có ngữ âm cơ mà

a: \(P=4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)

\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)

\(=\dfrac{\left(3^{32}-1\right)\left(3^{32}+1\right)}{2}=\dfrac{3^{64}-1}{2}\)

b: \(Q=\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)}{5^2-1}\)

\(=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)}{5^2-1}\)

\(=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)}{5^2-1}\)

\(=\dfrac{\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)}{5^2-1}\)

\(=\dfrac{\left(5^{32}-1\right)\left(5^{32}+1\right)}{24}=\dfrac{5^{64}-1}{24}\)

\(a.\left(x^2+5x+6\right)\left(x^2-15x+56\right)-144\\ =\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)-144\\ =\left[\left(x+2\right)\left(x-7\right)\right]\left[\left(x+3\right)\left(x-8\right)\right]-144\\ =\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144\\ =\left(x^2-5x-19+5\right)\left(x^2-5x-19-5\right)-144\\ =\left(x^2-5x-19\right)^2-5^2-144\\ =\left(x^2-5x-19\right)^2-169\\ =\left(x^2-5x+19\right)^2-13^2\\ =\left(x^2-5x-19-13\right)\left(x^2-5x-19+13\right)\\ =\left(x^2-5x-32\right)\left(x^2-5x-6\right)\\ =\left(x^2-5x-32\right)\left(x+1\right)\left(x-6\right)\) 

\(b.\left(x^2-11x+28\right)\left(x^2-7x+10\right)-72\\ =\left(x-4\right)\left(x-7\right)\left(x-5\right)\left(x-2\right)-72\\ =\left[\left(x-4\right)\left(x-5\right)\right]\left[\left(x-7\right)\left(x-2\right)\right]-72\\ =\left(x^2-9x+20\right)\left(x^2-9x+14\right)-72\\ =\left(x^2-9x+17+3\right)\left(x^2-9x+17-3\right)-72\\ =\left(x^2-9x+17\right)^2-3^2-72\\ =\left(x^2-9x+17\right)^2-81\\ =\left(x^2-9x+17\right)^2-9^2\\ =\left(x^2-9x+17-9\right)\left(x^2-9x+17+9\right)\\ =\left(x^2-9x+8\right)\left(x^2-9x+26\right)\\ =\left(x-1\right)\left(x-8\right)\left(x^2-9x+26\right)\)

1: DA=DK

=>ΔDAK cân tại D

=>\(\widehat{DAK}=\widehat{DKA}\)

mà \(\widehat{DKA}=\widehat{KAB}\)(hai góc so le trong, AB//DK)

nên \(\widehat{DAK}=\widehat{BAK}\)

=>AK là phân giác của góc BAD

2: ta có: CD=CK+KD

CD=AD+BC

Do đó: CK+KD=AD+BC

mà DA=DK

nên CK=CB

3: CK=CB

=>ΔCBK cân tại C

=>\(\widehat{CKB}=\widehat{CBK}\)

mà \(\widehat{CKB}=\widehat{ABK}\)(hai góc so le trong, AB//CK)

nên \(\widehat{ABK}=\widehat{CBK}\)

=>BK là phân giác của góc ABC

1:ΔABC cân tại A

mà AH là đường trung tuyến

nên AH\(\perp\)BC tại H

Xét tứ giác AHCD có

O là trung điểm chung của AC và HD

=>AHCD là hình bình hành

Hình bình hành AHCD có \(\widehat{AHC}=90^0\)

nên AHCD là hình chữ nhật

2: AHCD là hình chữ nhật

=>AD//HC và AD=HC

Ta có: AD//HC

=>AD//HB

Ta có: AD=CH

mà CH=HB

nên AD=HB

Xét tứ giác ADHB có

AD//HB

AD=HB

Do đó: ADHB là hình bình hành

3: \(CH=\dfrac{CB}{2}=3\left(cm\right)\)

AHCD là hình chữ nhật

=>\(S_{AHCD}=AH\cdot HC=4\cdot3=12\left(cm^2\right)\)

a: \(101^2=\left(100+1\right)^2=100^2+2\cdot100\cdot1+1^2\)

=10000+200+1

=10201

b: \(64^2+36^2+72\cdot64\)

\(=64^2+2\cdot64\cdot36+36^2\)

\(=\left(64+36\right)^2=100^2=10000\)

c: \(54^2+46^2-2\cdot54\cdot46=\left(54-46\right)^2=8^2=64\)

d: \(98\cdot102=\left(100-2\right)\left(100+2\right)=100^2-4=9996\)

\(\left(2x-3\right)^9-2x+3=0\)

=>\(\left(2x-3\right)^9-\left(2x-3\right)=0\)

=>\(\left(2x-3\right)\left[\left(2x-3\right)^8-1\right]=0\)

=>\(\left[{}\begin{matrix}2x-3=0\\\left(2x-3\right)^8-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\\left(2x-3\right)^8=1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\2x-3=1\\2x-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)

5 số =1,9,2,8,0

=(1+9).(2+8)+0

.=nhân