pm\sqrt{5}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{2}{3};-\sqrt{5};\sqrt{5}\right\}\)
\(2,-2x-\dfrac{2}{3}\left(\dfrac{3}{4}-\dfrac{1}{8}x\right)=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow-2x-\dfrac{1}{2}+\dfrac{1}{12}x=-\dfrac{1}{8}\)
\(\Leftrightarrow-2x+\dfrac{1}{12}x=-\dfrac{1}{8}+\dfrac{1}{2}\)
\(\Leftrightarrow-\dfrac{23}{12}=\dfrac{3}{8}\)
\(\Leftrightarrow x=-\dfrac{9}{46}\)
Vậy \(S=\left\{-\dfrac{9}{46}\right\}\)
\(3,\dfrac{1}{12}:\dfrac{4}{21}=3\dfrac{1}{2}:\left(3x-2\right)\)
\(\Leftrightarrow\dfrac{1}{12}.\dfrac{21}{4}=\dfrac{7}{2}.\dfrac{1}{3x-2}\)
\(\Leftrightarrow\dfrac{7}{16}=\dfrac{7}{6x-4}\)
\(\Leftrightarrow6x-4=7:\dfrac{7}{16}\)
\(\Leftrightarrow6x-4=16\)
\(\Leftrightarrow x=\dfrac{10}{3}\)
Vậy \(S=\left\{\dfrac{10}{3}\right\}\)
\(4,\dfrac{x-1}{x+2}=\dfrac{4}{5}\left(dk:x\ne-2\right)\)
\(\Rightarrow5\left(x-1\right)=4\left(x+2\right)\)
\(\Rightarrow5x-5=4x+8\)
\(\Rightarrow x=13\left(tmdk\right)\)
Vậy \(S=\left\{13\right\}\)
D
8
HN
27 tháng 12 2018
\(a,5^{x-2}=125\)
\(\Rightarrow5^{x-2}=5^3\)
\(\Rightarrow x-2=3\)
\(\Rightarrow x=5\)
\(b,3^{x+4}=243\)
\(\Rightarrow3^{x+4}=3^5\)
\(\Rightarrow x+4=5\)
\(\Rightarrow x=1\)
27 tháng 12 2018
\(5^{x-2}=125\)
\(\Rightarrow5^{x-2}=5^3\)
\(\Rightarrow x-2=3\)
\(\Rightarrow x=3+2\)
\(\Rightarrow x=5\)
HT
3
IA
31 tháng 10 2020
\(a,\frac{1}{3}x+0.25=\frac{5}{7}\)
\(\Leftrightarrow\frac{1}{3}x=\frac{13}{28}\)
\(\Leftrightarrow x=\frac{39}{28}\)
vậy...
\(b,\frac{11}{12}x+0,25=\frac{5}{6}\)
\(\Leftrightarrow\frac{11}{12}x=\frac{7}{12}\)
\(\Leftrightarrow x=\frac{7}{11}\)
vậy.....
\(c,\left(\frac{-1}{3}\right)^2+\frac{2}{3}x=\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{9}+\frac{2}{3}x=\frac{1}{4}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{5}{36}\)
\(\Leftrightarrow x=\frac{5}{24}\)
vậy......
\(d,\left(3x+2\right)^3=-\frac{8}{125}\)
\(\Leftrightarrow3x+2=-\frac{2}{5}\)
\(\Leftrightarrow3x=-\frac{12}{5}\)
\(\Leftrightarrow x=-\frac{4}{5}\)
vậy.......
31 tháng 10 2020
\(\frac{1}{3x}+0,25=\frac{5}{7}\)
\(\frac{1}{3x}+\frac{1}{4}=\frac{5}{7}\)
\(\frac{1}{3x}=\frac{13}{28}\)
\(3x=\frac{28}{13}\)
\(x=\frac{28}{39}\)
\(\frac{11}{12x}+0,25=\frac{5}{6}\)
\(\frac{11}{12x}+\frac{1}{4}=\frac{5}{6}\)
\(\frac{11}{12x}=\frac{7}{12}\)
\(x=\frac{11}{12}:\frac{7}{12}\)
\(x=\frac{7}{11}\)
\(\left(-\frac{1}{3}\right)^2+\frac{2}{3x}=\frac{1}{4}\)
\(\frac{1}{9}+\frac{2}{3x}=\frac{1}{4}\)
\(\frac{2}{3x}=\frac{5}{36}\)
\(x=\frac{2}{3}:\frac{5}{36}\)
\(x=\frac{5}{24}\)
\(\left(3x+2\right)^3=\left(-\frac{8}{125}\right)\)
\(\left(3x+2\right)^3=\left(-\frac{2}{5}\right)^3\)
\(\Rightarrow3x+2=-\frac{2}{3}\)
\(3x=-\frac{8}{3}\)
\(x=-\frac{9}{8}\)
FI
1
PT
24 tháng 8 2018
\(\left(3x-5\right)^8=\frac{1}{125}\left(5-3x\right)^{11}\)
\(\Leftrightarrow-\left(3x-5\right)^8=-\frac{1}{125}\left(3x-5\right)^{11}\)
\(\Leftrightarrow-1=-\frac{1}{125}\left(3x-5\right)^3\)
\(\Leftrightarrow\frac{1}{125}\left(3x+5\right)^3=1\)
\(\Leftrightarrow\left(3x-5\right)^3=125\)
\(\Leftrightarrow3x-5=\sqrt[3]{125}\)
\(\Leftrightarrow3x-5=5\)
\(\Leftrightarrow3x=10\)
\(\Leftrightarrow x=\frac{10}{3}\)
Vậy phương trình đã cho có tập nghiệm \(S=\left\{\frac{10}{3}\right\}\)
NM
0