1/Tìm x:
a) [(3^x . x + 2^4 .3) - 14] : 14 = 5
b) 2^2x+1 = 125
2) Tính:
abc + bca = ?
K
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18 tháng 1
b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)
=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\)
=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)
=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)
d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)
=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)
=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)
=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)
CM
5
27 tháng 7 2021
a) Ta có: \(\left(2x-5\right)^3=216\)
\(\Leftrightarrow2x-5=6\)
\(\Leftrightarrow2x=11\)
hay \(x=\dfrac{11}{2}\)
b) Ta có: \(2x-3⋮x+4\)
\(\Leftrightarrow-11⋮x+4\)
\(\Leftrightarrow x+4\in\left\{1;-1;11;-11\right\}\)
hay \(x\in\left\{-3;-5;7;-15\right\}\)
3 tháng 12 2023
Alo, sugeni two wai phem. Si ga no, you woo be the me that nas te, ai gi da
Y
31 tháng 5 2023
\(a,50\%x-0,2+x=\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{2}x-0,2+x=\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{2}x+x=\dfrac{4}{5}+0,2\)
\(\Leftrightarrow\dfrac{3}{2}x=\dfrac{4}{5}+\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{3}{2}x=1\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
\(b,\left(x-\dfrac{3}{4}\right):\dfrac{1}{2}+\dfrac{3}{2}=\dfrac{25}{2}\)
\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{25}{2}-\dfrac{3}{2}\)
\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{22}{2}\)
\(\Leftrightarrow x-\dfrac{3}{4}=11:2\)
\(\Leftrightarrow x=\dfrac{11}{2}+\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{25}{4}\)
CM
0
CM
0
O
25 tháng 3 2022
a) \(\dfrac{9}{7}\div x=\dfrac{3}{5}\)
\(x=\dfrac{9}{7}\div\dfrac{3}{5}\)
\(x=\dfrac{15}{7}\)
b) \(x+\dfrac{1}{3}=\dfrac{8}{6}-\dfrac{1}{2}\)
\(x+\dfrac{1}{3}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}-\dfrac{1}{3}\)
\(x=\dfrac{1}{2}\)
2 tháng 7 2021
a) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Leftrightarrow6x=-20\)
hay \(x=-\dfrac{10}{3}\)
c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)
\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)
\(\Leftrightarrow17x=17\)
hay x=1
1.
b) Sai đề
2) abc+bca=a100+b10+c+b100+c10+a=(a+b)100+(b+c)10+c+a
Thanks nhiều nha mình sẽ coi lại