Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
BÀI 1:
\(\dfrac{a}{k}=\dfrac{x}{a}\Rightarrow a^2=kx\)
\(\dfrac{b}{k}=\dfrac{y}{b}\Rightarrow b^2\)=ky
Vay \(\dfrac{a^2}{b^2}=\dfrac{kx}{ky}=\dfrac{x}{y}\)
ĐKXĐ: \(b,d\ne0,c\ne\pm d\)
Áp dụng t/c dtsbn:
\(\dfrac{a^{2k}+b^{2k}}{c^{2k}+d^{2k}}=\dfrac{a^{2k}-b^{2k}}{c^{2k}-d^{2k}}=\dfrac{a^{2k}+b^{2k}+a^{2k}-b^{2k}}{c^{2k}+d^{2k}+c^{2k}-d^{2k}}=\dfrac{2a^{2k}}{2c^{2k}}=\dfrac{a^{2k}}{c^{2k}}\left(1\right)\)
\(\dfrac{a^{2k}+b^{2k}}{c^{2k}+d^{2k}}=\dfrac{a^{2k}-b^{2k}}{c^{2k}-d^{2k}}=\dfrac{a^{2k}+b^{2k}-a^{2k}+b^{2k}}{c^{2k}+d^{2k}-c^{2k}+d^{2k}}=\dfrac{2b^{2k}}{2d^{2k}}=\dfrac{b^{2k}}{d^{2k}}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{a^{2k}}{c^{2k}}=\dfrac{b^{2k}}{d^{2k}}\Rightarrow\dfrac{a^{2k}}{b^{2k}}=\dfrac{c^{2k}}{d^{2k}}\Rightarrow\dfrac{a}{b}=\pm\dfrac{c}{d}\left(đpcm\right)\)
theo bài ra ta có:
\(\frac{b+c+d}{a}=\frac{c+d+a}{b}=\frac{d+a+b}{c}=\frac{a+b+c}{d}=k\)
\(\Rightarrow\frac{b+c+d}{a}+1=\frac{c+d+a}{b}+1=\frac{d+a+b}{c}+1=\frac{a+b+c}{d}+1=k+1\) \(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}=k+1\)
vì a + b + c + d khác 0 => a = b = c = d
ta có:
\(\Rightarrow\frac{4a}{a}=\frac{4b}{b}=\frac{4c}{c}=\frac{4d}{d}=k+1\)
=> 4 = 4 = 4 = 4 = k + 1
=> k + 1 = 4
=> k = 3
vật k = 3
theo đầu bài
=>\(\dfrac{b+c+d}{a}\)=\(\dfrac{c+d+a}{b}\)=\(\dfrac{d+a+b}{c}\)=\(\dfrac{a+b+c}{d}\)=\(\dfrac{b+c+d+c+d+a+d+a+b+a+b+c}{a+b+c+d}\)=\(\dfrac{3\left[a+b+c+d\right]}{a+b+c+d}\)=>=3
=>k=3
b/ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\dfrac{a}{d}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
=> \(\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{c+d+b}\right)^3\) (2)Từ (1) và (2)=>đpcm
Vì `x` tỉ lệ thuận với `y` theo hệ số tỉ lệ `k`
`-> x=k*y`
Thay `x=12, y=-3`
`12=k*-3`
`-> k=12 \div (-3)`
`-> k=-4`
Vậy, hệ số tỉ lệ `k=-4.`
Xét các đáp án trên `-> D.`
$\href{https://hoc24.vn/vip/14874167551162}{\text{_Duy Nam_}}$
a/c=b/d=k
=>a=ck; b=dk
=>\(\dfrac{c\cdot a^2+d\cdot b^2}{c^3+d^3}\)
\(=\dfrac{c\cdot c^2k^2+d\cdot d^2k^2}{c^3+d^3}=k^2\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=\dfrac{b+c+d+c+d+a+d+a+b+a+b+c}{a+b+c+d}=\dfrac{\left(a+a+a\right)+\left(b+b+b\right)+\left(c+c+c\right)+\left(d+d+d\right)}{a+b+c+d}=\dfrac{3a+3b+3c+3d}{a+b+c+d}=\dfrac{3\left(a+b+c+d\right)}{a+b+c+d}=3\)
Vậy \(k=3\)
Theo tính chất dãy tỉ số bằng nhau ,ta có :
\(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=\dfrac{b+c+d+c+d+a+d+a+b+a+b+c}{a+b+c+d}\)
\(=\dfrac{3\left(a+b+c+d\right)}{a+b+c+d}=3\)
=> k = 3
sửa: \(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=k\)
giải:
\(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}\\ =\dfrac{b+c+d+c+d+a+d+a+b+a+b+c}{a+b+c+d}\\ =\dfrac{3\left(a+b+c+d\right)}{a+b+c+d}=3=k\)
vậy k=3
