\(3^{x+2}-5.3^x\\ =3^x\left(3^2-5\right)\\ =3^x.\left(9-5\right)\\ =4.3^x\)

bạn ơi mình thiếu =36

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

\(P\left(x\right)=5x^2+x+2=5\left(x^2+\dfrac{1}{5}x\right)+2\\ =5\left(x^2+2.x.\dfrac{1}{10}+\left(\dfrac{1}{10}\right)^2\right)-5.\left(\dfrac{1}{10}\right)^2+2\\ =5\left(x+\dfrac{1}{10}\right)^2+\dfrac{39}{20}\)

Nhận xét: \(\left(x+\dfrac{1}{10}\right)^2\ge0\forall x\inℝ\\ \Rightarrow5\left(x+\dfrac{1}{10}\right)^2\ge0\\ \Rightarrow P\left(x\right)=5\left(x+\dfrac{1}{10}\right)^2+\dfrac{39}{20}\ge\dfrac{39}{20}\)

\(Min_{P\left(x\right)}=\dfrac{39}{20}\) tại \(\left(x+\dfrac{1}{10}\right)^2=0\Leftrightarrow x+\dfrac{1}{10}=0\Leftrightarrow x=-\dfrac{1}{10}\)

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k=>\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\left(1\right)\\ \dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\dfrac{b^2}{d^2}\left(2\right)\)

Từ (1) và (2) => \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

\(1-2.\left(3-2\right)\)

\(=1-2.1\)

\(=1-2\)

\(=1+\left(-2\right)\)

\(=-1\)

1 - 2.(3 - 2)

= 1 - 2.1

= 1 - 2 

= -1

a ΔABC cân tại A

mà AH là đường cao

nên H là trung điểm của BC

Xét ΔIBC có

IH là đường cao

IH là đường trung tuyến

DO đó: ΔIBC cân tại I

b: Ta có: \(\widehat{ABI}+\widehat{IBC}=\widehat{ABC}\)(tia BI nằm giữa hai tia BA,BC)

\(\widehat{ACI}+\widehat{ICB}=\widehat{ACB}\)

mà \(\widehat{ABC}=\widehat{ACB};\widehat{IBC}=\widehat{ICB}\)

nên \(\widehat{ABI}=\widehat{ACI}\)

Xét ΔIBF và ΔICE có

\(\widehat{IBF}=\widehat{ICE}\)

IB=IC

\(\widehat{FIB}=\widehat{EIC}\)(hai góc đối đỉnh)

Do đó: ΔIBF=ΔICE

=>BF=CE

Xét ΔFBC và ΔECB có

FB=EC

\(\widehat{FBC}=\widehat{ECB}\)

BC chung

Do đó: ΔFBC=ΔECB

=>FC=EB

c: AF+FB=AB

AE+EC=AC

mà FB=EC và AB=AC

nên AF=AE

Xét ΔABC có \(\dfrac{AF}{AB}=\dfrac{AE}{AC}\)

nên FE//BC

a) \(\overline{aaa}=\overline{a00}+\overline{a0}+a=a.100+a.10+a.1\\ =a.\left(100+10+1\right)=a.111=a.37.3⋮3\) (dpcm)

b) \(\overline{ab}+\overline{ba}=\overline{a0}+b+\overline{b0}+a\\ =a.10+b+b.10+a\\ =a.\left(10+1\right)+b.\left(1+10\right)\\ =a.11+b.11\\ =11\left(a+b\right)⋮11\) (dpcm)

c) \(\overline{ab}-\overline{ba}=\overline{a0}+b-\left(\overline{b0}+a\right)\\ =a.10+b-b.10-a\\ =a.\left(10-1\right)+b.\left(1-10\right)\\ =a.9+b.\left(-9\right)\\ =9.\left(a-b\right)⋮9\) (dpcm)

d) \(\overline{abcabc}=\overline{abc000}+\overline{abc}\\ =\overline{abc}.1000+\overline{abc}.1\\ =\overline{abc}.1001=\overline{abc}.11.91⋮11\) (dpcm)