\(\dfrac{a+b}{c+d}=\dfrac{a-2b}{c-2d}\)

Suy ra: \(\left(a+b\right)\left(c-2d\right)=\left(c+d\right)\left(a-2b\right)\)

\(\Rightarrow a\left(c-2d\right)+b\left(c-2d\right)=c\left(a-2b\right)+d\left(a-2b\right)\)

\(\Rightarrow ac-2ad+bc-2bd=ac-2bc+ad-2bd\)

\(\Rightarrow ac-2ad+bc=ac-2bc+ad\)

\(\Rightarrow2ad+bc=2bc+ad\)

\(\Rightarrow2ad-ad=2bc-bc\)

\(\Rightarrow ad=bc\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\left(đpcm\right)\)

\(\dfrac{2x-y}{2}=\dfrac{x+2y}{3}\)

Tương đương: \(3\left(2x-y\right)=2\left(x+2y\right)\)

\(\Rightarrow6x-3y=2x+4y\)

\(\Rightarrow6x=2x+4y+3y\)

\(\Rightarrow6x=2x+7y\)

\(\Rightarrow7y=6x-2x\)

\(\Rightarrow7y=5x\)

Suy ra \(\dfrac{x}{7}=\dfrac{y}{5}\Leftrightarrow\dfrac{x}{y}=\dfrac{7}{5}\)

\(linh=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3}{\sqrt{x-3}}+\dfrac{4}{\sqrt{x}-3}\)

Nên \(4⋮\sqrt{x}-3\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)\)

\(Ư\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Nên \(\left[{}\begin{matrix}\sqrt{x}-3=1\\\sqrt{x}-3=-1\\\sqrt{x}-3=2\\\sqrt{x}-3=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=2\\\sqrt{x}=5\\\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=4\\x=25\\x=1\end{matrix}\right.\)

\(\left[{}\begin{matrix}\sqrt{x}-3=4\\\sqrt{x}-3=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=7\\\sqrt{x}=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=49\\x\in\varnothing\end{matrix}\right.\)

\(linh=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+....+\dfrac{100}{5^{100}}\)

\(5linh=5\left(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{100}{5^{100}}\right)\)

\(5linh=1+\dfrac{2}{5}+\dfrac{3}{5^2}+\dfrac{4}{5^3}+...+\dfrac{100}{5^{99}}\)

\(5linh-linh=\left(1+\dfrac{2}{5}+\dfrac{3}{5^2}+\dfrac{4}{5^3}+...+\dfrac{100}{5^{99}}\right)-\left(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{100}{5^{100}}\right)\)

\(4linh=1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{99}}-\dfrac{100}{5^{100}}\)

Đặt:

\(linh_2=1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{99}}\)

\(5linh_2=5\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{99}}\right)\)

\(5linh_2=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{98}}\)

\(5linh_2-linh_2=\left(5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{98}}\right)-\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{99}}\right)\)

\(4linh_2=5-\dfrac{1}{5^{99}}\)

\(linh=\dfrac{5}{4}-\dfrac{1}{5^{99}.4}\)

Nên \(4linh=\dfrac{5}{4}-\dfrac{1}{5^{99}.4}-\dfrac{100}{5^{100}}\)

Khi đó \(linh=\dfrac{5}{16}-\dfrac{1}{5^{99}.16}-\dfrac{100}{5^{100}.4}\)

\(3\left(t+2\right)^2+\left(2t-1\right)^2-7\left(t+3\right)\left(t-3\right)=36\)

\(\Rightarrow3\left(t^2+4t+4\right)+\left(4t^2-4t+1\right)-\left(7t+21\right)\left(t-3\right)=36\)

\(\Rightarrow3\left(t^2+4t+4\right)+\left(4t^2-4t+1\right)-7t\left(t-3\right)+21\left(t-3\right)=36\)

\(\Rightarrow3\left(t^2+4t+4\right)+\left(4t^2-4t+1\right)-7t^2+21t+21t-63=36\)

\(\Rightarrow3t^2+12t+12+4t^2-4t+1-7t^2+21t+21t-63=36\)

\(\Rightarrow\left(3t^2+4t^2-7t^2\right)+\left(12t-4t+21t+21t\right)+\left(12+1-63\right)=36\)

\(\Rightarrow50t-50=36\)

\(\Rightarrow50t=50+36\Leftrightarrow50t=86\)

\(\Rightarrow t=\dfrac{86}{50}=\dfrac{43}{25}\)

\(linh=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}\)

\(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}\\\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}\\.............\\\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}\end{matrix}\right.\)

Suy ra:

\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+....+\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{99}}>\dfrac{99}{\sqrt{100}}\)

\(linh=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+.....+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}>\dfrac{99}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}\)

\(\)\(linh>10\left(đpcm\right)\)

Bài này ko phải 100 nhé

coi cach hack di vui lam

Re write:

1) I can sing and dance

\(\Rightarrow\) I am able to sing and dance

2) Watching TV is very interesting

\(\Rightarrow\)It's interesing to watch TV

3) Hoa goes to school on foot(sửa)

\(\Rightarrow\) Hoa walks to school

4) I like English(viết 2 câu)

-I enjoy English

-I am interested in English

5)This is the first time he went abroad.

\(\Rightarrow\) He hasn't went abroad before

6)The last time she kissed me was 5 months ago.

\(\Rightarrow\) She haven't kissed me for 5 months

You/free/this/Sunday evening ?

\(\Rightarrow\) Will you free in this Sunday evening?