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Đặt:\(7a=3b=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{k}{7}\\b=\dfrac{k}{3}\end{matrix}\right.\)
\(\Rightarrow\dfrac{k}{7}.\dfrac{k}{3}=20\Rightarrow\dfrac{k^2}{21}=20\Rightarrow k^2=420\Rightarrow k=\pm\sqrt{420}\)
Xét: \(k=\sqrt{420}\)
\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{\sqrt{420}}{7}\\b=\dfrac{\sqrt{420}}{3}\end{matrix}\right.\)
Xét: \(k=-\sqrt{420}\)
\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{-\sqrt{420}}{7}\\b=\dfrac{-\sqrt{420}}{3}\end{matrix}\right.\)
b) Dựa vào tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)
\(=\dfrac{a+b-c}{2+3-4}=\dfrac{100}{1}=100\)
\(\Rightarrow\left\{{}\begin{matrix}a=100.2=200\\b=100.3=300\\c=100.4=400\end{matrix}\right.\)
c) Đặt: \(\dfrac{a}{4}=\dfrac{b}{7}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=4k\\b=7k\end{matrix}\right.\)
\(\Rightarrow4k.7k=112\)
\(\Rightarrow28k^2=112\)
\(k^2=4\Rightarrow k=\pm2\)
Xét: \(k=2\)
\(\Rightarrow\left\{{}\begin{matrix}a=2.4=8\\b=2.7=14\end{matrix}\right.\)
Xét:\(k=-2\)
\(\Rightarrow\left\{{}\begin{matrix}a=-2.4=-8\\c=-2.7=-14\end{matrix}\right.\)
\(\text{a) }7a=3b\text{ và }ab=20\\ \text{Đặt }7a=3b=k\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{7}k\\b=\dfrac{1}{3}k\end{matrix}\right.\left(1\right)\\ \text{Từ }\left(1\right)\text{ suy ra : }\\ ab=20\\ \Leftrightarrow\left(\dfrac{1}{7}k\right)\left(\dfrac{1}{3}k\right)=20\\ \Leftrightarrow\left(\dfrac{1}{7}\cdot\dfrac{1}{3}\right)\left(k\cdot k\right)=20\\ \Leftrightarrow\dfrac{1}{21}k^2=20\\ \Leftrightarrow k^2=420\\ \Leftrightarrow k=\sqrt{420}\\ \text{Từ }k=\sqrt{420}\text{ suy ra : }\left\{{}\begin{matrix}a=\dfrac{1}{7}\cdot\sqrt{420}\\b=\dfrac{1}{3}\cdot\sqrt{420}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\dfrac{\sqrt{420}}{7}\\b=\dfrac{\sqrt{420}}{3}\end{matrix}\right.\\ \text{Vậy }a=\dfrac{\sqrt{420}}{7};b=\dfrac{\sqrt{420}}{3}\)
\(\text{b) }\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\text{ và }a+b-c=100\\ \text{ Theo bài ra ta có : }\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\\ a+b-c=100\\ \text{Áp dụng tính chất dãy tỉ số bằng nhau ta được : }\\ \dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{a+b-c}{2+3-4}=\dfrac{100}{1}=100\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=100\\\dfrac{b}{3}=100\\\dfrac{c}{4}=100\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=200\\b=300\\c=400\end{matrix}\right.\\ \text{Vậy }a=200;b=300;c=400\)
\(\text{c) }\dfrac{a}{4}=\dfrac{b}{7}\text{ và }ab=112\\ \text{Đặt }\dfrac{a}{4}=\dfrac{b}{7}=k\Rightarrow\left\{{}\begin{matrix}a=4k\\b=7k\end{matrix}\right.\left(1\right)\\ \text{Từ }\left(1\right)\text{ suy ra : }\\ ab=112\\ \Leftrightarrow4k\cdot7k=112\\ \Leftrightarrow28k^2=112\\ \Leftrightarrow k^2=4\\ \Leftrightarrow k=2\\ \text{Từ }k=2\Rightarrow\left\{{}\begin{matrix}a=4\cdot2\\b=7\cdot2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=8\\b=14\end{matrix}\right.\\ \text{Vậy }a=8;b=14\)
Giải:
a) Theo đề ra, ta có:
\(\dfrac{a}{b}=\dfrac{5}{7}\) và \(a+b=72\) (Sửa x+y =72)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{5}{7}\Leftrightarrow\dfrac{a}{5}=\dfrac{b}{7}\)
\(\Leftrightarrow\dfrac{a}{5}=\dfrac{b}{7}=\dfrac{a+b}{5+7}=\dfrac{72}{12}=6\)
\(\Rightarrow\dfrac{a}{5}=6\Rightarrow a=6.5=30\)
\(\Rightarrow\dfrac{b}{7}=6\Rightarrow b=6.7=42\)
Vậy ...
