\(P=\left(\frac{2x+1}{\sqrt{x^3}-1}-\frac{\sqrt{x}}{x+\sqrt{x}+1}\right):\left(\frac{3}{\sqrt{x}-1}+\frac{2\sqrt{x}+5}{1-x}\right)\)

\(ĐKXĐ:x\ge0,x\ne1\)

\(P=\frac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\frac{\left(\sqrt{x}+1\right)-\left(2\sqrt{x}+5\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

b, \(x=\frac{8}{3-\sqrt{5}}=\frac{2\left(9-5\right)}{3-\sqrt{5}}=2\left(3+\sqrt{5}\right)\)

\(=5+2\sqrt{5}+1=\left(\sqrt{5}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{5}+1\)

\(\Rightarrow P=\frac{\sqrt{5}+1+1}{\sqrt{5}+1-2}=\frac{\sqrt{5}+2}{\sqrt{5}-1}\)

c, \(P=\frac{\sqrt{x}-2+3}{\sqrt{x}-2}=1+\frac{3}{\sqrt{x}-2}\in N\)\(\Rightarrow\frac{3}{\sqrt{x}-2}\in Z\)

\(\Rightarrow\sqrt{x}-2\inƯ\left(3\right)\)

\(\Rightarrow x\in\left\{9;25\right\}\)

Hoành độ giao điểm (P) ; (d) tm pt 

\(x^2-2x-3=0\)

ta có a - b + c = 1 + 2 - 3 = 0 

Vậy pt có 2 nghiệm 

x = -1 ; x = 3 

Với x = -1 => y = 1 

Với x = 3 => y = 9

Vậy (P) cắt (D) tại A(-1;1) ; B(3;9) 

Áp dụng BĐT phụ \(4xy\le\left(x+y\right)^2\le1\)\(\Leftrightarrow xy\le\frac{1}{4}\)

Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)

Có \(K=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)\(=x^2+2x.\frac{1}{x}+\frac{1}{x^2}+y^2+2y.\frac{1}{y}+\frac{1}{y^2}\)\(=x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}+4\)

Áp dụng BĐT Cô-si cho 2 số dương \(x^2\)và \(y^2\), ta có: \(x^2+y^2\ge2\sqrt{x^2y^2}=2xy\)

Tương tự, ta có \(\frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2}.\frac{1}{y^2}}=\frac{2}{xy}\)

Từ đó \(K\ge2xy+\frac{2}{xy}+4\)\(=32xy+\frac{2}{xy}-30xy+4\)

Áp dụng BĐT Cô-si cho 2 số dương \(32xy\)và \(\frac{2}{xy}\), ta có: \(32xy+\frac{2}{xy}\ge2\sqrt{32xy.\frac{2}{xy}}=16\)

Lại có \(xy\le\frac{1}{4}\Leftrightarrow-xy\ge-\frac{1}{4}\)nên \(K\ge16-\frac{30}{4}+4=\frac{25}{2}\)

Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)

Vậy GTNN của K là \(\frac{25}{2}\)khi \(x=y=\frac{1}{2}\)

\(K=x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+4=x^2+\dfrac{1}{16x^2}+y^2+\dfrac{1}{16y^2}+\dfrac{15}{16x^2}+\dfrac{15}{16y^2}+4\ge\dfrac{1}{2}+\dfrac{1}{2}+4+\dfrac{2.15}{16xy}=5+\dfrac{2.15}{16xy}\)

\(x+y\ge2\sqrt{xy};\Rightarrow2\sqrt{xy}\le x+y\le1\Rightarrow2\sqrt{xy}\le1\Leftrightarrow xy\le\dfrac{1}{4}\)

\(\Rightarrow K\ge5+\dfrac{2.15}{16.\dfrac{1}{4}}=\dfrac{25}{2}\)

Ta có \(P=2020+\sqrt{x^2-10x+26}\)\(=2020+\sqrt{\left(x^2-10x+25\right)+1}\)\(=2020+\sqrt{\left(x-5\right)^2+1}\)

Nhận thấy \(\left(x-5\right)^2\ge0\)\(\Leftrightarrow\left(x-5\right)^2+1\ge1\)\(\Leftrightarrow\sqrt{\left(x+5\right)^2+1}\ge1\)\(\Leftrightarrow A\ge2021\)

Dấu "=" xảy ra khi \(x-5=0\Leftrightarrow x=5\)

Vậy GTNN của P là 2021 khi \(x=5\)

đó nha 

ĐK: \(x\ge0,x\ne1,x\ne4\).

a) \(P=\left(\frac{2x+1}{\sqrt{x^3}-1}-\frac{\sqrt{x}}{x+\sqrt{x}+1}\right)\div\left(\frac{3}{\sqrt{x}-1}+\frac{2\sqrt{x}+5}{1-x}\right)\)

\(=\left(\frac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\div\left(\frac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

\(=\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\div\frac{3\sqrt{x}+3-2\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

\(=\frac{\sqrt{x}+1}{\sqrt{x}-2}\)

