Bài 6:
TA có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
...
\(\frac{1}{8^2}<\frac{1}{7\cdot8}=\frac17-\frac18\)
Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{8^2}<1-\frac12+\frac12-\frac13+\cdots+\frac17-\frac18\)
=>\(A<1-\frac18\)
=>A<1
Bài 4:
a: \(-0,6x-\frac73=5,4\)
=>\(-\frac35x=\frac{27}{5}+\frac73=\frac{81+35}{15}=\frac{116}{15}\)
=>\(x=\frac{116}{15}:\frac{-3}{5}=\frac{116}{15}\cdot\frac{-5}{3}=\frac{-116}{3\cdot3}=\frac{-116}{9}\)
b: \(2,8:\left(\frac15-3x\right)=1\frac25\)
=>\(2,8:\left(\frac15-3x\right)=1,4\)
=>\(\frac15-3x=2\)
=>\(3x=\frac15-2=-\frac95\)
=>\(x=-\frac95:3=-\frac35\)
c: \(\left(x-\frac58\right)\cdot\frac{5}{18}=\frac{-15}{36}\)
=>\(x-\frac58=-\frac{15}{36}:\frac{5}{18}=-\frac{5}{12}\cdot\frac{18}{5}=\frac{-18}{12}=-\frac32\)
=>\(x=-\frac32+\frac58=\frac{-12}{8}+\frac58=-\frac78\)
d: \(-65x+7,5=19-11:\frac14\)
=>\(-65x+7,5=19-11\cdot4=19-44=-25\)
=>-65x=-32,5
=>\(x=\frac{32.5}{65}=0,5\)
e: \(\frac57\cdot\frac{3x-8}{5}=3\frac89\)
=>\(\frac{3x-8}{7}=\frac{35}{9}\)
=>\(3x-8=\frac{35}{9}\cdot7=\frac{245}{9}\)
=>\(3x=\frac{245}{9}+8=\frac{245}{9}+\frac{72}{9}=\frac{317}{9}\)
=>\(x=\frac{317}{27}\)
f: \(\left(3\frac12+2x\right)\cdot2\frac23=5\frac13\)
=>\(\left(2x+\frac72\right)\cdot\frac83=\frac{16}{3}\)
=>\(2x+\frac72=2\)
=>\(2x=2-\frac72=-\frac32\)
=>\(x=-\frac32:2=-\frac34\)
g: \(2\frac15-\frac15:x+100\%=3\frac{1}{45}\)
=>\(\frac{11}{5}-\frac15:x+1=3+\frac{1}{45}\)
=>\(\frac15:x=\frac{16}{5}-3-\frac{1}{45}=\frac15-\frac{1}{45}=\frac{8}{45}\)
=>\(x=\frac15:\frac{8}{45}=\frac15\cdot\frac{45}{8}=\frac98\)
h: \(\frac{x-2}{4}=\frac{x+5}{3}\)
=>4(x+5)=3(x-2)
=>4x+20=3x-6
=>4x-3x=-6-20
=>x=-26