b) Theo đề ra, ta có:
\(\dfrac{a}{6}=\dfrac{b}{4}=\dfrac{c}{3}\) và \(a+b-c=21\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\Leftrightarrow\dfrac{a}{6}=\dfrac{b}{4}=\dfrac{c}{3}=\dfrac{a+b-c}{6+4-3}=\dfrac{21}{7}=3\)
\(\Rightarrow\dfrac{a}{6}=3\Rightarrow a=3.6=18\)
\(\Rightarrow\dfrac{b}{4}=3\Rightarrow b=3.4=12\)
\(\Rightarrow\dfrac{c}{3}=3\Rightarrow a=3.3=9\)
Vậy ...
c) Theo đề ra, ta có:\(\dfrac{12}{x}=\dfrac{3}{y}\) và \(x-y=36\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{12}{x}=\dfrac{3}{y}\Leftrightarrow\dfrac{x}{12}=\dfrac{y}{3}\)
\(\Leftrightarrow\dfrac{x}{12}=\dfrac{y}{3}=\dfrac{x-y}{12-3}=\dfrac{36}{9}=4\)
\(\Rightarrow\dfrac{x}{12}=4\Rightarrow x=12.4=48\)
\(\Rightarrow\dfrac{y}{3}=4\Rightarrow x=3.4=12\)
Vậy ...
d) Theo đề ra, ta có:
\(\dfrac{a}{2}=\dfrac{b}{5}=\dfrac{c}{7}\) và \(a+b-c=20\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\Leftrightarrow\dfrac{a}{2}=\dfrac{b}{5}=\dfrac{c}{7}=\dfrac{a+b-c}{2+5-7}=\dfrac{20}{0}=\varnothing\)
Đề câu này sai nhé!
Chúc bạn học tốt!
a) Áp dụng tính chất dãy tỉ số bằng nhau ,ta có :
\(\dfrac{a}{b}=\dfrac{5}{7}\Leftrightarrow\dfrac{a}{5}=\dfrac{b}{7}=\dfrac{a+b}{5+7}=\dfrac{72}{12}=6\)
\(\Rightarrow\left\{{}\begin{matrix}a=5.6=30\\b=7.6=42\end{matrix}\right.\)
b) Áp dụng tính chất dãy tỉ số bằng nhau ,ta có :
\(\dfrac{a}{6}=\dfrac{b}{4}=\dfrac{c}{3}=\dfrac{a+b-c}{6+4-3}=\dfrac{21}{7}=3\)
\(\Rightarrow\left\{{}\begin{matrix}a=6.3=18\\b=4.3=12\\c=3.3=9\end{matrix}\right.\)
c) Áp dụng tính chất dãy tỉ số bằng nhau ,ta có :
\(\dfrac{12}{x}=\dfrac{3}{y}\Leftrightarrow\dfrac{x}{12}=\dfrac{y}{3}=\dfrac{x-y}{12-3}=\dfrac{36}{9}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=12.4=48\\y=3.4=12\end{matrix}\right.\)
d) Áp dụng tính chất dãy tỉ số bằng nhau ,ta có :
\(\dfrac{a}{2}=\dfrac{b}{5}=\dfrac{c}{7}=\dfrac{a+b-c}{2+5-7}=\dfrac{20}{0}\) (Vô lý)
=> Không thể làm
1. Tính:
a. \(\dfrac{\text{−1 }}{\text{4 }}+\dfrac{\text{5 }}{\text{6 }}=\dfrac{-3}{12}+\dfrac{10}{12}=\dfrac{7}{12}\)
b. \(\dfrac{\text{5 }}{\text{12 }}+\dfrac{\text{-7 }}{8}=\dfrac{10}{24}+\dfrac{-21}{24}=\dfrac{-11}{24}\)
c. \(\dfrac{-7}{6}+\dfrac{-3}{10}=\dfrac{-35}{30}+\dfrac{-9}{30}=\dfrac{-44}{30}=\dfrac{-22}{15}\)
d.\(\dfrac{-3}{7}+\dfrac{5}{6}=\dfrac{-18}{42}+\dfrac{35}{42}=\dfrac{17}{42}\)
2. Tính :
a. \(\dfrac{2}{14}-\dfrac{5}{2}=\dfrac{2}{14}-\dfrac{35}{14}=\dfrac{-33}{14}\)
b.\(\dfrac{-13}{12}-\dfrac{5}{18}=\dfrac{-39}{36}-\dfrac{10}{36}=\dfrac{49}{36}\)
c.\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)
d. \(0,6--1\dfrac{2}{3}=\dfrac{6}{10}--\dfrac{5}{3}=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9}{15}+\dfrac{25}{15}=\dfrac{34}{15}\)
3. Tính :
a.\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)
b.\(\dfrac{-6}{9}-\dfrac{12}{16}=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8}{12}-\dfrac{9}{12}=\dfrac{-17}{12}\)
c. \(\dfrac{-3}{7}-\dfrac{-2}{11}=\dfrac{-3}{7}+\dfrac{2}{11}=\dfrac{-33}{77}+\dfrac{14}{77}=\dfrac{-19}{77}\)
d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{1}+\dfrac{1}{10}\)
\(=\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
Chế Kazuto Kirikaya thử tham khảo thử đi !!!
Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya
d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
a) Ta có:
+) a/2=b/3
=>a=2b/3
+) b/5=c/4
=>c=4b/5
Lại có:
a-b+c=49
=> 2b/3 -b + 4b/5 =49
=> 7b/15==49
=> b= 105
Khi đó:
+) a=2b/3=2.105/3=70
+)c=4b/5=4.105/5=84
Vậy a=70; b=105; c=84...
chúc bạn học tốt
a: a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}=\dfrac{a}{a-b}\)
b: \(\dfrac{a}{b}=\dfrac{bk}{b}=k\)
\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k=\dfrac{a}{b}\)
c \(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{k}{3k+1}\)
\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{k}{3k+1}=\dfrac{a}{3a+b}\)
d: \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2=\dfrac{ac}{bd}\)
Mấy bài dễ tự làm nhé:D
1)
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)
Ta có điều phải chứng minh
\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)
Ta có điều phải chứng minh
a,
\(a+b=-9\\ b+c=2\\ c+a=-3\\ \Rightarrow a+b+b+c+c+a=\left(-9\right)+2+\left(-3\right)\\ 2a+2b+2c=-10\\ 2\left(a+b+c\right)=-10\\ a+b+c=-5\\ a+b=-9\\ \Rightarrow a+b+c=-5\Leftrightarrow\left(-9\right)+c=-5\Rightarrow c=4\\ b+c=2\\ \Rightarrow a+b+c=-5\Leftrightarrow a+2=-5\Rightarrow a=-7\\ c+a=-3\\ \Rightarrow a+b+c=-5\Leftrightarrow\left(-3\right)+b=-5\Rightarrow b=-2\)
Vậy \(a=-7;b=-2;c=5\)
b,
\(a+b=\dfrac{1}{2}\\ b+c=\dfrac{3}{4}\\ c+a=\dfrac{-5}{6}\\ \Rightarrow a+b+b+c+c+a=\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{-5}{6}\\ 2a+2b+2c=\dfrac{6}{12}+\dfrac{9}{12}+\dfrac{-10}{12}\\ 2\left(a+b+c\right)=\dfrac{5}{12}\\ a+b+c=\dfrac{5}{24}\\ a+b=\dfrac{1}{2}\\ \Rightarrow a+b+c=\dfrac{5}{24}\Leftrightarrow\dfrac{1}{2}+c=\dfrac{5}{24}\Rightarrow c=\dfrac{-7}{24}\\ b+c=\dfrac{3}{4}\\ \Rightarrow a+b+c=\dfrac{5}{24}\Leftrightarrow a+\dfrac{3}{4}=\dfrac{5}{24}\Rightarrow a=\dfrac{-13}{24}\\ a+c=\dfrac{-5}{6}\\ \Rightarrow a+b+c=\dfrac{5}{24}\Leftrightarrow b+\dfrac{-5}{6}=\dfrac{5}{24}\Rightarrow b=\dfrac{25}{24}\)
Vậy \(a=\dfrac{-13}{24};b=\dfrac{25}{24};c=\dfrac{-7}{24}\)
c,
\(a+b=2\\ b+c=6\\ c+a=3\\ \Rightarrow a+b+b+c+c+a=2+6+3\\ 2a+2b+2c=11\\ 2\left(a+b+c\right)=11\\ a+b+c=5,5\\ a+b=2\\ \Rightarrow a+b+c=5,5\Leftrightarrow2+c=5,5\Rightarrow c=3,5\\ b+c=6\\ \Rightarrow a+b+c=5,5\Leftrightarrow a+6=5,5\Rightarrow a=-0,5\\ c+a=3\\ \Rightarrow a+b+c=5,5\Leftrightarrow b+3=5,5\Rightarrow b=2,5\)
Vậy \(a=-0,5;b=2,5;c=3,5\)
d,
\(a+b=\dfrac{5}{6}\\ b+c=\dfrac{3}{4}\\ c+a=\dfrac{5}{3}\\ \Rightarrow a+b+b+c+c+a=\dfrac{5}{6}+\dfrac{3}{4}+\dfrac{5}{3}\\ 2a+2b+2c=\dfrac{10}{12}+\dfrac{9}{12}+\dfrac{20}{12}\\ 2\left(a+b+c\right)=\dfrac{13}{4}\\ a+b+c=\dfrac{13}{8}\\ a+b=\dfrac{5}{6}\\ \Rightarrow a+b+c=\dfrac{13}{8}\Leftrightarrow\dfrac{5}{6}+c=\dfrac{13}{8}\Rightarrow c=\dfrac{19}{24}\\ b+c=\dfrac{3}{4}\\ \Rightarrow a+b+c=\dfrac{13}{8}\Leftrightarrow a+\dfrac{3}{4}=\dfrac{13}{8}\Rightarrow a=\dfrac{7}{8}\\ c+a=\dfrac{5}{3}\\ \Rightarrow a+b+c=\dfrac{13}{8}\Leftrightarrow b+\dfrac{5}{3}=\dfrac{13}{8}\Rightarrow b=\dfrac{-1}{24}\)
