Cho các số thực x,y,z tM xy/x+ý=yz/ý+z=xz/x+z tính M=x2+y2+z2/xy+yz+xz
Cho x, y, z là các số khác 0 và x2 = yz, y2=xz, z2=xy
Tính \(\dfrac{\left(x+y+z\right)^{2022}}{x^{337}y^{674}z^{1011}}\)
Lời giải:
Từ điều kiện đề bài suy ra:
$\frac{x}{y}=\frac{y}{z}=\frac{z}{x}$
$\Rightarrow (\frac{x}{y})^3=(\frac{y}{z})^3=(\frac{z}{x})^3=\frac{x}{y}.\frac{y}{z}.\frac{z}{x}=1$
$\Rightarrow \frac{x}{y}=\frac{y}{z}=\frac{z}{x}=1$
$\Rightarrow x=y=z$.
Do đó:
$\frac{(x+y+z)^{2022}}{x^{337}.y^{674}.z^{1011}}=\frac{(3x)^{2022}}{x^{337}.x^{674}.x^{1011}}=\frac{3^{2022}.x^{2022}}{x^{2022}}=3^{2022}$
Lời giải:
Từ điều kiện đề bài suy ra:
$\frac{x}{y}=\frac{y}{z}=\frac{z}{x}$
$\Rightarrow (\frac{x}{y})^3=(\frac{y}{z})^3=(\frac{z}{x})^3=\frac{x}{y}.\frac{y}{z}.\frac{z}{x}=1$
$\Rightarrow \frac{x}{y}=\frac{y}{z}=\frac{z}{x}=1$
$\Rightarrow x=y=z$.
Do đó:
$\frac{(x+y+z)^{2022}}{x^{337}.y^{674}.z^{1011}}=\frac{(3x)^{2022}}{x^{337}.x^{674}.x^{1011}}=\frac{3^{2022}.x^{2022}}{x^{2022}}=3^{2022}$
Cho x,y,z >0 t/m x2+y2+z2=3.
C/m \(\dfrac{x}{\sqrt[3]{yz}}+\dfrac{y}{\sqrt[3]{xz}}+\dfrac{z}{\sqrt[3]{xy}}\ge xy+yz+zx\)
Bạn tham khảo lời giải tại đây:
Cách khác:
Áp dụng BĐT AM-GM và BĐT Cauchy-Schwarz:
\(\sum \frac{x}{\sqrt[3]{yz}}\geq \sum \frac{x}{\frac{y+z+1}{3}}=3\sum \frac{x}{y+z+1}=3\sum \frac{x^2}{xy+xz+x}\)
\(\geq 3. \frac{(x+y+z)^2}{2(xy+yz+xz)+(x+y+z)}\)
Ta sẽ chứng minh: \(\frac{3(x+y+z)^2}{2(xy+yz+xz)+(x+y+z)}\geq xy+yz+xz(*)\)
Đặt $x+y+z=a$ thì $xy+yz+xz=\frac{a^2-3}{2}$
Bằng BĐT AM-GM dễ thấy $\sqrt{3}< a\leq 3$
BĐT $(*)$ trở thành:
$\frac{3a^2}{a^2+a-3}\geq \frac{a^2-3}{2}$
$\Leftrightarrow a^4+a^3-12a^2-3a+9\leq 0$
$\Leftrightarrow (a-3)(a+1)(a^2+3a-3)\leq 0$
Điều này đúng với mọi $\sqrt{3}< a\leq 3$
Do đó BĐT $(*)$ đúng nên ta có đpcm.
Dấu "=" xảy ra khi $x=y=z=1$
Cho x2 + y2 + z2 = 10. Tính:
P = ( xy + yz + xz)2 + ( x - yz)2 + ( y - xz)2 + ( z - xy)2
Đề lỗi công thức rồi. Bạn xem lại.
