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29 tháng 5 2017

\(\dfrac{x+16}{9}=\dfrac{y-25}{-16}=\dfrac{z+49}{25}\) (1)

Ta có: \(4x^3-3=29\)

\(\Rightarrow4x^3=32\Rightarrow x^3=8\)

\(\Rightarrow x=2\)

Thay \(x=2\) vào điều (1) ta có:

\(\dfrac{2+16}{9}=\dfrac{y-25}{-16}=\dfrac{z+49}{25}\)

\(\Rightarrow\dfrac{y-25}{-16}=\dfrac{z+49}{25}=\dfrac{18}{9}\)

\(\Rightarrow\dfrac{y-25}{-16}=\dfrac{z+49}{25}=2\)

\(\Rightarrow\left\{{}\begin{matrix}y-25=2.\left(-16\right)\\z+49=2.25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y-25=-32\\z+49=50\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=-7\\z=1\end{matrix}\right.\)

Vậy giá trị của biểu thức \(A=x+2y+3z\) là:

\(A=2+2.\left(-7\right)+3.1=2-14+3=-9\)

Chúc bạn học tốt!!!

15 tháng 3 2017

Ta có : \(4x^3-3=29\)

\(\Rightarrow4x^3=32\)

\(\Rightarrow x^3=8\)

\(\Rightarrow x=2\)

Thay x = 2 vào \(\dfrac{x+16}{9}=\dfrac{y-25}{-16}\) ta có :

\(\dfrac{2+16}{9}=\dfrac{y-25}{-16}\)

\(\Rightarrow2=\dfrac{y-25}{-16}\)

\(\Rightarrow y-25=-32\)

\(\Rightarrow y=-7\)

Thay \(y=-7\) vào \(\dfrac{y-25}{-16}=\dfrac{z+49}{25}\) ta có :

\(\dfrac{-7-25}{-16}=\dfrac{z+49}{25}\)

\(\Rightarrow2=\dfrac{z+49}{25}\)

\(\Rightarrow z+49=50\)

\(\Rightarrow z=1\)

Thay x = 2; y = -7; z = 1 vào biểu thức A ta có :

\(A=2+2.\left(-7\right)+3.1\)

\(A=-9\)

Vậy A = -9

2. Tham khảo thêm tại đây nha bạn

https://hoc24.vn/hoi-dap/question/417550.html

18 tháng 4 2017

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

\(=-5,13:\left(\dfrac{145}{28}-\dfrac{17}{9}.\dfrac{125}{100}+\dfrac{79}{63}\right)\)

\(=-5,13:\left(\dfrac{145}{28}-\dfrac{17}{9}.\dfrac{5}{4}+\dfrac{79}{63}\right)\)

\(=-5,13:\left(\dfrac{145}{28}-\dfrac{85}{36}+\dfrac{79}{63}\right)\)

\(=-5,13:\dfrac{57}{14}=-5,13:\dfrac{15}{57}\)

\(=\dfrac{-71,82}{57}=1,26\)

Vậy \(A=1,26\)

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

\(=\left(\dfrac{10}{3}.1,9+19,5:\dfrac{13}{3}\right).\left(\dfrac{62-12}{75}\right)\)

\(=\left(\dfrac{19}{3}+\dfrac{58,5}{13}\right).\dfrac{50}{75}\)

\(=\left(\dfrac{19}{3}+4,5\right).\dfrac{2}{3}\)

\(=\dfrac{32,5}{3}.\dfrac{2}{3}=\dfrac{65}{9}=7\dfrac{2}{9}\)

Vậy \(B=7\dfrac{2}{9}\)

21 tháng 10 2017

quá đúng và cũng quá chuẩn!!!!!!!!!yeueoeoyeu

2 tháng 4 2018

Ta có:

\(a^2+ab+\dfrac{b^2}{3}=c^2+\dfrac{b^2}{3}+a^2+ac+c^2\)

\(\Rightarrow a^2+ab+\dfrac{b^2}{3}=2c^2+\dfrac{b^2}{3}+a^2+ac\)

\(\Rightarrow ab=2c^2+ac\)

\(\Rightarrow ab+ac=2ac+2c^2\)

\(\Rightarrow a\left(b+c\right)=2c\left(a+c\right)\)

\(\Rightarrow\dfrac{2c}{a}=\dfrac{b+c}{a+c}\left(đpcm\right)\)

