Bài 8: Tính chất của dãy tỉ số bằng nhau

Ta có :

\(\dfrac{x}{\dfrac{2}{3}}=\dfrac{z}{0,5};\dfrac{z}{1}=\dfrac{y}{\dfrac{4}{7}}\)

\(\Leftrightarrow\)\(\dfrac{x}{\dfrac{16}{3}}=\dfrac{z}{4}=\dfrac{y}{\dfrac{16}{7}}\)

\(\Rightarrow\)\(\dfrac{z+y}{4+\dfrac{16}{7}}=\dfrac{66}{\dfrac{44}{7}}=10,5\)

[ \(\dfrac{z}{4}=10,5\Rightarrow z=42\) ]

[ \(\dfrac{y}{\dfrac{16}{7}}=10,5\Rightarrow y=24\) ]

[\(\dfrac{x}{\dfrac{16}{3}}=10,5\Rightarrow x=56\) ]

Vậy \(x+y+z=42+24+56=122\)

Ta có: \(3a=7b\Rightarrow\dfrac{a}{7}=\dfrac{b}{3}\)

\(4b=3c\Rightarrow\dfrac{b}{3}=\dfrac{c}{4}\)

Khi đó: \(\dfrac{a}{7}=\dfrac{b}{3}=\dfrac{c}{4}\)

\(\Rightarrow\dfrac{a}{7}=\dfrac{4b}{12}=\dfrac{5c}{20}\)

Áp dụng tc dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{7}=\dfrac{4b}{12}=\dfrac{5c}{20}=\dfrac{a+4b-5c}{7+12-20}=\dfrac{-30}{-1}=30\)

Do \(\left\{{}\begin{matrix}\dfrac{a}{7}=30\\\dfrac{4b}{12}=30\\\dfrac{5c}{20}=30\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=210\\b=90\\c=120\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}a=210\\b=90\\c=120\end{matrix}\right.\).

Từ đề ta có: \(\dfrac{a}{7}=\dfrac{b}{3}=\dfrac{c}{4}\)

Áp dụng t/c của dãy tỉ số = nhau ta có:

\(\dfrac{a}{7}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{4b}{12}=\dfrac{5c}{20}=\dfrac{a+4b-5c}{7+12-20}=\dfrac{-30}{-1}=30\)

\(\Rightarrow\left\{{}\begin{matrix}a=30\cdot7\\b=30\cdot3\\c=30\cdot4\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=210\\b=90\\c=120\end{matrix}\right.\)

Vậy..........

\(\text{Theo bài ra ta có:}\)

\(3a=7b\Rightarrow\dfrac{3a}{7}=b\Rightarrow\dfrac{a}{7}=\dfrac{b}{3}\\ 4b=3c\Rightarrow\dfrac{4b}{3}=c\Rightarrow\dfrac{b}{3}=\dfrac{c}{4}\\ \Rightarrow\dfrac{a}{7}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a}{7}=\dfrac{4b}{12}=\dfrac{5c}{20}\)

\(a+4b-5c=-30\)

\(\text{Áp dụng tính chất dãy tỉ số bằng nhau ta được : }\)

\(\dfrac{a}{7}=\dfrac{4b}{12}=\dfrac{5c}{20}=\dfrac{a+4b-5c}{7+12-20}=\dfrac{-30}{-1}=30\) \(\left(1\right)\)

\(\text{Từ}\) \(\left(1\right)\) \(\text{suy ra : }\) \(\left\{{}\begin{matrix}\dfrac{a}{7}=30\Rightarrow a=210\\\dfrac{b}{3}=30\Rightarrow b=90\\\dfrac{c}{4}=30\Rightarrow c=120\end{matrix}\right.\)

\(\text{Vậy}\) \(a=210\\ b=90\\ c=120\)

\(\)

surf trc khi hỏi

surf trc khi hỏi

Bài 1:

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

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

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

Chúc bạn học tốt!!!

