Đặt \(a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac{1}{z}\Rightarrow xyz=1\) và \(x;y;z>0\)

Gọi biểu thức cần tìm GTNN là P, ta có:

\(P=\dfrac{1}{\dfrac{1}{x^3}\left(\dfrac{1}{y}+\dfrac{1}{z}\right)}+\dfrac{1}{\dfrac{1}{y^3}\left(\dfrac{1}{z}+\dfrac{1}{x}\right)}+\dfrac{1}{\dfrac{1}{z^3}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)}\)

\(=\dfrac{x^3yz}{y+z}+\dfrac{y^3zx}{z+x}+\dfrac{z^3xy}{x+y}=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)

\(P\ge\dfrac{\left(x+y+z\right)^2}{y+z+z+x+x+y}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\)

\(P_{min}=\dfrac{3}{2}\) khi \(x=y=z=1\) hay \(a=b=c=1\)

Đặt \(a = \frac{1}{x} ; b = \frac{1}{y} ; c = \frac{1}{z} \Rightarrow x y z = 1\) và \(x ; y ; z > 0\)

Gọi biểu thức cần tìm GTNN là P, ta có:

\(P = \frac{1}{\frac{1}{x^{3}} \left(\right. \frac{1}{y} + \frac{1}{z} \left.\right)} + \frac{1}{\frac{1}{y^{3}} \left(\right. \frac{1}{z} + \frac{1}{x} \left.\right)} + \frac{1}{\frac{1}{z^{3}} \left(\right. \frac{1}{x} + \frac{1}{y} \left.\right)}\)

\(= \frac{x^{3} y z}{y + z} + \frac{y^{3} z x}{z + x} + \frac{z^{3} x y}{x + y} = \frac{x^{2}}{y + z} + \frac{y^{2}}{z + x} + \frac{z^{2}}{x + y}\)

\(P \geq \frac{\left(\left(\right. x + y + z \left.\right)\right)^{2}}{y + z + z + x + x + y} = \frac{x + y + z}{2} \geq \frac{3 \sqrt[3]{x y z}}{2} = \frac{3}{2}\)

\(P_{m i n} = \frac{3}{2}\) khi \(x = y = z = 1\) hay \(a = b = c = 1\)

theo đề ta có: \(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\)

\(\Rightarrow x^2+y^2+z^2+2\cdot\left(xy+yz+zx\right)=0\)

\(\Rightarrow x^2+y^2+z^2=-2\left(xy+yz+xz\right)\left(1\right)\)

ta co: \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

mà x + y + z = 0

\(\Rightarrow x^3+y^3+z^3-3xyz=0\Rightarrow x^3+y^3+z^3=3xyz\left(2\right)\)

a. VT = \(\left(x^2+y^2+z^2\right)^2=x^4+y^4+z^4+2\cdot\left(x^2y^2+y^2z^2+x^2z^2\right)\)

ta có: \(\left(xy+yz+zx\right)^2=\left(x^2y^2+y^2z^2+x^2z^2\right)+2xyz\cdot\left(x+y+z\right)\)

vì x+y+z=0 nên: \(\left(xy+yz+zx\right)^2=\left(x^2y^2+y^2z^2+x^2z^2\right)\)

từ (1) ta có: \(\left(x^2+y^2+z^2\right)^2=\left\lbrack-2\left(xy+yz+zx\right)^{}\right\rbrack^2\) (*)

\(=4\cdot\left(xy+yz+zx\right)^2=4\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)\)

ta có: \(4\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)=x^4+y^4+z^4+2\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)\)

mà: \(2\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)=x^4+y^4+z^4\)

thay vào (*) ta được:

\(\left(x^2+y^2+z^2\right)^2=\left(x^4+y^4+z^4\right)+2\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)\)

\(=x^4+y^4+z^4+x^4+y^4+z^4=2\cdot\left(x^4+y^4+z^4\right)=VP\)

⇒ đpcm

b. \(VT=5\cdot\left(x^3+y^3+z^3\right)\left(x^2+y^2+z^2\right)\)

\(=5\cdot\left(3xyz\right)\left(x^2+y^2+z^2\right)\)

\(=15xyz\cdot\left(x^2+y^2+z^2\right)\) (3)

\(x+y+z=0\Rightarrow x+y=-z\)

\(x^5+y^5+z^5=x^5+y^5+\left\lbrack-\left(x+y\right)\right\rbrack^5=x^5+y^5-\left(x+y\right)^5\)

