Lời giải:
$\frac{1}{2^x}+\frac{1}{2^{x+4}}=17$
$\frac{16}{2^{x+4}}+\frac{1}{2^{x+4}}=17$
$\frac{17}{2^{x+4}}=17$
$2^{x+4}=1$
$x+4=0$
$x=-4$
Lời giải:
$\frac{1}{2^x}+\frac{1}{2^{x+4}}=17$
$\frac{16}{2^{x+4}}+\frac{1}{2^{x+4}}=17$
$\frac{17}{2^{x+4}}=17$
$2^{x+4}=1$
$x+4=0$
$x=-4$
a) Cho \(M=\dfrac{42-x}{x-15}\) . Tìm số nguyên x để m đạt giá trị nhỏ nhất .
b) Tìm x sao cho \(\left(\dfrac{1}{2}\right)^x+\left(\dfrac{1}{2}\right)^{x-4}=17\)
Tìm x.
\(1,\dfrac{3}{2}\left(x-\dfrac{1}{3}\right)-\dfrac{1}{2}\left(x+\dfrac{1}{2}\right)=\dfrac{1}{4}\)
\(2,3\left(x-2\right)-4\left(x+2\right)=x+2\)
\(3,4x\left(x-1\right)+4x-2\left(x+1\right)=-2\)
\(4,x\left(x+2\right)-3\left(x-1\right)=3\left(x+1\right)\)
Tìm x biết :
\(\left|x-\dfrac{1}{2}\right|+\left|x-\dfrac{1}{3}\right|+\left|x-\dfrac{1}{4}\right|+....+\left|x-\dfrac{1}{10}\right|=2x\)
Tim x, biet:
a, \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
b, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
c, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}\)\(=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
d, \(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
Help me!!!!!!!!!!
Can gap lam. Ai lam duoc cau no thi lam nha. Cam on nhieu truoc!!!!!!!!!!!!
a) Tìm tập hợp các số nguyên x, biết rằng\(4\dfrac{5}{9}:2\dfrac{5}{18}-7< x< \left(3\dfrac{1}{5}:3,2+4,5.1\dfrac{31}{45}\right):\left(-21\dfrac{1}{2}\right)\)
b) tìm x, biết \(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+\left|x+\dfrac{1}{12}\right|+\left|x+\dfrac{1}{20}\right|+....+\left|x+\dfrac{1}{110}\right|-11x\)
c)Tính gt biểu thức \(C=2x^3-5y^3+2015\) tại x,y thỏa mãn \(\left|x-1\right|+\left(y+2\right)^{20}=0\)
1) Tính
\(A=\dfrac{1}{13}+\dfrac{3}{13.23}+\dfrac{3}{23.33}+...+\dfrac{3}{2003.2013}\)
\(B=\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)....\left(\dfrac{1}{2018}-1\right)\)
2) Tìm x biết:
a) \(x^2-2x-15=0\)
b) \(\dfrac{3}{\left(x+2\right).\left(x+5\right)}+\dfrac{5}{\left(x+5\right).\left(x+10\right)}+\dfrac{7}{\left(x+10\right).\left(x+17\right)}=\dfrac{x+1}{\left(x+2\right).\left(x+17\right)}\)
3) Cho \(\dfrac{a}{b}=\dfrac{d}{c}\) . Chứng minh: \(\dfrac{a^2+d^2}{b^2+c^2}=\dfrac{ad}{bc}\)
4) Cho \(f\left(x\right)=x^{100}-x^{99}+...+x^2-x+1\)
\(g\left(x\right)=-x^{101}+x^{100}-x^{99}+...+x^2-x+1\)
Tính giá trị của hiệu \(f\left(x\right)-g\left(x\right)\) tại x=0,1
5) Cho tam giác ABC có \(\widehat{A}=\ge90\) ; \(M\in AB,N\in AC\)
Chứng minh: BC > MN
6) Cho tam giác ABC, M là trung điểm BC, biết \(\widehat{BAM}>\widehat{CAM}\) . So sánh B và C
1.Tìm x, biết.
a, \(|x|\) + x = \(\dfrac{1}{3}\)
b, \(|x+2|\) = x
c, \(\dfrac{2}{\left(x-1\right)\left(x-3\right)}\) + \(\dfrac{5}{\left(x-3\right)\left(x-8\right)}\) + \(\dfrac{12}{\left(x-8\right)\left(x-20\right)}\) - \(\dfrac{1}{x-20}\) = \(\dfrac{-3}{4}\)
d, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}\) + \(\dfrac{5}{\left(x+5\right)\left(x+10\right)}\) + \(\dfrac{7}{\left(x+10\right)\left(x+17\right)}\) = \(\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
e, \(\dfrac{x-1}{2009}\) + \(\dfrac{x-2}{2008}\) = \(\dfrac{x-3}{20007}\) + \(\dfrac{x-4}{2006}\)
Tìm x :
a) \(\dfrac{2x-3}{3}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\)
b) \(\dfrac{2}{3x}-\dfrac{3}{12}=\dfrac{4}{5}-\left(\dfrac{7}{x}-2\right)\)
c) \(\dfrac{x+2014}{2}+\dfrac{2x+4028}{7}=\dfrac{x+2009}{5}+\dfrac{x+2020}{6}\)
d)\(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+18\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
Help me , bn nào giải đc bài nò thì giải nha !!! =))
Tìm x:
\(3\left|x+4\right|-\left|2x+1\right|-5\left|x+3\right|+\left|x-9=5\right|\)
\(\left|x-2\right|+\left|x-3\right|+\left|2x-8\right|=9\\ \left|x+2\right|+\left|x+3\right|+\left|x+1\right|=4\\ \left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.13}\right|+...+\left|x+\dfrac{1}{397.401}\right|=101x\)