14 tháng 3 2020

Bái Phục , Mong ngài hãy nhận con làm đệ tử .haha

\(M=\dfrac{\dfrac{4}{3}+1}{\dfrac{5}{3}-1}=\dfrac{7}{3}:\dfrac{2}{3}=\dfrac{7}{2}\)

17 tháng 5 2017

Bài 1:

\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

\(=\dfrac{1}{99.97}-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{93.95}+\dfrac{1}{95.97}\right)\)

\(=\dfrac{1}{99.97}-\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{93.95}+\dfrac{2}{95.97}\right)\)

\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{93}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{97.99}-\dfrac{1}{2}.\dfrac{96}{97}\)

\(=\dfrac{1}{97.99}-\dfrac{48}{97}\)

Bạn tính nốt nhé

Bài 2, 3 bạn kiểm tra lại đề giúp mk

17 tháng 5 2017

Bài 1 :

\(\dfrac{1}{99.97}-\dfrac{1}{99.95}-\dfrac{1}{95.93}-......-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

\(=\dfrac{1}{97.99}-\left(\dfrac{1}{97.95}+\dfrac{1}{95.93}+...+\dfrac{1}{5.3}-\dfrac{1}{3.1}\right)\)

\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+...+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\)

\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{97.99}-\dfrac{48}{97}\)

\(=\dfrac{51}{97}\)

8 tháng 9 2018

\(a,A=\dfrac{7}{35}+\left(-1\dfrac{3}{4}+\dfrac{12}{7}\right)-\left(\dfrac{1}{4}-\dfrac{2}{7}-\dfrac{12}{35}\right)-\dfrac{3}{7}\)\(A=\dfrac{7}{35}-\dfrac{7}{4}+\dfrac{12}{7}-\dfrac{1}{4}+\dfrac{2}{7}+\dfrac{13}{35}-\dfrac{3}{7}\\ A=\left(\dfrac{7}{35}+\dfrac{13}{35}\right)-\left(\dfrac{7}{4}-\dfrac{1}{4}\right)+\left(\dfrac{12}{7}+\dfrac{2}{7}-\dfrac{3}{7}\right)\)

\(A=\dfrac{4}{7}-\dfrac{3}{2}+\dfrac{11}{7}\\ A=\left(\dfrac{4}{7}+\dfrac{11}{7}\right)-\dfrac{3}{2}\\ A=\dfrac{15}{7}-\dfrac{3}{2}=\dfrac{9}{14}\)

18 tháng 3 2017

\(4x^3-3=29\Rightarrow x^3=\dfrac{29+3}{4}=8\Rightarrow x=\sqrt[3]{8}=2\)

Thay số: \(\dfrac{x+16}{9}=\dfrac{2+16}{9}=2\)

Suy ra: \(y=\left(-16\right)\cdot2+25\Leftrightarrow y=-7\)\(z=25\cdot2-49\Leftrightarrow z=1\)

\(A=x+2y+3z\Leftrightarrow2+\left(-14\right)+3=-9\)

18 tháng 3 2017

\(4x^3-3=29\Rightarrow x^3=\dfrac{32}{4}=2^3\Rightarrow x=3\)

\(\dfrac{19}{9}=\dfrac{2y-2.25}{-32}=\dfrac{3z+49.3}{75}=\dfrac{2y+3z+49.3-25.2}{75-32}=\dfrac{2y+3z+97}{43}\)

\(\dfrac{\left(2y+3z+3\right)+94}{43}=\dfrac{19}{9}\) \(\Rightarrow\left(x+2y+3z\right)=\dfrac{43.19}{9}-94\)

15 tháng 9 2017

Ta có :

\(2x^3-1=15\)

\(\Leftrightarrow2x^3=16\)

\(\Leftrightarrow x^3=8\)

\(\Leftrightarrow x=2\)

Thay \(x=2\) zô : \(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)

\(\Leftrightarrow\dfrac{2+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)

\(\Leftrightarrow\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)

+) \(\dfrac{y-25}{16}=2\)

\(\Leftrightarrow y-25=32\)

\(\Leftrightarrow y=57\)

+) \(\dfrac{z+9}{25}=2\)

\(\Leftrightarrow z+9=50\)

\(\Leftrightarrow z=41\)

Ta có :

\(\left\{{}\begin{matrix}x=2\\y=57\\z=41\end{matrix}\right.\) \(\Leftrightarrow x+y+z=2+57+41=100\)