Vì AB//CD =>góc BAE=góc AED(so le trong);ABE=BEC(so le trong)

                                                                                                     Mà DAE=EAB(gt);ABE=EBC(gt)=>DAE=AED;EBC=BEC

                                                                                                     =>tam giác ADE cân tại D;tam giác BEC cân tại C

                                                                                                    =>AD=DE;BC=EC =>AD+BC=DE+EC

                                                                                                    =>AD+BC=CD(ĐPCM)

\(1,\left(\frac{a}{3}+4y\right)^2=\frac{a^2}{9}+\frac{8ay}{3}+16y^2\)

\(2,\)Bạn xem lại đề bài giùm mk nhé

\(\left(x^2+\frac{2}{5}y\right).\left(x^2-\frac{2}{5}y\right)=\left(x^2\right)^2-\left(\frac{2}{5}y\right)^2=x^4-\frac{4}{25}y^2\)

ai đó giúp mình giải bài này với

a

Xét  \(\Delta EBH\) và \(\Delta DHC\) có:

\(\widehat{EHB}=\widehat{DHC}\left(đ.đ\right)\)

\(\widehat{E}=\widehat{D}=90^0\)

\(\Rightarrow\Delta EBH~\Delta DHC\left(g.g\right)\)

b

\(\frac{S_{ABF}}{S_{ACF}}=\frac{\frac{AF\cdot BF}{2}}{\frac{AF\cdot CF}{2}}=\frac{BF}{CF}\)

Tuong tu ta co:

\(\frac{S_{ABD}}{S_{CBD}}=\frac{DA}{DC}\)

\(\frac{S_{BCE}}{S_{ACE}}=\frac{EB}{EA}\)

Nhan ve theo ve ta co dpcm

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

Bạn ơi xem lại phần b đề bài đi nhé

1,

\(\left(\frac{2}{3}x+y\right)^2=\left(\frac{2}{3}x\right)^2+2.\frac{2}{3}x.y+\left(y\right)^2=\frac{4}{9}x^2+\frac{4}{3}xy+y^2\)

\(\left(3a+\frac{1}{2}b\right)^2=\left(3a\right)^2+2.3a.\frac{1}{2}b+\left(\frac{1}{2}b\right)^2=9a^2+3ab+\frac{1}{4}b^2\)

2,

\(25a^2+4b^2+20ab=\left(5a\right)^2+\left(2b\right)^2+2.5a.2b=\left(5a+2b\right)^2\)

\(x^2+2x+1=\left(x\right)^2+2.x.1+\left(1\right)^2=\left(x+1\right)^2\)

\(9x^2+6x+1=\left(3x\right)^2+2.3x.1+\left(1\right)^2=\left(3x+1\right)^2\)

\(\left(2x+3y\right)^2+2.\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)

\(\left(a+b+c\right)^2+12=4\left(a+b+c\right)\)\(+2\left(ab+bc+ac\right)=0\)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac+12-4\left(a+b+c\right)-2\left(ab+bc+ac\right)=0\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+ac+bc\right)-2\left(ab+bc+ac\right)-4\left(a+b+c\right)+12=0\)

\(\Rightarrow a^2+b^2+c^2-4a-4b-4c+12=0\)

\(\Rightarrow\left(a^2-4a+4\right)+\left(b^2-4b+4\right)+\left(c^2-4c+4\right)=0\)

\(\Rightarrow\left(a-2\right)^2+\left(b-2\right)^2+\left(c-2\right)^2=0\)

Ta co: \(\left(a-2\right)^2\ge0\forall a\)

\(\left(b-2\right)^2\ge0\forall b\)

\(\left(c-2\right)^2\ge0\forall c\)

\(\Rightarrow\left(a-2\right)^2+\left(b-2\right)^2+\left(c-2\right)^2=0\Leftrightarrow\hept{\begin{cases}\left(a-2\right)^2=0\\\left(b-2\right)^2=0\\\left(c-2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-2=0\\b-2=0\\c-2=0\end{cases}\Leftrightarrow}a=b=c=2}\left(\right)\)

(đpcm) 

Mình nghĩ thế này nhé bạn! 

(a + b + c )² + 12 = 4 (a + b +c ) + 2(ab + bc +ac)

  \(\Leftrightarrow\)a² + b² + c² + 2ab + 2bc + 2ac + 12 = 4a + 4b + 4c + 2ab + 2ac + 2bc

\(\Leftrightarrow\) a² + b² + c² - 4a - 4b -4c +12 = 0

\(\Leftrightarrow\)a² - 4a + 4 + b² - 4b + 4 + c² - 4c + 4 =0

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

ta có (a-2 )² \(\ge0\forall a\)

          (b - 2 )² \(\ge0\forall b\)

            (c - 2 )² \(\ge0\forall c\)

mà (a-2)² + (b-2)² + (c-2)² = 0

\(\Rightarrow\hept{\begin{cases}\left(a-2\right)^2=0\\\left(b-2\right)^2=0\\\left(c-2\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a-2=0\\b-2=0\\c-2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=2\\b=2\\c=2\end{cases}\left(đpcm\right)}\)

vậy................... khi a=b = c =2

#mã mã#