PTTNT
a (x+a)(x+2a)(x+3a)(x+4a)+a^4
b (x^2+y^2+z^2)(x+y+z)^2+(xy+yz+zx)^2
K
Khách
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22 tháng 10 2021
\(a,=\left(xy-1-x-y\right)\left(xy-1+x+y\right)\\ b,Sửa:a^3+2a^2+2a+1\\ =a^3+a^2+a^2+a+a+1=\left(a+1\right)\left(a^2+a+1\right)\\ c,=1-4a^2-a\left(a^2-4\right)=1-4a^2-a^3+4a\\ =\left(1-a\right)\left(1+a+a^2\right)+4a\left(1-a\right)\\ =\left(1-a\right)\left(1+5a+a^2\right)\\ d,=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\\ =a^2\left(1-b\right)\left(1+b\right)+b\left(b-1\right)+a\left(b-1\right)\\ =\left(b-1\right)\left(-a^2-ab+b+a\right)\\ =\left(b-1\right)\left(b-1\right)\left(a+b\right)\left(1-a\right)\)
\(e,=x^2y+xy^2-yz\left(y+z\right)+x^2z-xz^2\\ =\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\\ =x^2\left(y+z\right)+x\left(y-z\right)\left(y+z\right)-yz\left(y+z\right)\\ =\left(y+z\right)\left(x^2+xy-xz-yz\right)\\ =\left(y+z\right)\left(x+y\right)\left(x-z\right)\)
\(f,=xyz-xy-yz-xz+x+y+z-1\\ =xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(x-1\right)\\ =\left(z-1\right)\left(xy-y-x+1\right)=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)
25 tháng 9 2018
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 \(a=x^2+y^2+z^2\); \(b=xy+yz+xz\)
\(\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(xy+yz+zx\right)^2\)
\(=a\left(a+2b\right)+b^2\)
\(=a^2+2ab+b^2=\left(a+b\right)^2\)
\(=\left(x^2+y^2+z^2+xy+yz+zx\right)^2\)
24 tháng 9 2019
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
20 tháng 10 2017
vì có 1 chút nhầm lẫn nên giờ mk mới ra mong bạn thứ lỗi
bài 1
\(\Leftrightarrow\frac{4a^4}{2a^3+2a^2b^2}+\frac{4b^4}{2b^3+2c^2b^2}+\frac{4c^4}{2c^3+2a^2c^2}\)
\(\ge\frac{\left(2a^2+2b^2+2c^2\right)^2}{2a^3+2b^3+2c^3+2a^2b^2+2c^2b^2+2a^2c^2}\)
\(\ge\frac{36}{a^4+a^2+b^4+b^2+c^4+c^2+2a^2b^2+2c^2b^2+2a^2c^2}\)
\(=\frac{36}{\left(a^2+b^2+c^2\right)^2+a^2+b^2+c^2}=3\ge a+b+c\)
Dấu bằng xảy ra khi \(a=b=c=1\)
KN
26 tháng 4 2020
Bài 2 là chuyên Bình Thuận, 2016-2017
Áp dụng bất đẳng thức Cauchy – Schwarz, ta có:
\(\frac{xy}{x^2+yz+zx}\le\frac{xy\left(y^2+yz+zx\right)}{\left(x^2+yz+zx\right)\left(y^2+yz+zx\right)}\le\frac{xy\left(y^2+yz+zx\right)}{\left(xy+yz+zx\right)^2}\)
Tương tự: \(\frac{yz}{y^2+zx+xy}\le\frac{xy\left(z^2+zx+xy\right)}{\left(xy+yz+zx\right)^2}\);\(\frac{zx}{z^2+xy+yz}\le\frac{zx\left(x^2+xy+yz\right)}{\left(xy+yz+zx\right)^2}\)
Cộng từng vế của 3 BĐT trên. ta được:
\(VT\le\frac{\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)}{\left(xy+yz+zx\right)^2}=\frac{x^2+y^2+z^2}{xy+yz+zx}\)
Đẳng thức xảy ra khi x = y = z
MT
8 tháng 10 2019
Hai câu đầu tham khảo
Câu hỏi của Bangtan Sonyeondan - Toán lớp 8 - Học toán với OnlineMath
MT
8 tháng 10 2019
c) \(E=\left(x+a\right)\left(x+2a\right)\left(a+3a\right)\left(x+4a\right)+a^4\)
\(=\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(a+3a\right)+a^4\)
\(=\left(x^2+5ax+4a^2\right)\left(a^2+5ax+6a^2\right)+a^4\)(1)
Đặt \(x^2+5ax+4a^2=t\)
\(\Rightarrow\left(1\right)=t\left(t+2a^2\right)+a^4\)
\(=t^2+2a^2t+a^4=\left(t+a^2\right)^2\)(2)
Mà \(x^2+5ax+4a^2=t\)
Nên \(\left(2\right)=\left(x^2+5ax+5a^2\right)^2\)
2 tháng 9 2018
\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)
\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)
\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)
TC
1
AH
Akai Haruma
Giáo viên
8 tháng 5 2023
Lời giải:
Ta có:
$xy+yz+xz=(x+y+z)^2-(x^2+y^2+z^2+xy+yz+xz)=1-\frac{2}{3}=\frac{1}{3}$
$\Rightarrow 3(xy+yz+xz)=1=(x+y+z)^2$
$\Leftrightarrow (x+y+z)^2-3(xy+yz+xz)=0$
$\Leftrightarrow x^2+y^2+z^2-xy-yz-xz=0$
$\Leftrightarrow 2(x^2+y^2+z^2-xy-yz-xz)=0$
$\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2=0$
Vì $(x-y)^2, (y-z)^2, (z-x)^2\geq 0$ với mọi $x,y,z$.
Do đó để tổng của chúng bằng $0$ thì $x-y=y-z=z-x=0$
$\Leftrightarrow x=y=z$
Khi đó:
$A=\frac{x}{x+x}+\frac{x}{x+x}+\frac{x}{x+x}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}$
a) \(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\)
\(=\left[\left(x+a\right)\left(x+4a\right)\right]\left[\left(x+2a\right)\left(x+3a\right)\right]+a^4\)
\(=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\)
\(=\left(x^2+5ax+5a^2\right)^2-\left(a^2\right)^2+a^4\)
\(=\left(x^2+5ax+5a^2\right)^2\)
b) \(\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(xy+yz+zx\right)^2\)
\(=\left(x^2+y^2+z^2\right)\left[x^2+y^2+z^2+2\left(xy+yz+zx\right)\right]+\left(xy+yz+zx\right)^2\)
\(=\left(x^2+y^2+z^2\right)^2+2\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)+\left(xy+yz+zx\right)^2\)
\(=\left(x^2+y^2+z^2+xy+yz+zx\right)^2\)