Câu hỏi của Nguyễn Thùy Dương

Question

1/Cho $\dfrac{a}{b}=\dfrac{c}{d}\left(b\ne0;d\ne0\right)$chứng tỏ rằng$\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{a.c}{b.d}$

2/Tìm x, y thỏa mãn:\(\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0...

Nguyễn Thị Bích Thủy · Accepted Answer

1. Đặt $\dfrac{a}{b}=\dfrac{c}{d}=k$
$\Rightarrow\left\{{}\begin{matrix}a=bk\c=dk\end{matrix}\right.$
$\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2$             $\left(1\right)$
$\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2$ $\left(2\right)$
Từ $\left(1\right)\text{và   (2)}$ $\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}$
2. $\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0$
$\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|\ge0\\left|\dfrac{2}{7}y+3\right|\ge0\end{matrix}\right.\Rightarrow\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|\ge0$
$\text{Mà   }\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0$
$\Rightarrow\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|=0\\left|\dfrac{2}{7}y+3\right|=0\end{matrix}\right.$
$\Rightarrow\left\{{}\begin{matrix}5-\dfrac{3}{4}x=0\\dfrac{2}{7}y+3=0\end{matrix}\right.$
$\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{4}x=5\\dfrac{2}{7}x=-3\end{matrix}\right.$
$\Rightarrow\left\{{}\begin{matrix}x=\dfrac{20}{3}\y=-\dfrac{21}{2}\end{matrix}\right.$
$\text{Vậy    }\left\{{}\begin{matrix}x=\dfrac{20}{3}\y=-\dfrac{21}{2}\end{matrix}\right.$Câu trả lời: 1. Đặt $\dfrac{a}{b}=\dfrac{c}{d}=k$
$\Rightarrow\left\{{}\begin{matrix}a=bk\c=dk\end{matrix}\right.$
$\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2$             $\left(1\right)$
$\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2$ $\left(2\right)$
Từ $\left(1\right)\text{và   (2)}$ $\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}$
2. $\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0$
$\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|\ge0\\left|\dfrac{2}{7}y+3\right|\ge0\end{matrix}\right.\Rightarrow\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|\ge0$
$\text{Mà   }\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0$
$\Rightarrow\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|=0\\left|\dfrac{2}{7}y+3\right|=0\end{matrix}\right.$
$\Rightarrow\left\{{}\begin{matrix}5-\dfrac{3}{4}x=0\\dfrac{2}{7}y+3=0\end{matrix}\right.$
$\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{4}x=5\\dfrac{2}{7}x=-3\end{matrix}\right.$
$\Rightarrow\left\{{}\begin{matrix}x=\dfrac{20}{3}\y=-\dfrac{21}{2}\end{matrix}\right.$
$\text{Vậy    }\left\{{}\begin{matrix}x=\dfrac{20}{3}\y=-\dfrac{21}{2}\end{matrix}\right.$

Nguyễn Thị Bích Thủy · Answer

3. $\dfrac{1}{2}a=\dfrac{2}{3}b=\dfrac{3}{4}c$

$\Rightarrow\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}$
$\text{Mà }a-b=15$
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
$\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}=\dfrac{a-b}{2-\dfrac{3}{2}}=\dfrac{15}{\dfrac{1}{2}}=30$
$\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=30\Rightarrow a=30.2=60\\dfrac{b}{\dfrac{3}{2}}=30\Rightarrow b=30.\dfrac{3}{2}=45\\dfrac{c}{\dfrac{4}{3}}=30\Rightarrow c=30.\dfrac{4}{3}=40\end{matrix}\right.$
$\text{Vậy }\left\{{}\begin{matrix}a=60\b=45\c=40\end{matrix}\right.$Câu trả lời: 3. $\dfrac{1}{2}a=\dfrac{2}{3}b=\dfrac{3}{4}c$

$\Rightarrow\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}$
$\text{Mà }a-b=15$
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
$\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}=\dfrac{a-b}{2-\dfrac{3}{2}}=\dfrac{15}{\dfrac{1}{2}}=30$
$\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=30\Rightarrow a=30.2=60\\dfrac{b}{\dfrac{3}{2}}=30\Rightarrow b=30.\dfrac{3}{2}=45\\dfrac{c}{\dfrac{4}{3}}=30\Rightarrow c=30.\dfrac{4}{3}=40\end{matrix}\right.$
$\text{Vậy }\left\{{}\begin{matrix}a=60\b=45\c=40\end{matrix}\right.$

Mashiro Shiina · Answer

Ủng hộ bài 4 đây :V

$M=3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}$

$M=3^x.3^1+3^x.3^2+3^x.3^3+...+3^x.3^{100}$

$M=3^x\left(3^1+3^2+3^3+...+3^{100}\right)$

Đặt: $T=3^1+3^2+3^3+...+3^{100}$

$T=\left(3^1+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)$

$T=1\left(3^1+3^2+3^3+3^4\right)+3^4\left(3^1+3^2+3^3+3^4\right)+...+3^{96}\left(3^1+3^2+3^3+3^4\right)$

$T=\left(1+3^4+...3^{96}\right)\left(3^1+3^2+3^3+3^4\right)=120\left(1+3^4+...+3^{96}\right)⋮120$

$\Rightarrow M⋮120\left(đpcm\right)$Câu trả lời: Ủng hộ bài 4 đây :V