5.

\(A=\dfrac{x}{x+\sqrt{x+yz}}+\dfrac{y}{y+\sqrt{y+zx}}+\dfrac{z}{z+\sqrt{z+xy}}\)

\(=\dfrac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}+\dfrac{y}{y+\sqrt{y\left(x+y+z\right)+zx}}+\dfrac{z}{z+\sqrt{z\left(x+y+z\right)+xy}}\)

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

\(=\dfrac{x\left(\sqrt{x^2+xy+yz+zx}-x\right)}{xy+yz+zx}+\dfrac{y\left(\sqrt{y^2+xy+yz+zx}-y\right)}{xy+yz+zx}+\dfrac{z\left(\sqrt{z^2+xy+yz+zx}-z\right)}{xy+yz+zx}\)

\(=\dfrac{x\sqrt{\left(x+y\right)\left(z+x\right)}-x^2}{xy+yz+zx}+\dfrac{y\sqrt{\left(x+y\right)\left(y+z\right)}-y^2}{xy+yz+zx}+\dfrac{z\sqrt{\left(z+x\right)\left(y+z\right)}-z^2}{xy+yz+zx}\)

Áp dụng BĐT \(ab\le\dfrac{a^2+b^2}{2}\) và BĐT \(a^2+b^2+c^2\ge ab+bc+ca\)

\(A=\dfrac{x\sqrt{\left(x+y\right)\left(z+x\right)}-x^2}{xy+yz+zx}+\dfrac{y\sqrt{\left(x+y\right)\left(y+z\right)}-y^2}{xy+yz+zx}+\dfrac{z\sqrt{\left(z+x\right)\left(y+z\right)}-z^2}{xy+yz+zx}\)

\(=\dfrac{x\sqrt{\left(x+y\right)\left(z+x\right)}+y\sqrt{\left(x+y\right)\left(y+z\right)}+z\sqrt{\left(z+x\right)\left(y+z\right)}-\left(x^2+y^2+z^2\right)}{xy+yz+zx}\)

\(\le\dfrac{x.\dfrac{2x+y+z}{2}+y.\dfrac{x+2y+z}{2}+z.\dfrac{x+y+2z}{2}-\left(x^2+y^2+z^2\right)}{xy+yz+zx}\)

\(=\dfrac{xy+yz+zx}{xy+yz+zx}=1\)

\(maxA=1\Leftrightarrow x=y=z=\dfrac{1}{3}\)

1.

a, \(A=(\dfrac{1}{2};2];B=[\dfrac{2}{3};+\infty)\)

b, \(A\cap B=\left[\dfrac{2}{3};2\right];A\cup B=\left(\dfrac{1}{2};+\infty\right)\)

2.

ĐK: \(x\ne2;x\ne-3\)

\(1+\dfrac{2}{x-2}=\dfrac{10}{x+3}-\dfrac{50}{\left(2-x\right)\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{2}{2-x}+\dfrac{10}{x+3}-\dfrac{50}{\left(2-x\right)\left(x+3\right)}=1\)

\(\Leftrightarrow\dfrac{2\left(x+3\right)}{\left(2-x\right)\left(x+3\right)}+\dfrac{10\left(2-x\right)}{\left(2-x\right)\left(x+3\right)}-\dfrac{50}{\left(2-x\right)\left(x+3\right)}=1\)

\(\Leftrightarrow\dfrac{-8x-24}{-x^2-x+6}=1\)

\(\Leftrightarrow-8x-24=-x^2-x+6\)

\(\Leftrightarrow x^2-7x-30=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=-3\left(l\right)\end{matrix}\right.\)

a, Có \(\dfrac{124-106}{2}+1=10\) số tự nhiên m thỏa mãn.

b, Có \(\dfrac{125-105}{5}+1=5\) số tự nhiên m thỏa mãn.

