a) \(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\)

\(=\left[\left(x+a\right)\left(x+4a\right)\right]\cdot\left[\left(x+2a\right)\left(x+3a\right)\right]+a^4\)

\(=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\)

\(=\left(x^2+5ax+5a^2-a^2\right)\left(x^2+5ax+5a^2+a^2\right)+a^4\)\

\(=\left(x^2+5ax+5a^2\right)^2-a^4+a^4\)

\(=\left(x^2+5ax+5a^2\right)^2\)

b) Đặt \(a=x^2+y^2+z^2\);     \(b=xy+yz+xz\)

\(\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(xy+yz+zx\right)^2\)

\(=a\left(a+2b\right)+b^2\)

\(=a^2+2ab+b^2=\left(a+b\right)^2\)

\(=\left(x^2+y^2+z^2+xy+yz+zx\right)^2\)

a) \left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4(x+a)(x+2a)(x+3a)(x+4a)+a4

=\left[\left(x+a\right)\left(x+4a\right)\right]\cdot\left[\left(x+2a\right)\left(x+3a\right)\right]+a^4=[(x+a)(x+4a)]⋅[(x+2a)(x+3a)]+a4

=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4=(x2+5ax+4a2)(x2+5ax+6a2)+a4

=\left(x^2+5ax+5a^2-a^2\right)\left(x^2+5ax+5a^2+a^2\right)+a^4=(x2+5ax+5a2−a2)(x2+5ax+5a2+a2)+a4\

=\left(x^2+5ax+5a^2\right)^2-a^4+a^4=(x2+5ax+5a2)2−a4+a4

=\left(x^2+5ax+5a^2\right)^2=(x2+5ax+5a2)2

b) Đặt a=x^2+y^2+z^2a=x2+y2+z2;     b=xy+yz+xzb=xy+yz+xz

\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(xy+yz+zx\right)^2(x2+y2+z2)(x+y+z)2+(xy+yz+zx)2

=a\left(a+2b\right)+b^2=a(a+2b)+b2

=a^2+2ab+b^2=\left(a+b\right)^2=a2+2ab+b2=(a+b)2

=\left(x^2+y^2+z^2+xy+yz+zx\right)^2=(x2+y2+z2+xy+yz+zx)2

câu a ta có : <MAE = 90

suy ra tam giác MAE là tam giác vuông :< AME + <MEA = 90 ĐỘ ( đ/lí tổng 3 góc áp dụng vào tam giác vuông )

gọi n là giao điểm của EH và CD

vì <MND =90 độ suy ra <NMD +<MPN=90độ

vì cùng phụ nhau với < m suy ra <MEA =<MDN

xét tam giác ACD và tam giác AME :

AD =AE (GT)

<MEA=<MDN (cmt)

<CAD =<MAE =90độ (do AC vuông góc với MB )

SUY RA TAM GIÁC ACD = TAM GIÁC AME(G.C.G)

mik chỉ làm đc câu a thôi nha

cảm ơn ạ

a) \(\left(4x^2-25\right)^2-9\left(2x-5\right)^2\)

\(=\left(4x^2-25\right)^2-\left(6x-15\right)^2\)

\(=\left(4x^2-25-6x+15\right)\left(4x^2-25+6x-15\right)\)

\(=\left(4x^2-6x-10\right)\left(4x^2+6x-40\right)\)

\(=\left(4x^2+4x-10x-10\right)\left(4x^2+16x-10x-40\right)\)

\(=\left[4x\left(x+1\right)-10\left(x+1\right)\right]\left[4x\left(x+4\right)-10\left(x+4\right)\right]\)

\(=\left(4x-10\right)\left(x+1\right)\left(4x-10\right)\left(x+4\right)\)

\(=\left(4x-10\right)^2\left(x+1\right)\left(x+4\right)\)

\(=4\left(2x-5\right)^2\left(x+1\right)\left(x+4\right)\)

b) \(a^6-a^4+2a^3+2a^2\)

\(=a^2\left(a^4-a^2+2a+2\right)\)

\(=a^2\left(a^4+a^3-a^3-a^2+2a+2\right)\)

\(=a^2\left[a^3\left(a+1\right)-a^2\left(a+1\right)+2\left(a+1\right)\right]\)

\(=a^2\left(a+1\right)\left(a^3-a^2+2\right)\)

( x² + 4x )² - 2( x² + 4x ) - 15

= x⁴ + 16x² - 2x² - 8x - 15

= x² + 14x² - 6x - 15

\(\left(x^2+4x\right)^2-2\left(x^2+4x\right)-15\)

\(=\left(x^2+4x\right)^2-2\left(x^2+4x\right)+1-16\)

\(=\left(x^2+4x-1\right)^2-4^2\)

\(=\left(x^2+4x-1-4\right)\left(x^2+4x-1+4\right)\)

\(=\left(x^2+4x-5\right)\left(x^2+4x+3\right)\)

đề hay nhỉ...

a+b=6 mà lại tìm x

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