Bài 3: Biểu đồ

Câu 2: 

a: Ta có: \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{378}{395}\\z=2004\end{matrix}\right.\)

b: \(\left|x-\dfrac{1}{2}\right|+\left|y+\dfrac{3}{2}\right|+\left|x-y-z-\dfrac{1}{2}\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+\dfrac{3}{2}=0\\x-y-z-\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{3}{2}\\z=\dfrac{3}{2}\end{matrix}\right.\)

\(1,\)

\(a,\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)

\(=\dfrac{11}{125}+\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)\)

\(=\dfrac{11}{125}+\left(\dfrac{-1}{2}\right)+\dfrac{1}{2}\)

\(=\dfrac{11}{125}\)

\(b,-1\dfrac{5}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)

\(=\dfrac{-12}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)

\(=-15.\left[\dfrac{12}{7}+\dfrac{2}{7}+\left(-5\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\right]\)

\(=-15.\left[2+\left(-5\right).\dfrac{1}{105}\right]\)

\(=-15.\left(2-\dfrac{1}{21}\right)\)

\(=-15.\dfrac{41}{21}=\dfrac{-615}{21}\)

\(2,\)

\(a,\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)

\(\Leftrightarrow\dfrac{11}{13}-\dfrac{5}{42}+x=\dfrac{-15}{28}+\dfrac{11}{13}\)

\(\Leftrightarrow x=\dfrac{-15}{28}+\dfrac{11}{13}-\dfrac{11}{13}+\dfrac{5}{42}\)

\(\Leftrightarrow x=\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\left(\dfrac{5}{42}+\dfrac{-15}{28}\right)\)

\(\Leftrightarrow x=\dfrac{5}{12}\)

Vậy \(x=\dfrac{5}{12}\)

\(b,\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|-3,75=-2,15\)

\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2,15+3,75=1,6=\dfrac{16}{10}=\dfrac{8}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{8}{5}\\x+\dfrac{4}{15}=\dfrac{-8}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{5}-\dfrac{4}{15}=\dfrac{4}{3}\\x=\dfrac{-8}{5}-\dfrac{4}{15}=\dfrac{-28}{15}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{4}{3};\dfrac{-28}{15}\right\}\)

\(c,7^{x+2}+2.7^{x-1}=345\)

\(\Leftrightarrow7^{x-1}.\left(7^3+2\right)=345\)

\(\Leftrightarrow7^{x-1}.\left(343+2\right)=345\)

\(\Leftrightarrow7^{x-1}.345=345\)

\(\Leftrightarrow7^{x-1}=345:345=1\)

\(\Leftrightarrow x-1=0\)

\(x=0+1=1\)

Vậy \(x=1\)

a) 25/6

b) -13,6651341

Đặt :

\(A=\dfrac{3}{9.14}+\dfrac{3}{14.19}+........+\dfrac{3}{\left(5n-1\right)\left(5n+4\right)}\)

\(\Leftrightarrow\dfrac{5}{3}A=\dfrac{5}{9.14}+\dfrac{5}{14.19}+........+\dfrac{5}{\left(5n-1\right)\left(5n+4\right)}\)

\(\Leftrightarrow\dfrac{5}{3}A=\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...........+\dfrac{1}{5n-1}-\dfrac{1}{5n+4}\)

\(\Leftrightarrow\dfrac{5}{3}A=\dfrac{1}{9}-\dfrac{1}{5n+4}\)

\(\Leftrightarrow A=\left(\dfrac{1}{9}-\dfrac{1}{5n+4}\right):\dfrac{5}{3}\)

\(\Leftrightarrow A=\left(\dfrac{1}{9}-\dfrac{1}{5n+4}\right).\dfrac{3}{5}\)

\(\Leftrightarrow A=\dfrac{1}{9}.\dfrac{3}{5}-\dfrac{1}{5n+4}.\dfrac{3}{5}\)

\(\Leftrightarrow A=\dfrac{1}{15}-\dfrac{1}{5n+4}.\dfrac{3}{5}< \dfrac{1}{15}\)

\(\Leftrightarrow A< \dfrac{1}{15}\left(đpcm\right)\)

là sao ko hỉu ak Khải Vũ tại vì mk ko có chơi mấy cái này

bao nhieu

???

kết quả bằng vô tận đấy bạn

55555555559999999999999999999999999999999910101010101010101010101010101010101010108.58.8.8.8.8.8.8.8.8.8.8.8.8.888..8.88.8.8.8.88..8.89999999999999999999999999999999999999999999999999999999

hì, bạn ghi sai đề ròi nhá

a) \(A=-5,13:\left(5\dfrac{5}{28}-1\dfrac{8}{9}.1,25+1\dfrac{16}{63}\right)\)

\(\Leftrightarrow A=-5,13:\left(5\dfrac{5}{28}-\dfrac{85}{36}+1\dfrac{16}{63}\right)\)

\(\Leftrightarrow A=-5,13:\left(\dfrac{355}{126}+1\dfrac{16}{63}\right)\)

\(\Leftrightarrow A=-5,13:\dfrac{57}{14}\)

\(\Leftrightarrow A=\dfrac{-63}{50}\)

b) \(B=\left(3\dfrac{1}{3}.1,9+19,5:4\dfrac{1}{3}\right).\left(\dfrac{62}{75}-\dfrac{4}{25}\right)\)

\(\Leftrightarrow B=\left(\dfrac{19}{3}+19,5.\dfrac{3}{13}\right).\dfrac{2}{3}\)

\(\Leftrightarrow B=\left(\dfrac{19}{3}+\dfrac{9}{2}\right).\dfrac{2}{3}\)

\(\Leftrightarrow B=\dfrac{65}{6}.\dfrac{2}{3}\)

\(\Leftrightarrow B=\dfrac{65}{9}\)

Theo định lí tổng 3 góc của 1 tam giác ta có:

C + A + B = 180⁰

C + 80⁰ + 60⁰ = 180⁰

C = 180⁰ - ( 80⁰ + 60⁰) = 40⁰

Vậy góc ACB có số đo là 40⁰

Vì AD là tia phân giác của BAC nên:

A₂ = A₁ = \(\dfrac{80}{2}=40^0\)

Vậy góc BAD có số đo là 40⁰

Theo định lí tổng 3 góc của 1 tam giác, ta có:

D₁ + C + A₁ = 180⁰

D₁ + 40⁰ + 40⁰ = 180⁰

D₁ = 180⁰ - ( 40⁰ + 40⁰ )

D₁ =100⁰

Vậy góc ADC có số đo là 100⁰

1) Phân số đầu nhân 2.

_ Phân số thứ 2 nhân 3, p/s thứ 3 giữ nguyên.

_ Lấy phân số đầu + p/s thứ 2 - p/s thứ 3.

_ Dựa vào dãy tỉ số bằng nhau tìm x, y, z.

2) \(x-y-z=0\Rightarrow x=y+z\)

Khi đó thay vào B được:

\(B=\left(1-\dfrac{z}{y+z}\right)\left(1-\dfrac{y+z}{y}\right)\left(1+\dfrac{y}{z}\right)\)

\(=\dfrac{y}{y+z}.\dfrac{z}{y}.\dfrac{y+z}{z}\)

\(=1\)

Vậy B = 1.