a, m ⋮ 2 ⇒ m có tận cùng là chẵn

⇒m ∈ {106; 108; 110; 112; 114; 116; 118; 120; 122; 124}

b, m ⋮ 5 ⇒ m có tận cùng là 0 hoặc 5

⇒ m ∈ {105; 110; 115; 120; 125}

c, m ⋮ 2 và ⋮ 5 ⇒ m có tận cùng là 0

⇒ m ∈ {110; 120}

a, m ⋮ 2 ⇒ m có tận cùng là chẵn

⇒m ∈ {106; 108; 110; 112; 114; 116; 118; 120; 122; 124}

b, m ⋮ 5 ⇒ m có tận cùng là 0 hoặc 5

⇒ m ∈ {105; 110; 115; 120; 125}

c, m ⋮ 2 và ⋮ 5 ⇒ m có tận cùng là 0

⇒ m ∈ {110; 120}

Nếu sai thì bạn thông cảm 

fv

C

c

`8,`

`a,`

`10x^2+3/4=0`

Vì \(x^2\ge0\text{ }\forall\text{ }x\)

`->`\(10x^2+\dfrac{3}{4}>0\text{ }\forall\text{ }x\)

`->` Đa thức vô nghiệm.

`b,`

`(x-1/2)^2+5=0`

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\text{ }\forall\text{ }x\)

`->`\(\left(x-\dfrac{1}{2}\right)^2+5>0\text{ }\forall\text{ }x\)

`->` Đa thức vô nghiệm.

`c,`

`y^6+2=0`

Vì \(y^6\ge0\text{ }\forall\text{ }x\)

`->`\(y^6+2>0\text{ }\forall\text{ }x\)

`->` Đa thức vô nghiệm.

`d,`

`x^2+2x+2=0`

Vì \(x^2\ge0\text{ }\forall\text{ }x\)

`->`\(x^2+2x+2>0\text{ }\forall\text{ }x\)

`->` Đa thức vô nghiệm.

\(B=\left(\dfrac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+3}{1-\sqrt{x}}\right).\left(\dfrac{x-1}{2x+\sqrt{x}-1}\right)\left(ĐKXĐ:x>0;x\ne1\right)\)\(=\left[\dfrac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\right].\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+2x-1}\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+2x-1}\)

\(=\dfrac{2\sqrt{x}+3}{\sqrt{x}-1}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+2x-1}\)

\(=\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+2x-1}\)

\(=\dfrac{2x+5\sqrt{x}+3}{\sqrt{x}+2x-1}\)

Vậy \(B=\dfrac{2x+5\sqrt{x}+3}{\sqrt{x}+2x-1}\) với \(x>0;x\ne1\)

Để \(B< 0\) thì \(\dfrac{2x+5\sqrt{x}+3}{\sqrt{x}+2x-1}< 0\)

\(\Rightarrow2x+5\sqrt{x}+3< \sqrt{x}+2x-1\)

\(\Leftrightarrow4\sqrt{x}< -4\Leftrightarrow\sqrt{x}< -1\Leftrightarrow x< 1\)

Vậy \(0< x< 1\) thì \(B< 0\)

a.

\(A\left(x\right)+B\left(x\right)+C\left(x\right)=\left(3x^6-5x^4+2x^2-7\right)+\left(8x^6+7x^4-x^2+11\right)+\left(x^6+7x^4-x^2+6\right)\)

\(=12x^6+9x^4+10\)

b.

\(A\left(x\right)-B\left(x\right)+C\left(x\right)=\left(3x^6-5x^4+2x^2-7\right)-\left(8x^6+7x^4-x^2+11\right)+\left(x^6+7x^4-x^2+6\right)\)

\(=-4x^6-5x^4+2x^2-12\)

c.

\(A\left(x\right)-B\left(x\right)-C\left(x\right)=\left(3x^6-5x^4+2x^2-7\right)-\left(8x^6+7x^4-x^2+11\right)-\left(x^6+7x^4-x^2+6\right)\)

\(=-6x^6-19x^4+4x^2-24\)