a) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)

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

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

\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\)

b) Ta có: \(x=3-2\sqrt{2}\)

\(=2-2\cdot\sqrt{2}\cdot1+1\)

\(=\left(\sqrt{2}-1\right)^2\)

Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(P=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\), ta được: 

\(P=\dfrac{\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)^2}}\)

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

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

\(=\dfrac{1}{\sqrt{2}-1}\)

\(=\sqrt{2}+1\)

Vậy: Khi \(x=3-2\sqrt{2}\) thì \(P=\sqrt{2}+1\)

Lời giải:

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

b. $B=3\Leftrightarrow \frac{\sqrt{x}}{1-\sqrt{x}}=3$

$\Rightarrow \sqrt{x}=3(1-\sqrt{x})$

$\Leftrightarrow 4\sqrt{x}=3\Leftrightarrow x=\frac{9}{16}$ (tm) 

c.

Khi $x=3-2\sqrt{2}=(\sqrt{2}-1)^2\Rightarrow \sqrt{x}=\sqrt{2}-1$

Khi đó:

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

a.

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

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

b.

\(P=\dfrac{B}{A}=\dfrac{x+3}{\sqrt{x}+1}:\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\left(x+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{x+3}{\sqrt{x}-1}=\dfrac{x-1+4}{\sqrt{x}-1}\)

\(=\sqrt{x}+1+\dfrac{4}{\sqrt{x}-1}\)\(=\sqrt{x}-1+\dfrac{4}{\sqrt{x}-1}+2\)

Theo BĐT AM - GM ta có: \(\sqrt{x}-1+\dfrac{4}{\sqrt{x}-1}\ge2\sqrt{\left(\sqrt{x}-1\right)\dfrac{4}{\sqrt{x}-1}}=4\)

\(\Rightarrow\dfrac{1}{P}\ge6\Rightarrow Min_{\dfrac{1}{P}}=6\)

Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{x}-1\right)^2=4\Rightarrow x=9\) (loại trường hợp \(\sqrt{x}-1=-2\))

Vậy GTNN của biểu thức \(\dfrac{1}{P}=6\) khi x = 9.

Câu 1:

a) ĐKXĐ: \(x>0;x\ne9\)

Với x=36 (thỏa mãn ĐKXĐ) thì A có giá trị :

\(A=\dfrac{\sqrt{36}+2}{1+\sqrt{36}}=\dfrac{6+2}{1+6}=\dfrac{8}{7}\)

b) Ta có: 

\(B=\left(\dfrac{2\sqrt{x}}{x-\sqrt{x}-6}+\dfrac{\sqrt{x}}{\sqrt{x}-3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}=\dfrac{x+4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)

c) Ta có:

\(P=A\cdot B=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}+4}{\sqrt{x}+2}=\dfrac{\sqrt{x}+4}{\sqrt{x}+1}=1+\dfrac{3}{\sqrt{x}+1}\)

Vì x là số nguyên lớn hơn 0 nên 

\(x\ge1\Rightarrow\sqrt{x}\ge1\Rightarrow\sqrt{x}+1\ge2>0\Rightarrow P\le1+\dfrac{3}{2}=\dfrac{5}{2}\)

Dấu bằng xảy ra khi x=1;

Gọi số sản phẩm dự định của xí nghiệp A và B lần lượt là x,y \(\left(x,y\in N;0< x,y< 720\right)\)

Vì tổng sản phẩm dự định là 720 nên ta có phương trình: \(x+y=720\left(1\right)\)

Vì thực tế , xí nghiệp A hoàn thành vượt mức 12% nên số sản phẩm xí nghiệp A thực tế là : \(112\%x=\dfrac{28}{25}x\)

Xí nghiệp B hoàn thành vượt mức 10% nên số sản phẩm xí nghiệp B thực tế là : \(110\%y=\dfrac{11}{10}y\)

Vì tổng số sản phẩm thực tế là 800 nên ta có phương trình: \(\dfrac{28}{25}x+\dfrac{11}{10}y=800\Leftrightarrow56x+55y=40000\left(2\right)\)

Từ (1)(2) ta có hệ phương trình:

\(\left\{{}\begin{matrix}x+y=720\\56x+55y=40000\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=720\\55\cdot720+x=40000\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=400\\y=320\end{matrix}\right.\left(t.m\right)\)

Vậy số sản phẩm 2 xí nghiệp làm theo kế hoạch lần lượt là 400 và 320 sản phẩm

1) Ta có phương trình:

\(3x^4-2x^2-40=0\Leftrightarrow\left(3x^4-12x^2\right)+\left(10x^2-40\right)=0\Leftrightarrow\left(x^2-4\right)\left(3x^2+10\right)=0\)

Mà \(3x^2+10\ge10>0\)

\(\Leftrightarrow x^2-4=0\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)

Vậy \(S=\left\{\pm2\right\}\) là tập nghiệm của phương trình

2)

