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\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)
\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow xy+yz+xz=0\)
A=\(xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}+\dfrac{3}{xyz}\right)=xyz.\dfrac{3}{xyz}=3\)
bạn tự chứng minh \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}=0\) nha
đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b;\dfrac{1}{z}=c\)
bài toán thành \(a^3+b^3+c^3-3abc=0\) nha
a: A=yx-4y-5x+20
=y(x-4)-5(x-4)
=(x-4)(y-5)
Khi x=14 và y=5,5 thì A=(14-4)(5,5-5)=0,5*10=5
b: \(B=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)
Khi x=5,2 và y=4,8 thì B=(5,2+4,8)(5,2-5)
=0,2*10=2
d: Khi x=5,75 và y=4,25 thì
D=5,75^3-5,75^2*4,25+4,25^3
=8087/64
c: \(D=xyz-xy-yz-xz+x+y+z-1\)
=xy(z-1)-yz+y-xz+z+x-1
=xy(z-1)-y(z-1)-z(x-1)+(x-1)
=(z-1)(xy-y)-(x-1)(z-1)
=(z-1)(xy-y-1)
=(11-1)(9*10-10-1)
=10*79=790
Bài 1:
Vì $x+y+z=1$ nên:
\(Q=\frac{x}{x+\sqrt{x(x+y+z)+yz}}+\frac{y}{y+\sqrt{y(x+y+z)+xz}}+\frac{z}{z+\sqrt{z(x+y+z)+xy}}\)
\(Q=\frac{x}{x+\sqrt{(x+y)(x+z)}}+\frac{y}{y+\sqrt{(y+z)(y+x)}}+\frac{z}{z+\sqrt{(z+x)(z+y)}}\)
Áp dụng BĐT Bunhiacopxky:
\(\sqrt{(x+y)(x+z)}=\sqrt{(x+y)(z+x)}\geq \sqrt{(\sqrt{xz}+\sqrt{xy})^2}=\sqrt{xz}+\sqrt{xy}\)
\(\Rightarrow \frac{x}{x+\sqrt{(x+y)(x+z)}}\leq \frac{x}{x+\sqrt{xy}+\sqrt{xz}}=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế suy ra:
\(Q\leq \frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+ \frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+ \frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)
Vậy $Q$ max bằng $1$
Dấu bằng xảy ra khi $x=y=z=\frac{1}{3}$
Bài 2:
Vì $x+y+z=1$ nên:
\(\text{VT}=\frac{1-x^2}{x(x+y+z)+yz}+\frac{1-y^2}{y(x+y+z)+xz}+\frac{1-z^2}{z(x+y+z)+xy}\)
\(\text{VT}=\frac{(x+y+z)^2-x^2}{(x+y)(x+z)}+\frac{(x+y+z)^2-y^2}{(y+z)(y+x)}+\frac{(x+y+z)^2-z^2}{(z+x)(z+y)}\)
\(\text{VT}=\frac{(y+z)[(x+y)+(x+z)]}{(x+y)(x+z)}+\frac{(x+z)[(y+z)+(y+x)]}{(y+z)(y+x)}+\frac{(x+y)[(z+x)+(z+y)]}{(z+x)(z+y)}\)
Áp dụng BĐT AM-GM:
\(\text{VT}\geq \frac{2(y+z)\sqrt{(x+y)(x+z)}}{(x+y)(x+z)}+\frac{2(x+z)\sqrt{(y+z)(y+x)}}{(y+z)(y+x)}+\frac{2(x+y)\sqrt{(z+x)(z+y)}}{(z+x)(z+y)}\)
\(\Leftrightarrow \text{VT}\geq 2\underbrace{\left(\frac{y+z}{\sqrt{(x+y)(x+z)}}+\frac{x+z}{\sqrt{(y+z)(y+x)}}+\frac{x+y}{\sqrt{(z+x)(z+y)}}\right)}_{M}\)
Tiếp tục AM-GM cho 3 số trong ngoặc lớn, suy ra \(M\geq 3\)
Do đó: \(\text{VT}\geq 2.3=6\) (đpcm)
Dấu bằng xảy ra khi $3x=3y=3z=1$
Bài này ez thôi, làm mãi rồi.
Theo đề bài, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
=>\(\dfrac{xy+yz+xz}{xyz}=0\)
=> xy+yz+zx=0
=> \(\left\{{}\begin{matrix}xy=-yz-zx\\yz=-xy-zx\\zx=-xy-yz\end{matrix}\right.\)
Ta có: x2+2yz=x2+yz-xy-zx=(x-y)(x-z)
y2+2xz=y2+xz-xy-yz=(x-y)(z-y)
z2+2xy=z2+xy-yz-xz=(x-z)(y-z)
=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
Bài 1:
a: \(A=\dfrac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)
\(=\dfrac{\left(x+1\right)\left(x^3+1\right)}{\left(x^2-x+1\right)\left(x^2+1\right)}=\dfrac{\left(x+1\right)^2}{x^2+1}\)
Để A=0 thì x+1=0
hay x=-1
b: \(B=\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}=\dfrac{x^2-4}{x^2-9}\)
Để B=0 thi (x-2)(x+2)=0
=>x=2 hoặc x=-2