\(=x^5+y^5-\left(x^5+5y^4+10x^3y^2+10x^2y^3+5xy^4+y^5\right)\)

\(=-5x^4y-10x^3y^2-10x^2y^3-5xy^4\)

\(=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)

\(=-5xy\left\lbrack x^3+y^3+2xy\left(x+y\right)\right\rbrack\)

\(=-5xy\left\lbrack\left(x+y\right)^3-3xy\left(x+Y\right)+2xy\left(x+y\right)\right\rbrack\)

\(=-5xy\left\lbrack\left(x+Y\right)^3-xy\left(x+y\right)\right\rbrack\)

\(=-5xy\left(x+Y\right)\left\lbrack\left(x+y\right)^2-xy\right\rbrack\)

vì x+y=-z nên ta có:

\(x^5+y^5+z^5=-5xy\left(-z\right)\left\lbrack\left(-z\right)^2-xy\right\rbrack=5xyz\left(x^2-zy\right)\)

mặt khác \(x+y=-z\Rightarrow\left(x+y\right)^2=z^2\Rightarrow x^2+y^2+2xy=z^2\)

\(x^2+y^2+z^2=x^2+y^2+\left(x+y\right)^2\)

\(=x^2+y^2+x^2+2xy+y^2=2\cdot\left(x^2+xy+y^2\right)\)

\(z^2-xy=\left(x+y\right)^2-xy=x^2+2xy+y^2-xy=x^2+xy+y^2\)

vậy \(x^5+y^5+z^5=5xyz\cdot\left(x^2+xy+y^2\right)=\frac52xyz\left(x^2+y^2+z^2\right)\)

\(\Rightarrow2\cdot\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)

⇒ \(6\cdot\left(x^5+y^5+z^5\right)=15xyz\left(x^2+y^2+z^2\right)\) (4)

từ (3) và (4) ⇒ VT = VP

câu c: phần này đã được chứng minh nằm trong câu b nha bạn

theo đề ta có: \(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\)

\(\Rightarrow x^2+y^2+z^2+2\cdot\left(xy+yz+zx\right)=0\)

\(\Rightarrow x^2+y^2+z^2=-2\left(xy+yz+xz\right)\left(1\right)\)

ta co: \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

mà x + y + z = 0

\(\Rightarrow x^3+y^3+z^3-3xyz=0\Rightarrow x^3+y^3+z^3=3xyz\left(2\right)\)

a. VT = \(\left(x^2+y^2+z^2\right)^2=x^4+y^4+z^4+2\cdot\left(x^2y^2+y^2z^2+x^2z^2\right)\)

ta có: \(\left(xy+yz+zx\right)^2=\left(x^2y^2+y^2z^2+x^2z^2\right)+2xyz\cdot\left(x+y+z\right)\)

vì x+y+z=0 nên: \(\left(xy+yz+zx\right)^2=\left(x^2y^2+y^2z^2+x^2z^2\right)\)

từ (1) ta có: \(\left(x^2+y^2+z^2\right)^2=\left\lbrack-2\left(xy+yz+zx\right)^{}\right\rbrack^2\) (*)

\(=4\cdot\left(xy+yz+zx\right)^2=4\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)\)

ta có: \(4\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)=x^4+y^4+z^4+2\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)\)

mà: \(2\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)=x^4+y^4+z^4\)

thay vào (*) ta được:

\(\left(x^2+y^2+z^2\right)^2=\left(x^4+y^4+z^4\right)+2\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)\)

\(=x^4+y^4+z^4+x^4+y^4+z^4=2\cdot\left(x^4+y^4+z^4\right)=VP\)

⇒ đpcm

b. \(VT=5\cdot\left(x^3+y^3+z^3\right)\left(x^2+y^2+z^2\right)\)

\(=5\cdot\left(3xyz\right)\left(x^2+y^2+z^2\right)\)

\(=15xyz\cdot\left(x^2+y^2+z^2\right)\) (3)

\(x+y+z=0\Rightarrow x+y=-z\)

\(x^5+y^5+z^5=x^5+y^5+\left\lbrack-\left(x+y\right)\right\rbrack^5=x^5+y^5-\left(x+y\right)^5\)

\(=x^5+y^5-\left(x^5+5y^4+10x^3y^2+10x^2y^3+5xy^4+y^5\right)\)

