\(=\dfrac{600x}{x}-\dfrac{1800}{x}+10x-30=600\)
\(=600-\dfrac{1800}{x}+\dfrac{10x^2}{x}-30=600\)
\(\Rightarrow\dfrac{1800}{x}+\dfrac{10x^2}{x}-30=0\)
=> \(1800+10x^2-30=0\)
=> \(10x^2+1770=0\)
=> 10x^2 = -1770
=> x^2 = \(-\dfrac{1770}{10}=-177\)
=> x = \(-\sqrt{177}\)
Tìm x
\(a,3-x=x+1,8\)
\(b,2x-5=7x+35\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(k,7x^2-11=6x^2-2\)
\(m,5\left(x+3.2^3\right)=10^2\)
\(n,\dfrac{4}{9}-(\dfrac{1}{6^2})=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(b,\dfrac{x}{2}-\left(\dfrac{3x}{5}-\dfrac{13}{5}\right)=-\left(\dfrac{7}{5}+\dfrac{7}{10}.x\right)\)
\(c,\dfrac{2x-3}{3}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\)
\(d,\dfrac{2}{3x}-\dfrac{3}{12}=\dfrac{4}{5}-\left(\dfrac{7}{x}-2\right)\)
\(e,2\left(x-1\right)=\left(x-1\right)^2\)
Tìm x:
1) \(\text{(x−1):0,16=−9:(1−x)}\)
2) \(\left(\left|x\right|-\dfrac{3}{2}\right)\left(2x^2-10\right)=0\)
3)\(8\sqrt{x}=x^2\left(x\ge0\right)\)
Có mấy giá trị của x thỏa mãn bt dưới đây:
\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
Tìm x, biết:
a) \(\left(-\dfrac{5}{9}\right)^{10}\) : x = \(\left(\dfrac{-5}{9}\right)^8\)
b) x ; \(\left(\dfrac{-5}{9}\right)^8\) = \(\left(\dfrac{-9}{5}\right)^8\)
c) x3 = -8
Tìm x :
1) \(\left(-0,75x+\dfrac{5}{2}\right).\dfrac{4}{7}-\left(-\dfrac{1}{3}\right)=-\dfrac{5}{6}\)
2) \(\left(4x-9\right)\left(2,5+\dfrac{-7}{3}x\right)=0\)
3) \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
4)\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\dfrac{-64}{125}\)
a/ \(\dfrac{-3}{5}\) - x = \(\dfrac{21}{10}\)
b/ x : \(\dfrac{2}{9}\) = \(\dfrac{9}{2}\)
c/ \(\dfrac{x}{9}\) = \(\dfrac{5}{3}\)
d/ x : \(\left(\dfrac{2}{5}\right)^3\)= \(\left(\dfrac{5}{2}\right)^3\)
Tìm x biết:
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{x\left(2x+1\right)}=\dfrac{1}{10},\left(x\inℕ^∗\right)\)
Giải chi tiết giúp mik nha.
tìm x biết \(\left|x-\dfrac{1}{3}\right|+\left|x-\dfrac{1}{15}\right|+\left|x-\dfrac{1}{35}\right|+\left|x-\dfrac{1}{63}\right|+...+\left|x-\dfrac{1}{399}\right|=-11x\)
\(x^2-19=5.9;\left(2x+1\right)^3=-0,001;\left(\dfrac{5}{6}\right)^{2x-1}=\left(\dfrac{5}{6}\right)^5;\left(\dfrac{1}{3}x-\dfrac{2}{3}\right)^3=27;\left(\dfrac{1}{32}\right)^x=\left(\dfrac{1}{2}\right)^{15}\)
