Bài 2:
a.
$2689-x=23\times 15$
$2689-x=345$
$x=2689-345=2344$
b. Không nhìn thấy đoạn cuối.
e.
$x-0,134=2,107+0,51.1,3$
$x-0,134=2,77$
$x=0,134+2,77$
$x=2,904$
g.
$\frac{3}{5}.x=\frac{1}{5}+\frac{2}{3}$
$\frac{3}{5}.x=\frac{13}{15}$
$x=\frac{13}{9}$
l.
$71+65.4=\frac{y+140}{y}+260$
$331=\frac{y+140}{y}+260$
$\frac{y+140}{y}=71$
$y+140=71y$
$140=70y$
$y=2$
Bài 3.
1. Ta có:
$(4,5+5,7+a):3=5,3$
$4,5+5,7+a=15,9$
$10,2+a=15,9$
$a=5,7$
2.
$\frac{3}{5}.\frac{a}{b}=\frac{1}{5}+\frac{2}{3}$
$\frac{3}{5}.\frac{a}{b}=\frac{13}{15}$
$\frac{a}{b}=\frac{13}{15}:\frac{3}{5}$
$\frac{a}{b}=\frac{13}{9}$
3.
$\frac{2}{5}< \frac{x}{6}< frac{8}{10}$
$\frac{2}{5}< \frac{x}{6}< frac{4}{5}$
$\frac{12}{30}< \frac{5x}{30}< frac{24}{30}$
$\Rightarrow 12< 5x< 24$
Vì $x$ tự nhiên nên $x\in \left\{3;4\right\}$
4.
Số số hạng: $\frac{2x-1-1}{2}+1=x$ (số hạng)
$1+3+5+....+(2x-1)=1$
$\frac{(2x-1+1).x}{2}=1$
$2x^2=2$
$x^2=1$
$x=1$
Bài 4:
a.
$(x+1)+(x+2)+....+(x+100)=5060$
$100x+(1+2+3+...+100)=5060$
$100x+\frac{100.101}{2}=5060$
$100x+5050=5060$
$100x=10$
$x=\frac{1}{10}$
b.
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right).x=\frac{33}{100}\)
\(\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\right).x=\frac{33}{100}\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right).x=\frac{33}{100}\)
\(\left(1-\frac{1}{100}\right).x=\frac{33}{100}\)
\(\frac{99}{100}.x=\frac{33}{100}\)
\(x=\frac{1}{3}\)
c. Không rõ đề.
d.
\(1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x(x+1)}=\frac{3980}{1991}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{x(x+1)}=\frac{1990}{1991}\)
\(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x(x+1)}=\frac{1990}{1991}\)
\(\frac{1}{2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{(x+1)-x}{x(x+1)}=\frac{1990}{1991}\)
\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1990}{1991}\)
\(\frac{1}{2}+\frac{1}{2}-\frac{1}{x+1}=\frac{1990}{1991}\)
\(\frac{1}{x+1}=\frac{1}{1991}\Rightarrow x=1990\)
