Bài 7: Tỉ lệ thức

\(\dfrac{x}{y}=\dfrac{7}{20}\Rightarrow\dfrac{x}{7}=\dfrac{y}{20}\) (1)

\(\dfrac{y}{z}=\dfrac{5}{8}\Rightarrow\dfrac{y}{5}=\dfrac{z}{8}\Rightarrow\dfrac{y}{20}=\dfrac{z}{32}\) (2)

Từ (1) và (2) suy ra \(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}\)

\(\Rightarrow\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}=\dfrac{2x+5y-2z}{14+100-64}=\dfrac{100}{50}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.7=14\\y=2.20=40\\z=2.32=64\end{matrix}\right.\)

Vậy...

Ta có : \(\dfrac{x}{y}\) = \(\dfrac{7}{20}\) \(\Rightarrow\dfrac{x}{7}=\dfrac{y}{20}\) ( 1)

Ta có : \(\dfrac{y}{z}=\dfrac{5}{8}\) \(\Rightarrow\dfrac{y}{5}=\dfrac{z}{8}\)

\(\Rightarrow\dfrac{y}{5}.\dfrac{1}{4}=\dfrac{z}{8}.\dfrac{1}{4}\)

\(\Rightarrow\dfrac{y}{20}=\dfrac{z}{32}\) (2)

Từ (1) và (2)

\(\Rightarrow\) \(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}\)

Đặt \(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}=k\)

\(\Rightarrow x=7k\) ; \(y=20k\) ; \(z=32k\)

Thay \(x=7k\) ; \(y=20k\) ; \(z=32k\) vào \(2x+5y-2z=100\)

\(\Rightarrow2.\left(7k\right)+5.\left(20k\right)-2.\left(32k\right)\) \(=100\)

\(\Rightarrow\)\(14k+100k-64k=100\)

\(\Rightarrow k.\left(14+100-64\right)=100\)

\(\Rightarrow k.50=100\)

\(\Rightarrow k=100:50\) \(\Rightarrow k=2\)

\(\Rightarrow x=7k=7.2=14\)

\(\Rightarrow y=20k=20.2=40\)

\(\Rightarrow z=32k=32.2=64\)

Vậy \(x=14\) ; \(y=40\) ;\(z=64\)

bạn ơi \(k\in Q\) nhé

Câu 2 :
\(x-y=7\)
\(\Rightarrow x=7+y\)
*)
\(B=\dfrac{3\left(7+y\right)-7}{2\left(7+y\right)+y}-\dfrac{3y+7}{2y+7+y}\)
\(=\dfrac{21+3y-7}{14+3y}-\dfrac{3y+7}{3y+7}\)
\(=\dfrac{14y+3y}{14y+3y}-1\)
\(=1-1\)
\(=0\)
Vậy B = 0

2/ Ta có :

\(B=\dfrac{3x-7}{2x+y}-\dfrac{3y+7}{2y+x}\)

\(=\dfrac{3x-\left(x-y\right)}{2x+y}-\dfrac{3y+\left(x-y\right)}{2y+x}\)

\(=\dfrac{3x-x+y}{2y+x}-\dfrac{3y+x-y}{2y+x}\)

\(=\dfrac{2x+y}{2x+y}-\dfrac{2y+x}{2y+x}\)

\(=1-1=0\)

Bài 3:
\(a+b+c=2018\text{ và }\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\)
\(2018\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{1}{2}.2018\)
\(\Rightarrow\dfrac{2018}{a+b}+\dfrac{2018}{b+c}+\dfrac{2018}{a+c}=1009\)
\(\Rightarrow\dfrac{\left(a+b\right)+c}{a+b}+\dfrac{a+\left(b+c\right)}{b+c}+\dfrac{b+\left(a+c\right)}{a+c}=1009\)
\(\Rightarrow1+\dfrac{c}{a+b}+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}+1=1009\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1009-3\)
\(\Rightarrow S=1006\)
Vậy \(S=1006\)

a) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có:

