Bài 9: Số thập phân hữu hạn. Số thập phân vô hạn tuần hoàn

a: \(=\dfrac{5}{2}-\dfrac{563}{165}-\dfrac{4}{3}+\dfrac{1}{3}\left(1-3-\dfrac{1}{2}\right)\)

\(=\dfrac{-247}{110}+\dfrac{1}{3}\cdot\dfrac{-5}{2}=\dfrac{-508}{165}\)

b: \(=\dfrac{4}{9}+\dfrac{1219}{990}-\dfrac{13}{99}=\dfrac{139}{90}\)

Bài 2: 

a: x<a<y

nên \(a\in\left\{21,37;21,4\right\}\)

b: x<a<y

nên \(a\in\left\{-25,381;-25\right\}\)

Giải:

\(\dfrac{bz-xy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)

\(\Rightarrow\dfrac{a\left(bz-cy\right)}{a^2}=\dfrac{b\left(cx-az\right)}{b^2}=\dfrac{c\left(ay-bx\right)}{c^2}\)

\(\Rightarrow\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-cbx}{c^2}\)

\(\Rightarrow\dfrac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}\)

\(=\dfrac{0}{a^2+b^2+c^2}\)

\(=0\)

Ta có: \(\dfrac{bz-cy}{a}\)

\(\Rightarrow bz-cy=0\)

\(\Rightarrow\dfrac{z}{c}=\dfrac{b}{y}\)\(\left(1\right)\)

Ta có:\(\dfrac{cx-az}{b}=0\)

\(\Rightarrow cx-az=0\)

\(\Rightarrow\dfrac{x}{a}=\dfrac{z}{c}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra:

\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)(đpcm)

Mình đã làm ở đây: Câu hỏi của Huyền Trang Tiến Tài

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\(\text{a) }0.3\left(12\right)=\dfrac{312-3}{990}=\dfrac{103}{330}\)

\(\text{b) }2.0\left(33\right)=2\dfrac{033-0}{990}=2\dfrac{33}{990}=3\dfrac{1}{30}=\dfrac{91}{30}\)

a) 18/5

b)66

You học tham số rồi hả ( Trước khi giải bài này you lên google tìm định nghĩa tham số hay trong sách j đó đi )

\(P\left(1\right)=1^2+2m.1+m^2\)
\(Q\left(0\right)=0+\left(2m^3+1\right).0+m^2=m^2\)
\(P\left(1\right)=Q\left(0\right)\)
\(\Rightarrow1+2m+m^2=m^2\)
\(\Rightarrow1+2m=0\)
\(\Rightarrow2m=-1\)

a: \(=\dfrac{5}{2}-\dfrac{563}{165}-\dfrac{4}{3}+\dfrac{1}{3}\cdot\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{7}{2}\right)\)

\(=\dfrac{-247}{110}+\dfrac{1}{3}\cdot\dfrac{-5}{2}=\dfrac{-247}{110}+\dfrac{-5}{6}=\dfrac{-508}{165}\)

b: \(=\left[\dfrac{5}{9}\cdot\dfrac{2}{9}\right]:\left(\dfrac{10}{3}\cdot\dfrac{25}{23}\right)-\dfrac{22}{15}\cdot\dfrac{3}{4}\)

\(=\dfrac{10}{18}:\dfrac{250}{69}-\dfrac{66}{60}\)

\(=\dfrac{10}{18}\cdot\dfrac{69}{250}-\dfrac{11}{10}\)

\(=\dfrac{-71}{75}\)

Giải:

a) Có: \(0,\left(37\right)=0,373737373737...\)

\(0,\left(62\right)=0,626262626262...\)

\(\Leftrightarrow0,\left(37\right)+0,\left(62\right)=0,99999999999...\)

Mà \(0,9999999999999...\simeq1\)

Hay \(0,\left(9\right)=1\)

Vậy \(0,\left(37\right)+0,\left(62\right)=1\).

b) \(0,\left(33\right).3=0,99999...=0,\left(9\right)=1\)

Vậy \(0,\left(33\right).3=1\).

Chúc bạn học tốt!!!

\(a)0,\left(37\right)=0,37373737....\)

\(0,\left(62\right)=0,62626262....\)\(\Leftrightarrow0,\left(37\right)+0,\left(62\right)=0,99999999....\)

Mà \(0,99999999....\simeq1\)

hoặc \(0,\left(9\right)\simeq1\)

\(\Rightarrow0,\left(37\right)+\left(0,62\right)=1\)

\(b)0,\left(33\right).3=1\)

\(\Leftrightarrow0,99999999....=0,\left(9\right)\simeq1\)

\(\Rightarrow0,\left(33\right).3=1\)

Chúc bạn học tốt!

==" ko hỉu 2 người kia giải sao nữa ???

Ta có:

\(0,\left(37\right)=\dfrac{37}{99}\)(cái này vì 0,0(1) = 1/99 nhé, mk khỏi cm)

\(0,\left(62\right)=\dfrac{62}{99}\)

\(\Rightarrow0,\left(37\right)+0,\left(62\right)=\dfrac{37}{99}+\dfrac{62}{99}=\dfrac{99}{99}=1\left(đpcm\right)\)

b, Tương tự:

\(0,\left(33\right)=\dfrac{33}{99}\Rightarrow0,\left(33\right).3=\dfrac{33}{99}.3=1\left(đpcm\right)\)

\(\dfrac{3}{7}\approx0,43;\dfrac{15}{21}\approx0,71;\dfrac{-4}{5}=-0,8;\dfrac{-7}{13}=-0,54\)

2,\(-2,54=\dfrac{-127}{50};-2,03=\dfrac{-203}{100};0,25=\dfrac{1}{4}\)