1. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=1\end{matrix}\right.\). Cmr: \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}+\frac{bc}{\sqrt{\left(1-a\right)^2\left(1+a\right)}}+\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\le\frac{3\sqrt{2}}{8}\)

2. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c\le1\end{matrix}\right.\). Cmr: \(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab\left(a+b\right)}+\frac{1}{bc\left(b+c\right)}+\frac{1}{ac\left(a+c\right)}\ge\frac{87}{2}\)

3. \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca=2abc\end{matrix}\right.\). Cmr: \(\frac{1}{a\left(2a-1\right)^2}+\frac{1}{b\left(2b-1\right)^2}+\frac{1}{c\left(2c-1\right)^2}\ge\frac{1}{2}\)

4. \(\left\{{}\begin{matrix}x,y,z>0\\x+y+z=2015\end{matrix}\right.\). Tìm min \(A=\frac{x^4+y^4}{x^3+y^3}+\frac{y^4+z^4}{y^3+z^3}+\frac{z^4+x^4}{z^2+x^2}\)

Mn giúp mk với ạ! Thanks nhiều

1. Giải hpt: \(\left\{{}\begin{matrix}x+y+z=0\\2x+3y+z=0\\\left(x+1\right)^2+\left(y+2\right)^2+\left(z+3\right)^2=26\end{matrix}\right.\)

2. Cho x,y,z là nghiệm của hpt : \(\left\{{}\begin{matrix}\frac{x}{3}+\frac{y}{12}-\frac{z}{4}=1\\\frac{x}{10}+\frac{y}{5}+\frac{z}{3}=1\end{matrix}\right.\) . Tính \(A=x+y+z\)

Ta có: \(\frac{a}{3}=\frac{b}{4};\frac{b}{2}=\frac{c}{5}\Rightarrow\frac{a}{6}=\frac{b}{12}=\frac{c}{20}=\frac{a+b+c}{6+12+20}=\frac{3}{38}\)

=> \(a=\frac{3}{38}.3=\frac{9}{19}\)

      \(b=\frac{3}{38}.12=\frac{18}{19}\)

       \(c=\frac{3}{38}.20=\frac{30}{19}\)

Vậy a = 9/19 ; b = 18/19 ; c = 30/19

đáp án là 1789

số thứ nhất là :15228:[(15510-15228):6]=324

ban tu ke so do nhe

so tuoi cua me la

32:4.5=40

sô tuoi cua cn la

40-32=8

so bi la

25 +25+25+8=83(vien)

      dap so 83 vien

bạn có bị làm sao ko mà đăng linh tinh như thế