Tìm x để biểu thức sau đạt GTNN
M= \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|\left|x+5\right|\)
Tìm GTNN của E=\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+...+\left|x-n\right|\) \(\left(n>5;n\in N\right)\)
Tìm x,biết
a, \(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
Với x ∉ -2,-5,-10,-17
b,\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
Với x∉1,3,8,20
c,\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
tìm x;y
b \(\left|x-y\right|+\left|y+\dfrac{9}{25}=0\right|\)
c \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)
a \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=\dfrac{-1}{4}-y\)
d \(\left|x\left(x^2-\dfrac{5}{4}\right)\right|=x\) \(\left(x\ge0\right)\)
e \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
2. Tìm x biết:
a)2(x+2)(x+4)\dfrac{2}{\left(x+2\right)\left(x+4\right)} + 4(x+4)(x+8)\dfrac{4}{\left(x+4\right)\left(x+8\right)} + 6(x+8)(x+14)\dfrac{6}{\left(x+8\right)\left(x+14\right)} = x(x+2)(x+14)\dfrac{x}{\left(x+2\right)\left(x+14\right)}
b)x2023\dfrac{x}{2023} + x+12022\dfrac{x+1}{2022} x+22021\dfrac{x+2}{2021} +...+ x+20221\dfrac{x+2022}{1} + 2023 = 0.
Gíup mình giải 2 bài này với!
Cảm ơn các bạn rất nhiều!!!
1) \(\left|3x+2\right|=\left|x+1\right|\)
2) \(\left|\left(x+2\right)\times x\right|=\left|x+2\right|\)
3) \(\left|2x+3\right|=x+1\)
4) \(\left|4x+5\right|+3.x=7\)
Tìm x, y, z:
a) \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2005\right|=0\)
b) \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\) = 0
c) \(\dfrac{16}{2^x}=1\)
d) \(\left(2x-1\right)^3=-27\)
e) \(\left(x-2\right)^2=1\)
f) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{25}\)
g) \(\left(x-1\right)^2=\left(x-1\right)^6\)
Tìm x để biểu thức B=\(\left|x-5\right|-\left|x-6\right|\) đạt GTNN
C= \(\left|x-2015\right|+\left|x+2016\right|+\left|x-2017\right|\) đạt GTNN
\(\left(\dfrac{1}{2}\right)^3.\left[\left(\dfrac{1}{2}\right)^X\right]^X-\dfrac{5}{8}=\left(\dfrac{1}{2}\right)^4.\left(-9\right)\)