Xét phương trình bậc 2 ẩn x :

\(x^2+\left(m-1\right)x-m^2-2=0\left(1\right)\)

Có hệ số: \(a=1;b=m-1;c=-m^2-2\)

\(\Rightarrow ac=-m^2-2\le-2< 0\)

Suy ra (1) có 2 nghiệm trái dấu \(x_1,x_2\) với mọi m thỏa mãn:

\(\left\{{}\begin{matrix}x_1+x_2=1-m\\x_1x_2=-m^2-2\end{matrix}\right.\left(2\right)\)

Đặt \(\left(\dfrac{x_1}{x_2}\right)^3=-a\left(a>0\right)\Rightarrow\left(\dfrac{x_2}{x_1}\right)^3=-\dfrac{1}{a}\) (do x₁,x₂ là 2 số trái dấu)

\(\Rightarrow T=-\left(a+\dfrac{1}{a}\right)\)

Áp dụng bất đẳng thức Cô-si cho 2 số dương \(a\) và \(\dfrac{1}{a}\) ta có:

\(a+\dfrac{1}{a}\ge2\sqrt{a\cdot\dfrac{1}{a}}=2\)

\(\Rightarrow T\le-2\)

Dấu "=" xảy ra \(\Leftrightarrow a=\dfrac{1}{a}\Leftrightarrow a=1\left(a>0\right)\Leftrightarrow x_1=-x_2\)

(2) trở thành: \(\left\{{}\begin{matrix}m-1=0\\x_1^2=m^2+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=1\\x_1^2=3\left(t.m\right)\end{matrix}\right.\)

Vậy T đạt giá trị nhỏ nhất là -2 tại m=1

a) \(ĐKXĐ:x>0\)

\(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\)

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

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

\(\Leftrightarrow A=x-\sqrt{x}\)

b) Để A = 0

\(\Leftrightarrow x-\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

vậy ...

a) \(M=\left(\dfrac{2x+3\sqrt{x}}{x\sqrt{x}+1}+\dfrac{1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\left(x>0\right)\)

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

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

\(=\dfrac{x+4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{x-\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+4}{\sqrt{x}+1}\)

b) Ta có: \(\sqrt{x}+4>\sqrt{x}+1\Rightarrow\dfrac{\sqrt{x}+4}{\sqrt{x}+1}>1\)

c) \(\dfrac{\sqrt{x}+4}{\sqrt{x}+1}=1+\dfrac{3}{\sqrt{x}+1}\)

Ta có: \(\left\{{}\begin{matrix}3>0\\\sqrt{x}+1>0\end{matrix}\right.\Rightarrow1+\dfrac{3}{\sqrt{x}+1}>1\Rightarrow M>1\)

Lại có: \(\sqrt{x}+1>1\left(x>0\right)\Rightarrow\dfrac{3}{\sqrt{x}+1}< 3\Rightarrow1+\dfrac{3}{\sqrt{x}+1}< 4\Rightarrow M< 4\)

\(\Rightarrow1< M< 4\Rightarrow M\in\left\{2;3\right\}\)

\(M=2\Rightarrow1+\dfrac{3}{\sqrt{x}+1}=2\Rightarrow\dfrac{3}{\sqrt{x}+1}=1\Rightarrow\sqrt{x}+1=3\)

\(\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

\(M=3\Rightarrow1+\dfrac{3}{\sqrt{x}+1}=3\Rightarrow\dfrac{3}{\sqrt{x}+1}=2\Rightarrow2\sqrt{x}+2=3\)

\(\Rightarrow2\sqrt{x}=1\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)

a: Ta có: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)

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

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

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

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

b: Ta có: \(\left(\sqrt{x}+1\right)\cdot A=x\)

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

\(\Leftrightarrow x-2\sqrt{x}+1=0\)

\(\Leftrightarrow x=1\left(loại\right)\)

\(a,E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\dfrac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\left(x>0;x\ne1\right)\\ E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\dfrac{x}{\sqrt{x}-1}\\ b,E>1\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\\ \Leftrightarrow\sqrt{x}-1>0\left[x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\right]\\ \Leftrightarrow x>1\left(tm\right)\)

\(c,E=\dfrac{x}{\sqrt{x}-1}=\dfrac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}\\ E=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\ge2\sqrt{\dfrac{\sqrt{x}-1}{\sqrt{x}-1}}+2=2+2=4\\ E_{min}=4\Leftrightarrow\sqrt{x}-1=1\Leftrightarrow x=4\)

a: M=A:B

\(=\dfrac{x+\sqrt{x}+10-\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{1}=\dfrac{x+7}{\sqrt{x}+3}\)

b: \(M=\dfrac{x-9+16}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}\)

=>\(M=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6>=2\sqrt{16}-6=2\)

Dấu = xảy ra khi (căn x+3)^2=16

=>căn x+3=4

=>x=1