cho x+y+z=a
x2+y2+z2=b
\(\dfrac{1}{\text{x
}}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{c}\)
Tính xy+yz+xz, x3+y3+z3
\(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2xz=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)
\(\Rightarrow2\left(xy+yz+xz\right)=\left(x+y+z\right)^2+\left(x^2+y^2+z^2\right)\)
\(\Rightarrow2\left(xy+yz+xz\right)=a^2+b\)
\(\Rightarrow xy+yz+xz=\dfrac{a^2+b}{2}\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{c}\Rightarrow\dfrac{xy+yz+xz}{xyz}=\dfrac{1}{c}\)
\(\Rightarrow xyz=c\left(xy+yz+xz\right)\)
\(\Rightarrow xyz=\dfrac{\left(a^2+b\right)c}{2}\)
\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(\Rightarrow x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz\)
\(\Rightarrow x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-\left(xy+yz+xz\right)\right)+3xyz\)
\(\Rightarrow x^3+y^3+z^3=a\left(b-\dfrac{a^2+b}{2}\right)+3\dfrac{\left(a^2+b\right)c}{2}\)
\(\Rightarrow x^3+y^3+z^3=a\dfrac{\left(b-a^2\right)}{2}+3\dfrac{\left(a^2+b\right)c}{2}\)
1). x2y2(y-x)+y2z2(z-y)-z2x2(z-x)
2)xyz-(xy+yz+xz)+(x+y+z)-1
3)yz(y+z)+xz(z-x)-xy(x+y)
4)2a2b+4ab2-a2c+ac2-4b2c+2bc2-4abc
5)y(x-2z)2+8xyz+x(y-2z)2-2z(x+y)2
6)8x3(y+z)-y3(z+2x)-z3(2x-y)
7) (x2+y2)3+(z2-x2)3-(y2+z2)3
1). x2y2(y-x)+y2z2(z-y)-z2x2(z-x)
2)xyz-(xy+yz+xz)+(x+y+z)-1
3)yz(y+z)+xz(z-x)-xy(x+y)
4)2a2b+4ab2-a2c+ac2-4b2c+2bc2-4abc
5)y(x-2z)2+8xyz+x(y-2z)2-2z(x+y)2
6)8x3(y+z)-y3(z+2x)-z3(2x-y)
7) (x2+y2)3+(z2-x2)3-(y2+z2)3
bn gõ bài trong công thức trực quan ik, khó nhìn lắm, ko làm đc
1) \(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^3-x^3y^2+y^2z^3-y^3z^2-z^2x^2\left(z-x\right)\)
\(=\left(y^2z^3-x^3y^2\right)-\left(y^3z^2-x^2y^3\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(z^3-x^3\right)-y^3\left(z^2-x^2\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(z-x\right)\left(z^2+zx+x^2\right)-y^3\left(z-x\right)\left(z+x\right)-z^2x^2\left(z-x\right)\)
\(=\left(z-x\right)\left[y^2\left(z^2+zx+x^2\right)-y^3\left(z+x\right)-z^2x^2\right]\)
\(=\left(z-x\right)\left[\left(y^2z^2+xy^2z+x^2y^2\right)-\left(y^3z+xy^3\right)-z^2x^2\right]\)
\(=\left(z-x\right)\left(y^2z^2+xy^2z+x^2y^2-y^3z-xy^3-z^2x^2\right)\)
\(=\left(z-x\right)\left[\left(y^2z^2-y^3z\right)-\left(x^2z^2-x^2y^2\right)+\left(xy^2z-xy^3\right)\right]\)
\(=\left(z-x\right)\left[y^2z\left(z-y\right)-x^2\left(z^2-y^2\right)+xy^2\left(z-y\right)\right]\)
\(=\left(z-x\right)\left[y^2z\left(z-y\right)-x^2\left(z-y\right)\left(z+y\right)+xy^2\left(z-y\right)\right]\)
\(=\left(z-x\right)\left(z-y\right)\left[y^2z-x^2\left(z+y\right)+xy^2\right]\)
\(=\left(z-x\right)\left(z-y\right)\left(y^2z-x^2z-x^2y+xy^2\right)\)
\(=\left(z-x\right)\left(z-y\right)\left[\left(y^2z-x^2z\right)-\left(x^2y-xy^2\right)\right]\)
\(=\left(z-x\right)\left(z-y\right)\left[z\left(y^2-x^2\right)-xy\left(x-y\right)\right]\)
\(=\left(z-x\right)\left(z-y\right)\left[z\left(y-x\right)\left(y+x\right)+xy\left(y-x\right)\right]\)
\(=\left(z-x\right)\left(z-y\right)\left(y-x\right)\left[z\left(y+x\right)+xy\right]\)
\(=\left(z-x\right)\left(z-y\right)\left(y-x\right)\left(yz+xz+xy\right)\)
2) \(xyz-\left(xy+yz+xz\right)+\left(x+y+z\right)-1\)
\(=xyz-xy-yz-xz+x+y+z-1\)
\(=\left(xyz-xy\right)-\left(yz-y\right)-\left(xz-x\right)+\left(z-1\right)\)
\(=xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)\)
\(=\left(z-1\right)\left(xy-y-x+1\right)\)
\(=\left(z-1\right)\left[\left(xy-y\right)-\left(x-1\right)\right]\)
\(=\left(z-1\right)\left[y\left(x-1\right)-\left(x-1\right)\right]\)
\(=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)
Cho x+y+z=4 xy+xz+xt+yz+yt+zt=1 tìm GTNN của x2+y2+z2+t2
phân tích a)(x-y)3+(y-z)3+(z-x)3
b)x.(y2-z2)+y.(z2-x2)+z.(x2-y2)
c)xy.(x-y)-xz.(x+z)-yz.(zx-y+z)
d)x.(y+z)2+y.(z-x)2+z.(x+y)2-4xyz
a)(x-y)3+(y-z)3+(z-x)3
=3(x-y+y-z+z-x)=3
b)nhân vào là rồi đối trừ là hết luôn ( nhưng là mũ 2 hay nhân 2 v mk là theo nhân 2 nhé]