\(=-5x^4y-10x^3y^2-10x^2y^3-5xy^4\)

\(=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)

\(=-5xy\left\lbrack x^3+y^3+2xy\left(x+y\right)\right\rbrack\)

\(=-5xy\left\lbrack\left(x+y\right)^3-3xy\left(x+Y\right)+2xy\left(x+y\right)\right\rbrack\)

\(=-5xy\left\lbrack\left(x+Y\right)^3-xy\left(x+y\right)\right\rbrack\)

\(=-5xy\left(x+Y\right)\left\lbrack\left(x+y\right)^2-xy\right\rbrack\)

vì x+y=-z nên ta có:

\(x^5+y^5+z^5=-5xy\left(-z\right)\left\lbrack\left(-z\right)^2-xy\right\rbrack=5xyz\left(x^2-zy\right)\)

mặt khác \(x+y=-z\Rightarrow\left(x+y\right)^2=z^2\Rightarrow x^2+y^2+2xy=z^2\)

\(x^2+y^2+z^2=x^2+y^2+\left(x+y\right)^2\)

\(=x^2+y^2+x^2+2xy+y^2=2\cdot\left(x^2+xy+y^2\right)\)

\(z^2-xy=\left(x+y\right)^2-xy=x^2+2xy+y^2-xy=x^2+xy+y^2\)

vậy \(x^5+y^5+z^5=5xyz\cdot\left(x^2+xy+y^2\right)=\frac52xyz\left(x^2+y^2+z^2\right)\)

\(\Rightarrow2\cdot\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)

⇒ \(6\cdot\left(x^5+y^5+z^5\right)=15xyz\left(x^2+y^2+z^2\right)\) (4)

từ (3) và (4) ⇒ VT = VP

câu c: phần này đã được chứng minh nằm trong câu b nha bạn

\({x^2} = {4^2} + {2^2} = 20 \Rightarrow x = 2\sqrt 5 \)

\({y^2} = {5^2} - {4^2} = 9 \Leftrightarrow y = 3\)

\({z^2} = {\left( {\sqrt 5 } \right)^2} + {\left( {2\sqrt 5 } \right)^2} = 25 \Rightarrow z = 5\)

\({t^2} = {1^2} + {2^2} = 5 \Rightarrow t = \sqrt 5 \)

Vì MN // BC theo Talet ta có:

\(\dfrac{y}{20}\) =  \(\dfrac{10}{15}\)  = \(\dfrac{x}{12}\) => x = \(\dfrac{10}{15}\) . 12 = 8;   y = \(\dfrac{10}{15}\) . 20 = \(\dfrac{40}{3}\)

x = 8 cm; y = 13,(3) cm

a: Xét ΔMNQ có KP//NQ

nên \(\frac{MP}{MQ}=\frac{KP}{NQ}\)

=>\(\frac{x}{8}=\frac36=\frac12=\frac48\)

=>x=4

a: Xét ΔKAD và ΔBDA có

\(\hat{KAD}=\hat{BDA}\) (hai góc so le trong, AK//BD)

AD chung

\(\hat{KDA}=\hat{BAD}\) (hai góc so le trong, AB//CD)

Do đó: ΔKAD=ΔBDA

=>KA=BD

mà BD=AC

nên AK=AC

=>ΔAKC cân tại A

b: ΔAKC cân tại A

=>\(\hat{AKC}=\hat{ACK}\)

mà \(\hat{AKC}=\hat{BDC}\) (hai góc đồng vị, BD//AK)

nên \(\hat{BDC}=\hat{ACD}\)

Xét ΔBDC va ΔACD có

BD=AC

\(\hat{BDC}=\hat{ACD}\)

CD chung

Do đó: ΔBDC=ΔACD

=>\(\hat{BCD}=\hat{ADC}\)

=>ABCD là hình thang cân

a: ta có: EI⊥BF

AC⊥BF

Do đó: EI//AC

=>\(\hat{IEB}=\hat{ACB}\) (hai góc đồng vị)

mà \(\hat{ABC}=\hat{ACB}\) (ΔABC cân tại A)

nên \(\hat{KBE}=\hat{IEB}\)

Xét ΔKBE vuông tại K và ΔIEB vuông tại I có

BE chung

\(\hat{KBE}=\hat{IEB}\)

Do đó: ΔKBE=ΔIEB

=>EK=BI

b: Điểm D ở đâu vậy bạn?