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

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

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: Ta có: \(\widehat{bMB}=\widehat{NMC}\)(hai góc đối đỉnh)

mà \(\widehat{bMB}=50^0\)

nên \(\widehat{NMC}=50^0\)

Ta có: \(\widehat{MNC}+\widehat{aNC}=180^0\)(hai góc kề bù)

=>\(\widehat{MNC}+110^0=180^0\)

=>\(\widehat{MNC}=70^0\)

Xét ΔMNC có \(\widehat{NMC}+\widehat{MNC}+\widehat{C}=180^0\)

=>\(\widehat{C}+50^0+70^0=180^0\)

=>\(\widehat{C}=60^0\)

b: Ta có: \(\widehat{NMB}+\widehat{NMC}=180^0\)(hai góc kề bù)

=>\(\widehat{NMB}+50^0=180^0\)

=>\(\widehat{NMB}=130^0\)

Ta có: MN//AB

=>\(\widehat{CMN}=\widehat{CBA}\)(hai góc đồng vị)

=>\(\widehat{CBA}=50^0\)

BN là phân giác của góc CBA

=>\(\widehat{NBM}=\dfrac{\widehat{ABC}}{2}=25^0\)

Xét ΔNMB có \(\widehat{NMB}+\widehat{BNM}+\widehat{NBM}=180^0\)

=>\(\widehat{MNB}=180^0-130^0-25^0=25^0\)

c: BN là phân giác của góc CBA

=>\(\widehat{ABN}=\dfrac{\widehat{ABC}}{2}=25^0\)

Xét ΔABC có \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)

=>\(\widehat{BAN}+60^0+50^0=180^0\)

=>\(\widehat{BAN}=70^0\)

Xét ΔBAN có \(\widehat{BAN}+\widehat{ABN}+\widehat{ANB}=180^0\)

=>\(\widehat{ANB}=180^0-75^0-25^0=85^0\)

Số đó là: 

\(6,4:32\%=20\)

5/8 của số đó là:

\(20\times\dfrac{5}{8}=12,5\)

ĐS: ... 

Số đó là :

\(6,4:32\%=20\)

\(\dfrac{5}{8}\) số đó là:

\(20\times\dfrac{5}{8}=12,5\)

a: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC=\sqrt{5^2+12^2}=13\left(cm\right)\)

Xét ΔABC có AD là phân giác

nên \(\dfrac{BD}{AB}=\dfrac{CD}{AC}\)

=>\(\dfrac{BD}{5}=\dfrac{CD}{12}\)

mà BD+CD=BC=13cm

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{BD}{5}=\dfrac{CD}{12}=\dfrac{BD+CD}{5+12}=\dfrac{13}{17}\)

=>\(BD=\dfrac{13}{17}\cdot5=\dfrac{65}{17}\left(cm\right);CD=\dfrac{13}{17}\cdot12=\dfrac{156}{17}\left(cm\right)\)

b: Xét ΔCDE vuông tại D và ΔCAB vuông tại A có

\(\widehat{DCE}\) chung

Do đó: ΔCDE~ΔCAB

=>\(k=\dfrac{CD}{CA}=\dfrac{156}{17}:12=\dfrac{13}{17}\)

c: ΔCDE~ΔCAB

=>\(\dfrac{CD}{CA}=\dfrac{CE}{CB}\)

=>\(\dfrac{CD}{CE}=\dfrac{CA}{CB}\)

Xét ΔCDA và ΔCEB có

\(\dfrac{CD}{CE}=\dfrac{CA}{CB}\)

\(\widehat{C}\) chung

Do đó: ΔCDA~ΔCEB

=>\(\dfrac{DA}{EB}=\dfrac{CA}{CB}\)

=>\(DA\cdot CB=BE\cdot AC\)

d: ΔCDE~ΔCAB

=>\(\dfrac{DE}{AB}=\dfrac{CD}{CA}\)

=>\(\dfrac{DE}{5}=\dfrac{156}{17}:12=\dfrac{13}{17}\)

=>\(DE=\dfrac{13}{17}\cdot5=\dfrac{65}{17}\left(cm\right)\)

Xét tứ giác ABDE có \(\widehat{EAB}+\widehat{EDB}=90^0+90^0=180^0\)

nên ABDE là tứ giác nội tiếp

=>\(\widehat{DEB}=\widehat{DAB}=45^0\)

Xét ΔDEB vuông tại D có \(\widehat{DEB}=45^0\)

nên ΔDEB vuông cân tại D

ΔBDE vuông cân tại D

=>\(S_{BDE}=\dfrac{1}{2}\cdot DB\cdot DE=\dfrac{1}{2}\cdot DB^2=\dfrac{1}{2}\cdot\left(\dfrac{65}{17}\right)^2=\dfrac{1}{2}\cdot\dfrac{4225}{289}=\dfrac{4225}{578}\left(cm^2\right)\)

Số bị chia là \(17\cdot50+2=852\)

=>Chọn A

Hiệu của hai số bằng 2/3 số bé

=>Số lớn=5/3 số bé

Tổng số phần bằng nhau là 5+3=8(phần)

SỐ lớn là 1888:8x5=1180

Số bé là 1880-1180=708