\(\Leftrightarrow x\cdot\dfrac{2}{9}+\dfrac{1213}{990}=\dfrac{382}{165}\)
=>x*2/9=1079/990
=>x=1079/220
Tìm x biết:
\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=5x-1\)-1
\(\left|x+1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)
\(\left|x+1\right|-\left|3x+2\right|=x+2\)
tim x
\(a,\left(x-1\right)^2=\left|\frac{1}{4}-\frac{1}{2}-\frac{3}{4}\right|\)
b , \(\left(x^x-8\right).\left(x^2-15\right)< 0\)
Tim x : \(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
tìm x,biết:
a)\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
b)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
c)\(\left(x+2\right)^2=\frac{38}{25}+\frac{9}{10}-\frac{11}{15}+\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950}\)
giúp tớ với,huhu
Tim x biet
\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(\right)x+10}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
Tìm x, biết:
a)\(3\left|x+4\right|-\left|2x+1\right|-5\left|x+3\right|+\left|x-9\right|=5\)
b)\(\left|\frac{11}{5}-x\right|+\left|x+\frac{1}{5}\right|+\frac{41}{5}=1,2\)
c)\(2\left|x+\frac{7}{2}\right|+\left|x\right|-\frac{7}{2}=\left|\frac{11}{5}-x\right|\)
Tìm x
\(a,\left|x.\left(x-7\right)\right|=x\)
\(b,\left|x-1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)
\(c,7^{x+2}+2.7^{x-1}=345\)
tim x
\(\left|x-1\right|+\left|x-2\right|+\left|y-3\right|+\left|x-4\right|=3\)
tim GTNN
\(P=\left(x+3y-5\right)^2-6xy+26\)
\(Q=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
tim x
a, \(\left(x+1\right)\left(2-x\right)\left(3+x^2\right)\ge0\)
b, \(\left(2x+1\right)\left(x+3\right)\left(x-5\right)>0\)
c,\(\frac{\left(3x-6\right)\left(x-4\right)}{x+3}< 0\)
