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dễ mà
a, tách ra (đừng có ghi từ này vào nha)
(ab+bc+ca)^2=a^2b^2+b^2c^2+c^2a^2
Vì a^2b^2+b^2c^2+c^2a^2=a^2b^2+b^2c^2+c^2a^2
=>(ab+bc+ca)^2=a^2b^2+b^2c^2+c^2a^2
.
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2\)
Do đó: \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
b: \(\left(\dfrac{a-b}{c-d}\right)^4=\left(\dfrac{bk-b}{dk-d}\right)^4=\left(\dfrac{b}{d}\right)^4\)
\(\dfrac{a^4+b^4}{c^4+d^4}=\dfrac{b^4k^4+b^4}{d^4k^4+d^4}=\dfrac{b^4}{d^4}\)
Do đó: \(\left(\dfrac{a-b}{c-d}\right)^4=\dfrac{a^4+b^4}{c^4+d^4}\)
các bn giúp mình nhé!
mk ko chép đề mà tách luôn nha
M = x2x2 + x2x2 + x2y2 + x2y2 + x2y2 + y2y2 + y2
= ( x2x2 + x2y2 ) + ( x2x2 + x2y2 ) + ( x2y2 + y2y2 ) + y2
= x2( x2 + y2 ) + x2( x2 + y2 ) + y2( x2 + y2 ) + y2
= ( x2 + y2 ) (x2 + x2 + y2 ) + y2
= 1( x2 + 1) + y2
= x2 + y2 +1 = 2
a, 128 = 122.4 = (122)4 = 1444
812 = 83.4 = (83)4 = 5124
Vì 5124 > 1444
=> 812 > 128
b, (-5)39 = (-5)3.13 = [(-5)3]13 = (-125)13 = -12513
(-2)91 = (-2)7.13 = [(-2)7]13 = (-128)13 = -12813
Có 12513 < 12813
=> -12513 > -12813
=> (-5)39 > (-2)91
Có
cảm ơn
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow a=bk;c=bk\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k.k=k^2\) \(\left(1\right)\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}k^2\)(2)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
\(\Rightarrow\dfrac{ac}{a^2+c^2}=\dfrac{bd}{b^2+d^2}\)
Ta có:
\(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
\(c^2=b.d\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\)
Do đó:\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
Do đó:\(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{b}\left(đpcm\right)\)
\(\left\{{}\begin{matrix}b^2=ac\\c^2=bd\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{b}{c}\\\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
Vậy \(\dfrac{a}{b}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
\(\rightarrowđpcm\)
Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk,c=dk\)
a) Ta có:
\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2}{d^2}\) (1)
\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2}{d^2}\) (2)
Từ (1) và (2) suy ra \(\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Vậy \(\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b) Ta có:
\(\frac{a^4+b^4}{c^4+d^4}=\frac{\left(bk\right)^4+b^4}{\left(dk\right)^4+d^4}=\frac{b^4.k^4+b^4}{d^4.k^4+d^4}=\frac{b^4.\left(k^4+1\right)}{d^4.\left(k^4+1\right)}=\frac{b^4}{d^4}\) (1)
\(\frac{\left(a+b\right)^4}{\left(c+d\right)^4}=\frac{\left(bk+b\right)^4}{\left(dk+d\right)^4}=\frac{\left[b\left(k+1\right)\right]^4}{\left[d\left(k+1\right)\right]^4}=\frac{b^4}{d^4}\) (2)
Từ (1) và (2) suy ra \(\frac{a^4+b^4}{c^4+d^4}=\frac{\left(a+b\right)^4}{\left(c+d\right)^4}\)
Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
\(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)(đpcm)