Vậy \(a=\dfrac{7}{8};b=\dfrac{-1}{24};c=\dfrac{19}{24}\)
\(\left\{{}\begin{matrix}a+b=-9\\b+c=2\\c+a=-3\end{matrix}\right.\)
\(\Rightarrow a+b+b+c+c+a=\left(-9\right)+2+\left(-3\right)\)
\(\Rightarrow2a+2b+2c=-10\)
\(\Rightarrow2\left(a+b+c\right)=-10\)
\(\Rightarrow a+b+c=-5\)
\(\Rightarrow\left\{{}\begin{matrix}c=-5-9=-14\\a=-5-2=-7\\b=-5-\left(-3\right)=-2\end{matrix}\right.\)
Sửa câu a:
(x - 2)2 - 36 = 0
(x - 2 - 6)(x - 2 + 6) = 0
(x - 8)(x + 4)= 0
\(\Leftrightarrow \begin{bmatrix} x - 8= 0 & & \\ x + 4 = 0 & & \end{bmatrix}\)
\(\Leftrightarrow \begin{bmatrix} x = 8 & & \\ x = - 4 & & \end{bmatrix}\)
pn bỏ dấu ngoặc bên phải nhé
Vậy x = 8; x = - 4
2:
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)
\(\Rightarrow\dfrac{a+5}{b+6}=\dfrac{a-5}{b-6}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a+5}{b+6}=\dfrac{a-5}{b-6}=\dfrac{a+5-a+5}{b+6-b+6}=\dfrac{10}{12}=\dfrac{5}{6}=\dfrac{a+5+a-5}{b+6+b-6}=\dfrac{2a}{2b}=\dfrac{a}{b}\)
Từ đó suy ra \(\dfrac{a}{b}=\dfrac{5}{6}\)
\(\RightarrowĐPCM\)
\(a,Tacó:\\ \dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}=\dfrac{a^3}{2^3}=\dfrac{a\cdot a\cdot a}{2\cdot2\cdot2}=\dfrac{a\cdot b\cdot c}{2\cdot3\cdot5}=\dfrac{810}{30}=27\\ \Rightarrow\left\{{}\begin{matrix}a=27\cdot2=54\\b=27\cdot3=81\\c=27\cdot5=135\end{matrix}\right.\\ Vậy...\)
Các câu khác cx cùng dạng tương tự bn tự làm nha!
a, \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}\) và a . b . c = 810
Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}=k\)
=> \(\left\{{}\begin{matrix}a=2k\\b=3k\\c=5k\end{matrix}\right.\)
Mà a . b . c = 810
=> 2k . 3k . 5k = 810
=> 30\(k^3\) = 810
=> \(k^3=810:30\)
=> \(k^3=27\)
=> \(k^3=3^3\)
=> k = 3
=> \(a=2.3=6\)
\(b=3.3=9\)
\(c=5.3=15\)
Vậy .....
b, \(\dfrac{a}{4}=\dfrac{b}{3}=\dfrac{c}{9}\)và a - 3b + 4c = 62
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{4}=\dfrac{b}{3}=\dfrac{c}{9}=\dfrac{a-3b+4c}{4-3.3+4.9}=\dfrac{62}{31}=2\)
=> \(\dfrac{a}{4}=2\Rightarrow a=8\)
\(\dfrac{b}{3}=2\Rightarrow b=6\)
\(\dfrac{c}{9}=2\Rightarrow c=18\)
Vậy .......
ý a) sao đang \(a,b,c\) lại thành \(x,y,z\) ? :DD??
b: Đặt \(\dfrac{a}{5}=\dfrac{b}{7}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5k\\b=7k\end{matrix}\right.\)
Ta có: ab=140
nên \(35k^2=140\)
\(\Leftrightarrow k^2=4\)
Trường hợp 1: k=2
\(\Leftrightarrow\left\{{}\begin{matrix}a=5k=10\\b=7k=14\end{matrix}\right.\)
Trường hợp 2: k=-2
\(\Leftrightarrow\left\{{}\begin{matrix}a=5k=-10\\b=7k=-14\end{matrix}\right.\)