ĐKXĐ: x>=-0,46
\(\sqrt{5x+2,3}=3\)
=>5x+2,3=9
=>5x=6,7
=>x=1,34
Đúng 0
Bình luận (0)
Các câu hỏi tương tự
root(2x + 2, 3) + root(2x + 1, 3) = root(2x ^ 2, 3) + root(2x ^ 2 + 1, 3)
11 tháng 5 2021 lúc 21:01
Cho `2x^3=3y^3=4z^3`
`CMR:(\root{3}{2x^2+3y^2+4z^2})/(\root{3}{2}+\root{3}{3}+\root{3}{4})=1`
Giúp!
$x=\root(3)(22\sqrt(2+)25-\root(3)(22\sqrt(2))- 25)$
root(x ^ 2 + 2, 3) = 3
3 * root(x - 2, 3) = - 6
`(\root[3]{26+15\sqrt{3}}-\sqrt{3})/(\sqrt{6-2\sqrt{5}-\sqrt{5})`
[4x/(x^2+x+3)]+[5x/(x^2-5x+3)]=-3/2
\(\frac{3}{5x-1}-\frac{2}{3-5x}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\)
tìm x biết a (x-2)^2+(x+3)^2-4(x+1)=5
b (2x-3)(2x+3)-(x-1)^2-3x(x-5)=-44
c (5x+1)^2-(5x+3)(5x-3)=30
d (x+3)^2+(x-2)(x+2)-2(x-1)^2=7