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3) áp dụng đẳng thức \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
<=>\(1-3xyz=1\left(1-xy-yz-zx\right)\)
<=>\(3xyz=xy+yz+zx\)
mặt khác ta có \(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx=1\)
<=>\(1+2xy+2yz+2zx=1\)
<=> \(xy+yz+zx=0\)
do đó 3xyz=0<=> \(\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}\)
lần lượt thay x;y;z vào hệ ta có các cặp nghiệm (x;y;z)=(0;0;1),(0;1;0),(1;0;0)
do đó x^2017+y^2017+z^2017=1
bài 28
\(P=\frac{\left[a^2-\left(b+c\right)^2\right]\left(a+b-c\right)}{\left(a+b+c\right)\left[\left(a-c\right)^2-b^2\right]}\)
=>\(P=\frac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a-c-b\right)\left(a-c+b\right)}\)
=>\(P=1\)
Bài 30 phải là xy+y+x=3.
Ta có: xy+y+x=3 => (x+1)(y+1)=4(1)
yz+y+z=8 => (y+1)(z+1)=9(2)
zx+x+z=15 => (x+1)(z+1)=16(3)
Nhân (1), (2) và (3) theo vế, ta có:
[(x+1)(y+1)(z+1)]2=576
=> (x+1)(y+1)(z+1)=24(I) hoặc (x+1)(y+1)(z+1)=-24(II)
Lần lượt thay (1),(2),(3) vào (I),(II), tính x,y,z.
Kết quả: P=43/6 hoặc P=-79/6
\(2x^2+2y^2=5xy\Leftrightarrow2x^2+2y^2-5xy=0\)
\(\Leftrightarrow\left(2x-y\right)\left(x-2y\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{y}{2}\\x=2y\end{cases}}\)
Mặt khác : x > y > 0 \(\Rightarrow x=2y\)
Ta có : \(E=\frac{x+y}{x-y}=\frac{2y+y}{2y-y}=\frac{3y}{y}=3\)
a) Dễ tự làm đi
b) Xét 1 + a2 = ab + bc + ca + a2
= b(c + a) + a(c + a)
= (c + a)(b + a)
Cmtt ta có : 1 + b2 = (c + b)(a + b)
1 + c2 = (b+c)( a + c)
Do đó : A = \(\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)\(=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)\left(c+b\right)\left(b+a\right)\left(c+a\right)\left(a+c\right)\left(b+c\right)}\)= 1
Xét a2 + 2bc - 1 = a2 + 2bc - ab - bc - ca
= a2 - ab + bc - ca
= a(a-b) - c(a-b)
= (a-b)(a-c)
Cmtt ta cũng có : b2 + 2ac - 1 = (b-c)(b-a)
c2 + 2ab - 1 = (c-a)(c-b)
Do đó : \(B=\frac{\left(a^2+2bc-1\right)\left(b^2+2ac-1\right)\left(c^2+2ba-1\right)}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}\)
\(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(b-a\right)\left(c-a\right)\left(c-b\right)}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}\)
= -1
10. a)
\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\Leftrightarrow\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\)
\(\Leftrightarrow\left(a+b\right)\left(x^4+y^4\right)=ab\left(x^2+y^2\right)^2\Leftrightarrow\left(bx^2-ay^2\right)^2=0\Leftrightarrow bx^2=ay^2\)
b) Từ \(ay^2=bx^2\Rightarrow\frac{y^2}{b}=\frac{x^2}{a}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\frac{x^{2008}}{a^{1004}}=\frac{1}{\left(a+b\right)^{1004}}\); \(\frac{y^{2008}}{b^{1004}}=\frac{1}{\left(a+b\right)^{1004}}\)
\(\Rightarrow\frac{x^{2008}}{a^{1004}}+\frac{y^{2008}}{b^{1004}}=\frac{2}{\left(a+b\right)^{1004}}\)
25. Ta có \(\left(ax+by+cz\right)^2=0\Leftrightarrow a^2x^2+b^2y^2+c^2z^2=-2\left(abxy+bcyz+acxz\right)\)
Xét mẫu số của P : \(bc\left(y-z\right)^2+ac\left(x-z\right)^2+ab\left(x-y\right)^2=bc\left(y^2-2yz+z^2\right)+ac\left(x^2-2xz+z^2\right)+ab\left(x^2-2xy+y^2\right)\)
\(=y^2bc-2bcyz+bcz^2+acx^2-2xzac+acz^2+abx^2-2abxy+aby^2\)
\(=y^2bc+bcz^2+acx^2+acz^2+abx^2+aby^2-2\left(abxy+xzac+bcyz\right)\)
\(=y^2bc+bcz^2+acx^2+acz^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2\)
\(=c\left(ax^2+by^2+cz^2\right)+b\left(ax^2+by^2+cz^2\right)+a\left(ax^2+by^2+cz^2\right)=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)\)
\(\Rightarrow P=\frac{ax^2+by^2+cz^2}{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}=\frac{1}{a+b+c}=\frac{1}{2007}\)
8. \(\frac{x^3}{a^3}+\frac{y^3}{b^3}=\left(\frac{x}{a}+\frac{y}{b}\right)^3-3.\frac{xy}{ab}\left(\frac{x}{a}+\frac{y}{b}\right)=1^3-3.\left(-2\right).1=7\)
18. Ta có : \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)
\(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2xyz\left(\frac{1}{abz}+\frac{1}{xbc}+\frac{1}{acy}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2xyz\left(\frac{ayz+bxz+cxy}{abcxyz}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)
19. Nhân cả hai vế của đẳng thức giả thiết với \(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\)được
\(\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\right)=0\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{b+c}{\left(c-a\right)\left(a-b\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}=0\)
Ta có ;
\(\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{b+c}{\left(c-a\right)\left(a-b\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}=\frac{\left(a+b\right)\left(a-b\right)+\left(b+c\right)\left(b-c\right)+\left(c+a\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
\(\Rightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
1) Thay xyz = 1 , ta có :
\(\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+xz}=\frac{z}{z+xz+xyz}+\frac{xz}{xz+xyz+xyz^2}+\frac{1}{1+z+xz}\)
\(=\frac{z}{z+xz+1}+\frac{xz}{xz+1+z}+\frac{1}{z+xz+1}=\frac{z+xz+1}{z+xz+1}=1\)
2) Phân tích A thành nhân tử được \(A=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)
Vì a + b + c = 0 nên A = 0
3) Phân tích A thành \(\frac{\left(b-a\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)