Nguyễn Thanh Hằng

khó lắm anh ơi =((

Trên BC lấy D sao cho AB = BD

Từ D kẻ DE vuông góc với AC

Xét \(\Delta ABD\) có :

\(AB=BD\)

\(\Leftrightarrow\Delta ABD\) cân tại B

\(\Leftrightarrow\widehat{BAD}=\widehat{BDA}\)

Lại có :

\(\widehat{BAD}+\widehat{DAC}=\widehat{BAC}\)

\(\Leftrightarrow\widehat{BDA}+\widehat{DAC}=\widehat{BAC}\)

Mà \(\widehat{BAC}=90^0\)

\(\Leftrightarrow\widehat{BDA}+\widehat{DAC}=90^0\left(1\right)\)

Xét \(\Delta DEA\) có : \(\widehat{E1}=90^0\)

\(\Leftrightarrow\widehat{ADE}+\widehat{CAD}=90^0\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\widehat{BAD}=\widehat{ADE}\)

Chứng minh : \(\Delta ABH=\Delta EAD\left(ch-gn\right)\)

\(\Leftrightarrow AH=AE\)

Xét \(\Delta DEC\) có : \(\widehat{E2}=90^0\)

\(EC< DC\)

\(\Leftrightarrow AE+EC+BD< DC+EC+BD\)

Mà \(AE+EC=AC;DC+BD=BC\)

\(\Leftrightarrow AC+BD< CE+BC\)

Mà \(BD=AB;CE=AH\)

\(\Leftrightarrow AC+AB< BC+AH\left(đpcm\right)\)

Ta có :

+) \(f\left(1\right)=0\)

\(\Leftrightarrow1^3+a.1^2+b.1-2=0\)

\(\Leftrightarrow1+a+b-2=0\)

\(\Leftrightarrow a+b-1=0\)

\(\Leftrightarrow a+b=1\left(1\right)\)

+) \(f\left(-1\right)=0\)

\(\Leftrightarrow\left(-1\right)^3+a.\left(-1\right)^2+b\left(-1\right)-2=0\)

\(\Leftrightarrow-1+a-b-2=0\)

\(\Leftrightarrow a-b-3=0\)

\(\Leftrightarrow a-b=3\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\left(a+b\right)+\left(a-b\right)=1+3\)

\(\Leftrightarrow2a=4\)

\(\Leftrightarrow a=2\)

\(\Leftrightarrow b=-1\)

Vậy ..

\(Q\left(-1\right)=0\)

\(\Leftrightarrow m.\left(-1\right)^3+2.m.\left(-1\right)-3=0\)

\(\Leftrightarrow-m-2m-3=0\)

\(\Leftrightarrow m\left(-1-2\right)-3=0\)

\(\Leftrightarrow m\left(-3\right)=3\)

\(\Leftrightarrow m=-1\)

Vậy ...

Áp dụng bất đẳng thức \(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\)

\(\dfrac{3^{11}+1}{3^{10}+1}< \dfrac{3^{11}+1+2}{3^{10}+1+2}=\dfrac{3^{11}+3}{3^{10}+3}=\dfrac{3\left(3^{10}+1\right)}{3\left(3^9+1\right)}=\dfrac{3^{10}+1}{3^9+1}\)

\(\Leftrightarrow\dfrac{3^{11}+1}{3^{10}+1}< \dfrac{3^{10}+1}{3^9+1}\)

\(\dfrac{x}{6}+\dfrac{x}{10}+\dfrac{x}{15}+........+\dfrac{x}{78}=\dfrac{220}{39}\)

\(\Leftrightarrow\dfrac{2x}{12}+\dfrac{2x}{20}+........+\dfrac{2x}{156}=\dfrac{220}{39}\)

\(\Leftrightarrow2x\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+..........+\dfrac{1}{12.13}\right)=\dfrac{220}{39}\)

\(\Leftrightarrow2x\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=\dfrac{220}{39}\)

\(\Leftrightarrow2x.\dfrac{10}{39}=\dfrac{220}{39}\)

\(\Leftrightarrow x.\dfrac{20}{39}=\dfrac{220}{39}\)

\(\Leftrightarrow x=11\)

Vậy ...

\(x\left(x-1\right)+15=0\)

\(\Leftrightarrow x^2-x+15=0\)

\(\Leftrightarrow x^2-\dfrac{1}{2}x-\dfrac{1}{2}x+15=0\)

\(\Leftrightarrow x\left(x-\dfrac{1}{2}\right)-\dfrac{1}{2}\left(x-\dfrac{1}{2}\right)+\dfrac{59}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{59}{4}=0\left(1\right)\)

Với mọi x ta có :

\(\left(x-\dfrac{1}{2}\right)^2\ge0\)

\(\dfrac{59}{4}>0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{59}{4}>0\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\) K tìm dc x

còn cái đề nào bá đạo hơn k -_-

okie, Văn hóa quá cơ ạ :)

Đăng topic ở chủ đề công dân mà chẳng thấy có chút công dân nào, tội.... :((

\(x^2-3x=0\)

\(\Leftrightarrow x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Vậy ..

Ta có :

\(A\left(-2\right)=B\left(\dfrac{1}{10}\right)\)

\(\Leftrightarrow\left(-2\right)^2-a\left(-2\right)+3=5.\dfrac{1}{10}-1\)

\(\Leftrightarrow4+2a+3=\dfrac{1}{2}-1\)

\(\Leftrightarrow2a+7=-\dfrac{1}{2}\)

\(\Leftrightarrow2a=-7,5\)

\(\Leftrightarrow a=-3,27\)

Vậy ..

Gọi \(d=ƯCLN\left(2n+1;3n+2\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}2n+1⋮d\\3n+2⋮d\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6n+3⋮d\\6n+4⋮d\end{matrix}\right.\)

\(\Leftrightarrow1⋮d\)

\(\Leftrightarrow d=1\)

\(\LeftrightarrowƯCLN\left(2n+1;3n+2\right)=1\)

\(\Leftrightarrow\dfrac{2n+1}{3n+2}\) toois giản