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Question

Giúp mình vớii:( gấp ạ :((

Nguyễn Lê Phước Thịnh · Accepted Answer

b: Đặt $\dfrac{a}{b}=\dfrac{c}{d}=k$

$\Leftrightarrow\left\{{}\begin{matrix}a=bk\c=dk\end{matrix}\right.$

Ta có: $\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}$

$=\dfrac{\left(bk+dk\right)^2}{\left(b+d\right)^2}$

$=k^2$(1)

Ta có: $\dfrac{a^2-c^2}{b^2-d^2}$

$=\dfrac{\left(bk\right)^2-\left(dk\right)^2}{b^2-d^2}$

$=k^2\left(2\right)$

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

Câu trả lời:

b: Đặt $\dfrac{a}{b}=\dfrac{c}{d}=k$

$\Leftrightarrow\left\{{}\begin{matrix}a=bk\c=dk\end{matrix}\right.$

Ta có: $\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}$

$=\dfrac{\left(bk+dk\right)^2}{\left(b+d\right)^2}$

$=k^2$(1)

Ta có: $\dfrac{a^2-c^2}{b^2-d^2}$

$=\dfrac{\left(bk\right)^2-\left(dk\right)^2}{b^2-d^2}$

$=k^2\left(2\right)$

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

Nguyễn Lê Phước Thịnh · Answer

Đặt $\dfrac{a}{b}=\dfrac{c}{d}=k$

$\Leftrightarrow\left\{{}\begin{matrix}a=bk\c=dk\end{matrix}\right.$

Ta có: $\dfrac{4a+3c}{4b+3d}=\dfrac{4bk+3dk}{4b+3d}=k$

$\dfrac{4a-3c}{4b-3d}=\dfrac{4bk-3dk}{4b-3d}=k$

Do đó: $\dfrac{4a+3c}{4b+3d}=\dfrac{4a-3c}{4b-3d}$

Câu trả lời:

Đặt $\dfrac{a}{b}=\dfrac{c}{d}=k$

$\Leftrightarrow\left\{{}\begin{matrix}a=bk\c=dk\end{matrix}\right.$

Ta có: $\dfrac{4a+3c}{4b+3d}=\dfrac{4bk+3dk}{4b+3d}=k$

$\dfrac{4a-3c}{4b-3d}=\dfrac{4bk-3dk}{4b-3d}=k$

Do đó: $\dfrac{4a+3c}{4b+3d}=\dfrac{4a-3c}{4b-3d}$

x	10	-2	-3	1	0	1.21	0.25
\(^{x^2}\)	100	4	9	1	0	1.4641	0.0625

1.44	-25	\(\dfrac{4}{9}\)
2.0736	625	\(\dfrac{16}{81}\)