Ta có:\(\frac{1}{ab}+\frac{1}{cd}\ge\frac{8}{\left(a+b\right)\left(c+d\right)}\Leftrightarrow\left(\frac{1}{ab}+\frac{1}{cd}\right)\left(a+b\right)\left(c+d\right)\ge8\)

Xét bất đẳng thức Cô si

\(\hept{\begin{cases}\frac{1}{ab}+\frac{1}{cd}\ge2\sqrt{\frac{1}{abcd}}\\a+b\ge2\sqrt{ab}\\c+d\ge2\sqrt{cd}\end{cases}}\)

\(\Rightarrow\left(\frac{1}{ab}+\frac{1}{cd}\right)\left(a+b\right)\left(c+d\right)\ge2\cdot\frac{1}{\sqrt{abcd}}\cdot2\sqrt{ab}\cdot2\sqrt{cd}\)

\(\Rightarrow\left(\frac{1}{ab}+\frac{1}{cd}\right)\left(a+b\right)\left(c+d\right)\ge8\left(đpcm\right)\)

Dấu "=" xảy ra khi\(\hept{\begin{cases}\frac{1}{ab}=\frac{1}{cd}\\a=b\\c=d\end{cases}}\Leftrightarrow a=b=c=d\)

ĐKXĐ: \(x\ne y,x\ne z,y\ne z\)

\(\dfrac{4}{\left(y-x\right)\left(z-x\right)}+\dfrac{1}{\left(y-x\right)\left(y-z\right)}+\dfrac{3}{\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{4\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}-\dfrac{x-z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}+\dfrac{3\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{4\left(y-z\right)-\left(x-z\right)+3\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=\dfrac{2x+y-3z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

4(y-z)+z-x+3(y-x)

=4y-4z+z-x+3y-3x

=7y-3z-4x

\(\Leftrightarrow x^2+2x+1+4-x^2=5\Leftrightarrow2x=0\Leftrightarrow x=0\)

\(\Leftrightarrow x^2+2x+1+4-x^2=5\)

=>2x+5=5

hay x=0

Answer:

Có \(\left|x\right|\ge0\Rightarrow\left|x\right|+16\ge0\)

\(\Rightarrow\frac{\left|x\right|+16}{-17}\le0\)

\(\frac{\left|x\right|+16}{-17}=\frac{\left|x\right|}{-17}+\frac{16}{-17}\)

Dấu "=" xảy ra khi \(\frac{\left|x\right|}{-17}=0\Rightarrow\left|x\right|=0.\left(-17\right)\Rightarrow x=0\)

Tìm GTLN mà bn

y=82,59 nha

~HT~

k cho mình nha

@@@@@@@@@@@@@

Ảo ma

\(\dfrac{1}{\left(x-y\right)\left(y-z\right)}+\dfrac{1}{\left(y-z\right)\left(z-x\right)}+\dfrac{1}{\left(z-x\right)\left(x-y\right)}\)(đk: \(x\ne y\ne z\))

\(=\dfrac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\dfrac{0}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)

Chắc đề là tính ha!

\(=\dfrac{x+y+y-z+x-y}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ =\dfrac{0}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ =0\\ Vậy.A=0\)

49 + 48 + 47 + ... + 1 + 0

= (49 + 0) . 50 : 2 = 1225

\(1+2+3+4+...+99+100\)

\(=\dfrac{\left(1+100\right).\left[\left(100-1\right):1+1\right]}{2}=5050\)

=5050

100+100:2+3+200:=353

= 253 nha