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Lời giải:
a) Vì $abc=1$ nên ta có:
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}=\frac{ac}{abc.+ac+c}+\frac{b.ac}{bc.ac+b.ac+ac}+\frac{c}{ac+c+1}\)
\(=\frac{ac}{1+ac+c}+\frac{1}{c+1+ac}+\frac{c}{ac+c+1}=\frac{ac+1+c}{ac+c+1}=1\)
(đpcm)
b)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\Rightarrow \left\{\begin{matrix} x=ka\\ y=kb\\ z=kc\end{matrix}\right.\)
\(x+y+z=ka+kb+kc=k(a+b+c)=k\)
\(x^2+y^2+z^2=k^2a^2+k^2b^2+k^2c^2=k^2(a^2+b^2+c^2)=k^2\)
\(\Rightarrow A=xy+yz+xz=\frac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\frac{k^2-k^2}{2}=0\)
Do \(xyz\ne0\) ta có:
\(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}=0\Leftrightarrow xyz\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)=0\Leftrightarrow x+y+z=0\)
Lại có: \(x^3+y^3+z^3=x^3+y^3+3x^2y+3y^2x-3xy\left(x+y\right)+z^3\)
\(=\left(x+y\right)^3+z^3-3xy\left(-z\right)=\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)+3xyz=3xyz\)
Vậy nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz\)
\(P=\dfrac{x^2}{yz}+\dfrac{y^2}{xz}+\dfrac{z^2}{xy}=\dfrac{x^3}{xyz}+\dfrac{y^3}{xyz}+\dfrac{z^3}{xyz}=\dfrac{x^3+y^3+z^3}{xyz}=\dfrac{3xyz}{xyz}=3\)
Bài 1:
Đặt \(\left(\frac{x}{y}; \frac{y}{z}; \frac{z}{x}\right)=(a,b,c)\Rightarrow abc=1\)
Khi đó:
\(A^2+B^2+C^2-ABC=(b+\frac{1}{b})^2+(c+\frac{1}{c})^2+(a+\frac{1}{a})^2-(a+\frac{1}{a})(b+\frac{1}{b})(c+\frac{1}{c})\)
\(=b^2+\frac{1}{b^2}+2+c^2+\frac{1}{c^2}+2+a^2+\frac{1}{a^2}+2-(ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab})(c+\frac{1}{c})\)
\(a^2+b^2+c^2+(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})+6-[abc+\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)+\left(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\right)+\frac{1}{abc}]\)
\(=a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+6-[1+\left(\frac{abc}{c^2}+\frac{abc}{a^2}+\frac{abc}{b^2}\right)+\left(\frac{a^2}{abc}+\frac{b^2}{abc}+\frac{c^2}{abc}\right)+1]\)
\(=a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+6-[1+(\frac{1}{c^2}+\frac{1}{b^2}+\frac{1}{a^2})+(a^2+b^2+c^2)+1]\)
\(=4\)
Câu 2:
Ta có:
\(xy+yz+xz+2xyz=\frac{ab}{(b+c)(c+a)}+\frac{bc}{(c+a)(a+b)}+\frac{ac}{(b+c)(a+b)}+\frac{2abc}{(a+b)(b+c)(c+a)}\)
\(=\frac{ab(a+b)}{(a+b)(b+c)(c+a)}+\frac{bc(b+c)}{(a+b)(b+c)(c+a)}+\frac{ac(a+c)}{(a+b)(b+c)(c+a)}+\frac{2abc}{(a+b)(b+c)(c+a)}\)
\(=\frac{ab(a+b)+bc(b+c)+ca(c+a)+2abc}{(a+b)(b+c)(c+a)}\)
\(=\frac{ab(a+b+c)+bc(b+c+a)+ca(c+a)}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a+b+c)(ab+bc)+ac(a+c)}{(a+b)(b+c)(c+a)}=\frac{(c+a)b(a+b+c)+ac(a+c)}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a+c)[b(a+b+c)+ac]}{(a+b)(b+c)(c+a)}=\frac{(a+c)[b(a+b)+c(a+b)]}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a+c)(b+c)(a+b)}{(a+b)(b+c)(c+a)}=1\)
1/ x + y + z = 3. Tìm Max P = xy + yz + xz
Ta có: (x - y)² ≥ 0 <=> x² - 2xy + y² ≥ 0 <=> x² + y² ≥ 2xy
hay 2xy ≤ x² + y² , dấu " = " xảy ra <=> x = y
tương tự:
+) 2yz ≤ y² + z²
+) 2xz ≤ x² + z²
cộng 3 vế của 3 bđt trên
--> 2xy + 2yz + 2xz ≤ 2(x² + y² + z²)
--> xy + yz + xz ≤ x² + y² + z²
--> xy + yz + xz + 2xy + 2yz + 2xz ≤ x² + y² + z² + 2xy + 2yz + 2xz
--> 3(xy + yz + xz) ≤ (x + y + z)²
--> 3(xy + yz + xz) ≤ 3²
--> xy + yz + xz ≤ 3
Vậy MaxP = 3 ; Dấu " = " xảy ra <=> x = y = z = 1
Ta có: (x - y)² ≥ 0 <=> x² - 2xy + y² ≥ 0 <=> x² + y² ≥ 2xy
hay 2xy ≤ x² + y² , dấu " = " xảy ra <=> x = y
tương tự:
+) 2yz ≤ y² + z²
+) 2xz ≤ x² + z²
cộng 3 vế của 3 bđt trên
--> 2xy + 2yz + 2xz ≤ 2(x² + y² + z²)
--> xy + yz + xz ≤ x² + y² + z²
--> xy + yz + xz + 2xy + 2yz + 2xz ≤ x² + y² + z² + 2xy + 2yz + 2xz
--> 3(xy + yz + xz) ≤ (x + y + z)²
--> 3(xy + yz + xz) ≤ 3²
--> xy + yz + xz ≤ 3
Vậy Max B = 3 ; Dấu " = " xảy ra <=> x = y = z = 1
Các thánh giúp e nha Ace Legona Nguyễn Huy Tú Toshiro Kiyoshi Phương An Akai Haruma @Nguyễn Vũ Phượng Thảo
dài đấy
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\\ < =>xy+yz+xz=0\\ < =>\left\{{}\begin{matrix}xy=-yz-xz\\yz=-xy-xz\\xz=-xy-yz\end{matrix}\right.\)
\(\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-xz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
cmtt
\(=>\left\{{}\begin{matrix}\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}\\\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\end{matrix}\right.\)
A = ...
= \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\)
=\(\dfrac{yz+xz+xy}{\left(x-y\right)\left(x-z\right)}\left(1\right)\)
mà xy + yz + xz = 0
=> (1) = 0
=> a = 0
Ta có: \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}=\dfrac{x+y+z}{1}\)
\(\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=\dfrac{x^2+y^2+z^2}{1}\)
\(\left(x+y+z\right)^2=x^2+y^2+z^2\)
\(\Rightarrow2\left(xy+yz+xz\right)=0\)
\(\Rightarrow xy+yz+xz=0\left(đpcm\right)\)
Chúc bạn học tốt!
arigatou Kaito Kid