Ta có: \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{y+t+x}=\dfrac{t}{y+x+z}\)

\(\Rightarrow\dfrac{x}{y+z+t}+1=\dfrac{y}{z+t+x}+1=\dfrac{z}{y+t+x}+1=\dfrac{t}{y+x+z}+1\)

\(\Rightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{y+t+x}=\dfrac{x+y+z+t}{y+x+z}\)+) Xét \(x+y+z+t=0\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(x+t\right)\\z+t=-\left(x+y\right)\\x+t=-\left(y+z\right)\end{matrix}\right.\)

\(\Rightarrow A=-1\)

+) Xét \(x+y+z+t\ne0\Rightarrow x=y=z=t\)

\(\Rightarrow A=1\)

Vậy A = -1 hoặc A = 1

Ta có:\(\dfrac{x}{y+z+t}+1=\dfrac{y}{z+t+x}+1=\dfrac{z}{y+t+x}+1=\dfrac{t}{y+x+z}+1\)\(\Rightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{t+x+y}=\dfrac{x+y+z+t}{x+y+z}\)

Nếu x+y+z+t\(\ne\)0 thì y+z+t=z+t+x=t+x+y=x+y+z

=>x=y=z=t nên P=1+1+1+1=4

Nếu X+y+z+t=0 thì P=-4

bn cần gấp ko mk lm cho

Mk sẽ làm theo đề bài mà bạn nói dưới bình luận câu trả lời của bạn @Hồng Phúc Nguyễn.

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (1)

a) Thay (1) vào đề:

\(VT=\dfrac{a+2006b}{a-2006b}=\dfrac{bk+2006b}{bk-2006b}=\dfrac{b\left(k+2006\right)}{b\left(k-2006\right)}=\dfrac{k+2006}{k-2006}\)

\(VP=\dfrac{c+2006d}{c-2006d}=\dfrac{dk+2006d}{dk-2006d}=\dfrac{d\left(k+2006\right)}{d\left(k-2006\right)}=\dfrac{k+2006}{k-2006}\)

\(\Rightarrow VT=VP\Leftrightarrow\dfrac{a+2006b}{a-2006b}=\dfrac{c+2006d}{c-2006d}.\)

b) Thay (1) vào đề:

\(VT=\dfrac{2006\left(a+c\right)}{2006a}=\dfrac{2006\left(bk+dk\right)}{2006bk}=\dfrac{bk+dk}{bk}=\dfrac{k\left(b+d\right)}{bk}=\dfrac{b+d}{b}\)

\(VP=\dfrac{b+d}{b}\)

\(\Rightarrow VT=VP\Leftrightarrow\dfrac{2006\left(a+c\right)}{2006a}=\dfrac{b+d}{b}\rightarrowđpcm\).

\(\dfrac{a+2006b}{a-2006b}=\dfrac{c+2006d}{c-2006d}\)

\(\Leftrightarrow\)(a+2006b)(c-2006d)=(c+2006d)(a-2006b)

a(c-2006d)+2006b(c-2006d)=c(a-2006b)+2006d(a-2006b)

ac-2006ad+2006bc-4024036bd=ac-2006bc+2006ad-4024036bd

(ac-2006ad+2006bc-402436bd)-(ac-2006bc+2006ad-4024036bd=0

Suy ra 2 đẳng thức trên =nhau

Đề là gì vậy bạn???????!!!!!!!!!!!!!!!!

Ta có:

\(x^2.y^2=144\)

=> \(\left(xy\right)^2=144\)

=>\(\left[{}\begin{matrix}xy=12\\xy=-12\end{matrix}\right.\)

Lại có:

\(\dfrac{x}{3}=\dfrac{y}{4}\)

=> \(\dfrac{x}{3}.\dfrac{y}{4}=\dfrac{x}{3}.\dfrac{x}{3}=\dfrac{y}{4}.\dfrac{y}{4}\)

=> \(\dfrac{xy}{12}=\dfrac{x^2}{9}=\dfrac{y^2}{16}\)

+/ xy = 12

=> \(1=\dfrac{x^2}{9}=\dfrac{y^2}{16}\)

=> \(\left\{{}\begin{matrix}x^2=9\\y^2=16\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=\pm3\\y=\pm4\end{matrix}\right.\)

+/ xy = -12

=> \(-1=\dfrac{x^2}{9}=\dfrac{y^2}{16}\)

=> \(\left\{{}\begin{matrix}x^2=-9\\y^2=-16\end{matrix}\right.\)( vô lí - loại)

Vậy \(\left(x;y\right)\in\left\{\left(-3;-4\right);\left(3;4\right)\right\}\)

Bài này sao ko ai làm nhỉ ?

Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\Rightarrow x=3k;y=4k\)

Ta có: \(x^2y^2=\left(xy\right)^2=\left(3k\cdot4k\right)^2=\left(12k^2\right)^2=144k^2=144\)

\(\Rightarrow k^2=1\Rightarrow k=\pm1\)

Xét \(k=1\) thì \(\left\{{}\begin{matrix}x=3k=3\cdot1=3\\y=4k=4\cdot1=4\end{matrix}\right.\)

Xét \(k=-1\) thì \(\left\{{}\begin{matrix}x=3k=3\cdot\left(-1\right)=-3\\y=4k=4\cdot\left(-1\right)=-4\end{matrix}\right.\)

Ta có: x/3=y/4; x^2.y^2=144

Đặt x/3=k=>x=3k

y/4=k=>y=4k

Mà x^2.y^2=144=> (3k)^2.(4k)^2=144

<=>9k^2.16k^2=144

<=> 144.k^4=144

<=> k^4=1

=> k=1 hoặc k=-1

Suy ra: x=3k=3 hoặc -3

y=4k=4 hoặc -4

a)

Theo đề ra, ta có:

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\) và \(a^2+3b^2-2c^2=-16\)

\(\Rightarrow\dfrac{a^2}{2^2}=\dfrac{3b^2}{3.3^2}=\dfrac{2c^2}{2.4^2}\)

Hay \(\Rightarrow\dfrac{a^2}{4}=\dfrac{3b^2}{27}=\dfrac{2c^2}{32}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{a^2}{4}=\dfrac{3b^2}{27}=\dfrac{2c^2}{32}=\dfrac{a^2+3b^2-2c^2}{4+27-32}=-\dfrac{16}{-1}=16\)

\(\Rightarrow\dfrac{a^2}{4}=16\Rightarrow a^2=4.16=64\Rightarrow a=\sqrt{64}=\left\{-8;8\right\}\)

\(\Rightarrow\dfrac{3b^2}{27}=16\Rightarrow b^2=\dfrac{27.16}{3}=144\Rightarrow b=\sqrt{144}=\left\{-12;12\right\}\)

\(\Rightarrow\dfrac{2c^2}{32}=16\Rightarrow c^2=\dfrac{32.16}{2}=256\Rightarrow c=\sqrt{256}=\left\{-16;16\right\}\)

Vậy ...

b)

Theo đề ra, ta có:

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\) và \(a^3+b^3+c^3=792\)

\(\Rightarrow\dfrac{a^3}{2^3}=\dfrac{b^3}{3^3}=\dfrac{c^3}{4^3}\)

Hay \(\dfrac{a^3}{8}=\dfrac{b^3}{27}=\dfrac{c^3}{64}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{a^3}{8}=\dfrac{b^3}{27}=\dfrac{c^3}{64}=\dfrac{a^3+b^3+c^3}{8+27+64}=\dfrac{792}{99}=8\)

\(\Rightarrow\dfrac{a^3}{8}=8\Rightarrow a^3=8.8=64\Rightarrow a=4\)

\(\Rightarrow\dfrac{b^3}{27}=8\Rightarrow b^3=8.27=216\Rightarrow b=6\)

\(\Rightarrow\dfrac{c^3}{64}=8\Rightarrow c^3=8.64=512\Rightarrow c=8\)

Vậy...

Chúc bạn học tốt!

Không cần nữa đâu

BT 8.4 :

a,Ta có: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{ab}{cd}\) = k

\(\Rightarrow a=bk;c=dk\)

Thay a = bk; c = dk vào VT ta được:

\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)

Thay a = bk; c = dk vào VP ta được:

\(\dfrac{ab}{cd}=\dfrac{bk\times b}{dk\times d}=\dfrac{b^2}{d^2}\)

\(\Rightarrow\) VT = VP

Vậy \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{ab}{cd}\)

b,Ta có \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}=k\)

\(\Rightarrow a=bk;c=dk\)

Thay a = bk; c = dk vào VT ta được:

\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{\left[b\times\left(k-1\right)\right]^2}{\left[d\times\left(k-1\right)\right]^2}=\dfrac{b^2}{d^2}\)

Thay a = bk; c = dk vào VP ta được:

\(\dfrac{ab}{cd}=\dfrac{bk\times b}{dk\times d}=\dfrac{b^2}{d^2}\)

tập 1 hay tập 2 z bn

Chép đề ra, ở đây không chỉ có học sinh lớp 7 thôi đâu !