\(x=\frac{8}{3-\sqrt{5}}=\frac{16}{6-2\sqrt{5}}=\frac{16}{5-2\sqrt{5}+1}=\left(\frac{4}{\sqrt{5}-1}\right)^2\)

\(\Rightarrow x=\sqrt{\left(\frac{4}{\sqrt{5}-1}\right)^2}=\left|\frac{4}{\sqrt{5}-1}\right|=\frac{4}{\sqrt{5}-1}\)

\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}=\frac{\frac{4}{\sqrt{5}-1}+1}{\frac{4}{\sqrt{5}-1}-2}=\frac{4+\sqrt{5}-1}{4-2\sqrt{5}+2}=\frac{3+\sqrt{5}}{6-2\sqrt{5}}=\frac{7+3\sqrt{5}}{4}\)

\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}=\frac{\sqrt{x}-2+3}{\sqrt{x}-2}=1+\frac{3}{\sqrt{x}-2}\)

Có \(-2\le\sqrt{x}-2< 0\)thì \(\frac{3}{\sqrt{x}-2}\le-\frac{3}{2}\)nên \(P\)không là số tự nhiên. 

Suy ra \(\sqrt{x}-2>0\Leftrightarrow x>4\)

\(P\)là số tự nhiên thì \(\frac{3}{\sqrt{x}-2}\)là số tự nhiên suy ra \(\sqrt{x}-2=\frac{3}{n}\)(với \(n\)là số tự nhiên) 

\(\Leftrightarrow x=\left(\frac{3}{n}+2\right)^2\).

a, Thay m = 7 ta được y = x + 7 - 1 = x + 6 

Hoành độ giao điểm tm pt 

\(x^2-x-6=0\)

\(\Delta=1-4\left(-6\right)=1+24=25>0\)

Vậy pt có 2 nghiệm pb 

hay (P) cắt (D) tại 2 điểm pb 

Với mọi số thực \(a_i\) , ta có:

\(\left(a_1-a_2\right)^2+\left(a_2-a_3\right)^2+...+\left(a_n-a_1\right)^2\ge0\)

\(\Leftrightarrow2\left(a_1^2+a_2^2+a_3^2+...+a_n^2\right)\ge2\left(a_1a_2+a_2a_3+...+a_na_1\right)\)

\(\Leftrightarrow a_1^2+a_2^2+...+a_n^2\ge a_1a_2+a_2a_3+...+a_na_1\) (đpcm)

ừa ae

(a₁ - a₂)² + (a₂ - a₃)² + ...+(a_r - a₁) \(\ge\) 0

\( \Leftrightarrow \) 2 (a₁² + a₂² + ...+ a_n² ) \(\ge\) 2 ( a₁ a₂ + a₂ a₃ +...+ a_n a₁ )

\( \Leftrightarrow\) a₁² + a₂²+...+ a_n² \(\ge\)  a₁ a₂ + a₂ a₃ +...+ a_n a₁  (ĐPCM)

Với mọi a;b;c;d;e ta có:

\(\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+\left(a-2e\right)^2\ge0\)

\(\Leftrightarrow4a^2+4b^2+4c^2+4d^2+4e^2\ge4ab+4ac+4ad+4ae\)

\(\Leftrightarrow a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\) (đpcm)

Dấu "=" xảy ra khi \(\dfrac{a}{2}=b=c=d=e\)

BĐT

\(\Leftrightarrow4a^2+4b^2+4c^2+4d^2+4e^2\ge4a\left(b+c+d+e\right)\)

\(\Leftrightarrow4a^2+4b^2+4c^2+4d^2+4e^2\ge4ab+4ac+4ad+4ae\)

\(\Leftrightarrow4a^2+4b^2+4c^2+4d^2+4e^2-\left(4ab+4ac+4ad+4ae\right)\ge0\)

\(\Leftrightarrow a^2-4ab+4b^2+a^2-4ac+4c^2+a^2-4ad+4d^2+a^2-4ae+4e^2\ge0\)

\(\Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+\left(a-2e\right)^2\ge0\), luôn đúng với \(\forall a,b,c,d,e\in R\)

Dấu "=" xảy ra khi và chỉ khi \(a=2b=2c=2d=2e\)

a, Với x >= 0 ; x khác 16 

\(A=\left(\frac{x+5\sqrt{x}-27+\left(3-\sqrt{x}\right)\left(\sqrt{x}+4\right)}{x-16}\right):\frac{1}{\sqrt{x}+4}\)

\(=\left(\frac{x+5\sqrt{x}-27+3\sqrt{x}+12-x-4\sqrt{x}}{x-16}\right):\frac{1}{\sqrt{x}+4}\)

\(=\left(\frac{4\sqrt{x}-15}{x-16}\right):\frac{1}{\sqrt{x}+4}=\frac{4\sqrt{x}-15}{\sqrt{x}-4}\)

b, Ta có \(B=-2A\Rightarrow\sqrt{x}-4=-\frac{8\sqrt{x}-30}{\sqrt{x}-4}\)

\(\Leftrightarrow x-8\sqrt{x}+16=-8\sqrt{x}+30\Leftrightarrow x-14=0\Leftrightarrow x=14\left(tm\right)\)