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b, \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=\left(x-y\right)^2\left(x-y\right)-\left(y-z\right)^2\left[\left(x-y\right)+\left(z-x\right)\right]+\left(z-x\right)^2\left(z-x\right)\)
\(=\left(x-y\right)^2\left(x-y\right)-\left(y-z\right)^2\left(x-y\right)-\left(y-z\right)^2\left(z-x\right)+\left(z-x\right)^2\left(z-x\right)\)
\(=\left(x-y\right)\left[\left(x-y\right)^2-\left(y-z\right)^2\right]-\left(z-x\right)\left[\left(y-z\right)^2-\left(z-x\right)^2\right]\)
\(=\left(x-y\right)\left(x-y-y+z\right)\left(x-y+y-z\right)-\left(z-x\right)\left(y-z-z+x\right)\left(y-z+z-x\right)\)
\(=\left(x-y\right)\left(x-2y+z\right)\left(x-z\right)-\left(z-x\right)\left(y-2z+x\right)\left(y-x\right)\)
\(=\left(x-y\right)\left(x-2y+z\right)\left(x-z\right)-\left(x-z\right)\left(y-2z+x\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(x-z\right)\left(x-2y+z-y+2z-x\right)\)
\(=\left(x-y\right)\left(x-z\right)\left(3z-3y\right)\)
\(=3\left(x-y\right)\left(x-z\right)\left(z-y\right)\)
c, \(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)-y^2z^2\left[\left(y-x\right)-\left(z-x\right)\right]-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)-y^2z^2\left(y-x\right)+y^2z^2\left(z-x\right)-z^2x^2\left(z-x\right)\)
\(=\left(x^2y^2-y^2z^2\right)\left(y-x\right)+\left(y^2z^2-z^2x^2\right)\left(z-x\right)\)
\(=y^2\left(x-z\right)\left(x+z\right)\left(y-x\right)+z^2\left(y-x\right)\left(x+y\right)\left(z-x\right)\)
\(=y^2\left(x-z\right)\left(x+z\right)\left(y-x\right)-z^2\left(y-x\right)\left(x+y\right)\left(x-z\right)\)
\(=\left(x-z\right)\left(y-x\right)\left[y^2\left(x+z\right)-z^2\left(x+y\right)\right]\)
\(=\left(x-z\right)\left(y-x\right)\left(y^2x+y^2z-z^2x-z^2y\right)\)
\(=\left(x-z\right)\left(y-x\right)\left[x\left(y^2-z^2\right)+yz\left(y-z\right)\right]\)
\(=\left(x-z\right)\left(y-x\right)\left[x\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)\right]\)
\(=\left(x-z\right)\left(y-x\right)\left(y-z\right)\left(xy+xz+yz\right)\)
d, \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
Câu a:
Xét tử số:
\(x^2(y-z)+y^2(z-x)+z^2(x-y)\)
\(=x^2(y-z)-y^2[(y-z)+(x-y)]+z^2(x-y)\)
\(=x^2(y-z)-y^2(y-z)-y^2(x-y)+z^2(x-y)\)
\(=(x^2-y^2)(y-z)-(y^2-z^2)(x-y)\)
\(=(x-y)(y-z)[(x+y)-(y+z)]=(x-y)(y-z)(x-z)\)
Xét mẫu số:
\(x^2y-x^2z+y^2z-y^3=x^2(y-z)-y^2(y-z)=(x^2-y^2)(y-z)\)
\(=(x-y)(x+y)(y-z)\)
Do đó:
\(\frac{x^2(y-z)+y^2(z-x)+z^2(x-y)}{x^2y-x^2z+y^2z-y^3}=\frac{(x-y)(y-z)(x-z)}{(x-y)(x+y)(y-z)}=\frac{x-z}{x+y}\)
Câu b:
Xét tử số:
\(x^5+x+1=x^5-x^2+x^2+x+1=x^2(x^3-1)+x^2+x+1\)
\(=x^2(x-1)(x^2+x+1)+(x^2+x+1)\)
\(=(x^2+x+1)(x^3-x^2+1)\)
Xét mẫu số:
\(x^3+x^2+x=x(x^2+x+1)\)
Do đó: \(\frac{x^5+x+1}{x^3+x^2+1}=\frac{(x^2+x+1)(x^3-x^2+1)}{x(x^2+x+1)}=\frac{x^3-x^2+1}{x}\)
Có:\(x+y+z=0\)
\(\Rightarrow\left(x+y+z\right) ^2=0\)
\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)
\(\Rightarrow x^2+y^2+z^2=-2\left(xy+yz+xz\right)\)
Có:
\(\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\)
\(=y^2-2yz+z^2+z^2-2xz+z^2+x^2-2xy+y^{^2}\)
\(=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\)
\(=2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2\)
\(=3\left(x^2+y^2+z^2\right)\)
\(\Rightarrow\dfrac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
\(=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}\)
\(=\dfrac{1}{3}\)
Quy đồng tính bình thường.
