Với x là số dương => (a-4)/5 >0 => a> 4
Với số âm thì ngược lại

A = 1/30 + 1/42 + 1/56 + 1/72 + 1/90

A = 1/5x6 + 1/6x7 + 1/7x8 + 1/8x9 + 1/9x10

A = 1/5 - 1/6+ 1/6 -1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10

A = 1/5 - 1/10

A = 1/10

\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)

\(\Rightarrow A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

\(\Rightarrow A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(\Rightarrow A=\frac{1}{5}-\frac{1}{10}\)

\(\Rightarrow A=\frac{2-1}{10}\)

\(\Rightarrow A=\frac{1}{10}\)

\(a,n^2=\frac{1}{16}\)

\(n^2=\left(\frac{1}{4}\right)^2\)

\(n=\left|\frac{1}{4}\right|\)

=>\(n=\frac{1}{4}\)hoặc \(n=-\frac{1}{4}\)

\(b,n^3.n^2=\frac{32}{243}\)

\(n^5=\frac{32}{243}\)

\(n^5=\left(\frac{2}{3}\right)^5\)

\(n=\frac{2}{3}\)

\(c,\left(x^2\right)^2=\frac{81}{16}\)

\(x^4=\left(\frac{3}{2}\right)^4\)

=>\(x=\left|\frac{3}{2}\right|\)

\(x=\frac{3}{2}\)hoặc \(x=-\frac{3}{2}\)

SSH: (399-1):2+1= 200
Neu chia moi nhom 4 so thi so cap so la:
200:4 = 50
Ta co:
B=1+3-5-7+9+11-13-15+...+393+395-397-399
B= (1+3-5-7)+(9+11-13-15)+...+(393+395-397-399)
B= -8 + -8 +...+ -8
B= -8 . 50
B= -400

SSH: (399-1):2+1= 200
Neu chia moi nhom 4 so thi so cap so la:
200:4 = 50
Ta co:
A=1+3-5-7+9+11-13-15+...+393+395-397-399
A= (1+3-5-7)+(9+11-13-15)+...+(393+395-397-399)
A= -8 + -8 +...+ -8
A= -8 . 50

A= -400

Đề bài thiếu điều kiện a<b hoặc x<y

Ta có:\(z< y\Rightarrow\frac{a}{m}< \frac{b}{m}\Rightarrow a< b\)\(z=\frac{a+b}{2m}>\frac{a+a}{2m}=\frac{2a}{2m}=\frac{a}{m}=x\)\(z=\frac{a+b}{2m}< \frac{b+b}{2m}=\frac{2b}{2m}=\frac{b}{m}=y\)

Vậy: x < y thì x < z < y