cho cái hình ms giải dc

( Tự vẽ hình )

a) Xét  \(\Delta ABE\)và  \(\Delta KCE\)có :

\(\widehat{CEK}=\widehat{BEA}\)( đối đỉnh )

\(CE=EB\left(gt\right)\)

\(\widehat{KCB}=\widehat{CBA}\left(DK//AB\right)\)

\(\Rightarrow\Delta ABE=\Delta KCE\left(g-c-g\right)\left(đpcm\right)\)

b)  \(\Rightarrow AE=EK\)

Xét \(\Delta ADK\)có AE = EK \(\Rightarrow DE\)là trung tuyến  \(\Delta ADK\)

Mà DE là đường phân giác  \(\Delta ADK\)

\(\Rightarrow\Delta ADK\)cân tại D ( đpcm )

c) \(\Rightarrow\)DE là đường cao \(\Delta ADK\)

\(\Rightarrow\widehat{AED}=90^o\left(đpcm\right)\)

mai mik giải cho. h trễ rồi

\(C=-\left(x^2-2xy+4y^2-2x-10y+8\right)\)

     \(=-\left[\left(x^2-2xy+y^2\right)-2\left(x-y\right)+1+\left(3y^2-9y+3\right)+4\right]\)

       \(=-\left[\left(x-y\right)^2-2\left(x-y\right)+1+3\left(y-1\right)^2+4\right]\)

      \(=-\left[\left(x-y-1\right)^2+3\left(y-1\right)^2+4\right]\)

      \(=-\left[\left(x-y-1\right)^2+3\left(y-1\right)^2\right]-4\le-4\)

          GTLN là -4    tại x=2;y=1

\(2x^4+\left(1-2x\right)^4=\frac{1}{27}\)

\(\Rightarrow2x^4+\left(1-2x\right)^4=\frac{1}{3}\cdot3\cdot\left[2x^4+\left(1-2x\right)^4\right]\)

\(\Rightarrow\frac{1}{3}\left(1^2+1^2+1^2\right)\left[x^4+x^4+\left(1-2x\right)^4\right]\)\(\ge\)\(\frac{1}{3}\left[x^2+x^2+\left(1-2x\right)^2\right]^2\)

\(\Rightarrow\frac{1}{3}\cdot\frac{1}{9}\left\{3.\left[x^2+x^2+\left(1-2x\right)^2\right]\right\}^2\)

\(\Rightarrow\frac{1}{27}\left\{\left(1^2+1^2+1^2\right)\left[x^2+x^2+\left(1-2x\right)^2\right]\right\}^2\)\(\ge\)\(\frac{1}{27}\left[x+x+\left(1-2x\right)\right]^4=\frac{1}{27}\)

Vậy phương trình có nghiệm khi dấu đẳng thức xảy ra 

\(\Leftrightarrow\hept{\begin{cases}x^2=x^2=\left(1-2x\right)^2\\x=x=1-2z\end{cases}}\Leftrightarrow x=\frac{1}{3}\)

KL : ................................................................... 

P/s : chưa chắc 

Xét : \(\frac{1}{\text{k}\left(\text{k}+1\right)}=\frac{1}{\text{k}}-\frac{1}{\text{k}+1}\)

\(\Leftrightarrow VT=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{n}-\frac{1}{n+1}\right)=1-\frac{1}{n+1}< 1\)

  \(\Rightarrow\) ĐPCM 

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n.\left(n+1\right)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)

\(=1-\frac{1}{n+1}\)

mà  \(n\inℕ^∗\Rightarrow\frac{1}{n+1}>0\)

\(\Rightarrow1-\frac{1}{n+1}< 1\)

                              ( đpcm )

Đây là toán lp 8 sao?

           ?_?

\(a,=64x^3-48x^2+12x-1-\left(64x^3+12x-48x^2-9\right)\)

       \(=\left(64x^3-64x^3\right)+\left(48x^2-48x^2\right)+\left(12x-12x\right)+\left(9-1\right)\)

        \(=8\)  => ko phụ thuộc vào biến x

\(b,=2\left(x+y\right)\left(x^2-xy+y^2\right)-3\left(x^2+y^2\right)\)

thay x+y=1 vào 

\(=2\left(x^2-xy+y^2\right)-3\left(x^2+y^2\right)\)

\(=2x^2-2xy+2y^2-3x^2-3y^2\)

\(=-\left(x^2+2xy+y^2\right)=-\left(x+y\right)^2=-1\)  =>ko phụ thuộc vào biến

\(c,=x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-1\right)\)

      \(=6x^2+2-6x^2+6=8\)

\(d,\frac{\left(2x+5\right)^2+\left(5x-2\right)^2}{x^2+1}=\frac{4x^2+20x+25+25x^2-20x+4}{x^2+1}=\frac{29\left(x^2+1\right)}{x^2+1}=29\)