\(A=\dfrac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+2\left(\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{z-x}\right)\)\(=\dfrac{2x^2+2y^2+2z^2-2xy-2yz-2xz}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\dfrac{2yz+2xz+2xy-2x^2-2y^2-2z^2}{ }\)
=0
a) A=(x+z)(y+t)
= xy+xt+zy+zt
Áp dụng bất đẳng thức cô si cho 2 số ta có
x2+y2 ≥ 2\(\sqrt{x^2y^2}\)
⇔x2+y2 ≥ 2xy
TT ta có
x2+t2 ≥ 2xt
y2+z2 ≥ 2yz
z2+t2 ≥ 2zt
cộng vế vs vế ta có
=> x2+y2+x2+t2+y2+z2+t2 ≥ 2xy+2xt+2yz+2zt
⇔ 2(x2+y2+z2+t2) ≥ 2(xy+xt+yz+zt)
⇔ 2 .1 ≥2 A
⇔ 1≥ A
⇔ A ≤ 1
=> Max A =1 dấu "=" xảy ra khi x=y=t=z= \(\pm\dfrac{1}{2}\)
Câu b)
Đây là bài toán quen thuộc của dạng toán xác định điểm rơi trong BĐT Cô-si:
Áp dụng BĐT Cô-si:
\(\frac{2}{3}x^2+\frac{2}{3}y^2\geq 2\sqrt{\frac{2}{3}x^2.\frac{2}{3}y^2}=\frac{4}{3}|xy|\geq \frac{4}{3}xy\)
\(\frac{1}{3}x^2+\frac{4}{3}t^2\geq 2\sqrt{\frac{1}{3}x^2.\frac{4}{3}t^2}=\frac{4}{3}|xt|\geq \frac{4}{3}xt\)
\(\frac{1}{3}y^2+\frac{4}{3}z^2\geq 2\sqrt{\frac{1}{3}y^2.\frac{4}{3}z^2}=\frac{4}{3}|yz|\geq \frac{4}{3}yz\)
\(\frac{2}{3}z^2+\frac{2}{3}t^2\geq 2\sqrt{\frac{2}{3}z^2.\frac{2}{3}t^2}=\frac{4}{3}|zt|\geq \frac{4}{3}zt\)
Cộng theo vế các BĐT thu được và rút gọn:
\(\Rightarrow x^2+y^2+2z^2+2t^2\geq \frac{4}{3}(xy+xt+yz+zt)\)
\(\Leftrightarrow \frac{4}{3}(xy+xt+yz+zt)\leq 1\)
\(\Leftrightarrow B=(x+z)(y+t)\leq \frac{3}{4}\) hay $B_{\max}=\frac{3}{4}$
Dấu bằng xảy ra khi \(x=y=2z=2t\Leftrightarrow (x,y,z,t)=\left(\frac{1}{\pm \sqrt{3}}; \frac{1}{\pm\sqrt{3}}; \frac{1}{\pm 2\sqrt{3}}; \frac{1}{\pm 2\sqrt{3}}\right)\)