\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (1)

\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (2)

Từ (1) và (2) suy ra \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)

b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=q\Rightarrow\left\{{}\begin{matrix}a=bq\\c=dq\end{matrix}\right.\)

Ta có:

\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bq+b}{dq+d}\right)^2=\left[\dfrac{b\left(q+1\right)}{d\left(q+1\right)}\right]^2=\dfrac{b}{d}\) (1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bq\right)^2+b^2}{\left(dq\right)^2+d^2}=\dfrac{b^2.q^2+b^2}{d^2.q^2+d^2}=\dfrac{b^2\left(q^2+1\right)}{d^2\left(q^2+1\right)}=\dfrac{b}{d}\) (2)

Từ (1) và (2) suy ra \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)

\(\dfrac{a}{b}=\dfrac{c}{d}\) => \(\dfrac{a}{c}=\dfrac{b}{d}\)

áp dụng tính chất dãy tỉ số = nhau ta có

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{2c-3d}\)

= \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\) (đpcm)

Còn 3 đến 4 cách nữa: áp dụng t/c của dãy tỉ số bằng nhau, t/c của tỉ lệ thức, áp dụng ĐNg,...

bạn ở yên khánh ninh bình ak đâu mà giống đề t zậy

a,x.(1/4 + 1/5) - (1/7 + 1/8) = 0

=> x.9/20-15/56=0

=> x . 9/20 = 15/56

=> x= 15/56. 20/9

=>x = 25/42

b, -3/4 - |4/5 - x| = -1

=> |4/5 - x| = -3/4-1

=> |4/5 - x| = -7/4

=> x \(\in\varnothing\)

áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

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

\(=\dfrac{a+b+c}{3a+3b+3c}=\dfrac{a+b+c}{3\left(a+b+c\right)}\)

mà các số \(\dfrac{a}{a+2b}=\dfrac{b}{b+2b}=\dfrac{c}{c+2a}\) là số nguyên dương nên a,b,c là các số nguyên dương

\(\Rightarrow\) (a+b+c)\(⋮\) 3 (ĐPCM)

giup mik vs!!!

Thầy phinit và các thầy cô ơi giúp em với!!!

a, ta có :

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{2b}{2d}\)

áp dụng tính chất dă y tỉ số bằng nhau ta có:

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{2b}{2d}=\dfrac{a+2b}{c+2d}=\dfrac{2a-b}{2c-d}\)

\(\Rightarrow\dfrac{a+2b}{c+2d}=\dfrac{2a-b}{2c-d}\Rightarrow\dfrac{a+2b}{2a-b}=\dfrac{c+2d}{2c-d}\) (ĐPCM)

b, ta có:

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3c}{3d}\)

áp dụng tính chất dă tỉ số bằng nhau ta có :

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3c}{3d}=\dfrac{a+3c}{b+3d}=\dfrac{a-c}{b-d}\)

\(\Rightarrow\dfrac{a+3c}{b+3d}=\dfrac{a-c}{b-d}\)

\(\Rightarrow\left(a+3c\right)\left(b-d\right)=\left(b+3d\right)\left(a-c\right)\) (ĐPCM)

+) Nếu \(a,b,c\ne0\) thì theo tính chất dãy tỉ số bằng nhau ta có :

\(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)

+) Nếu \(a+b+c=0\)

\(\Leftrightarrow b+c=-a;c+a=-b;a+b=-c\)

\(\Leftrightarrow\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=\dfrac{a}{-a}=\dfrac{b}{-b}=\dfrac{c}{-c}=-1\)

ta có

b²=ac\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\) (1)

c²=bd\(\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\) (2)

từ(1),(2)\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

\(\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}\)

áp dung tính chấ t dăy tỉ số bằng nhau ta có

\(\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\) (ĐPCM)