a: |2x-3|=1

=>2x-3=1 hoặc 2x-3=-1

=>x=1(nhận) hoặc x=2(loại)

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

b: ĐKXĐ: x<>-1; x<>2

\(B=\dfrac{2x^2-4x+3x+3-2x^2-1}{\left(x-2\right)\left(x+1\right)}=\dfrac{-x+2}{\left(x-2\right)\left(x+1\right)}=\dfrac{-1}{x+1}\)

1) Áp dụng bđt Cauchy cho 3 số dương ta có

 \(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)

\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)

\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)

Cộng (1);(2);(3) theo vế ta được

\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)

\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)

\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)

2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)

Dấu"=" khi a = 4b

nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)

Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)

Đặt \(\sqrt{a+b}=t>0\) ta được

\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)

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

Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)

nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)

khi đó a + b = 1

mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

Làm lại nha cái này đúng, kia sai nha=)

b)

Với \(\left\{{}\begin{matrix}x\ne3\\x\ne2\end{matrix}\right.\)

\(P=A-B=(\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)})+\dfrac{2x-1}{x-3}\\ =\left(\dfrac{2x-9-x^2-9}{\left(x-3\right)\left(x-2\right)}\right)+\dfrac{\left(2x-1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2}{\left(x-3\right)\left(x-2\right)}+\dfrac{2x^2-4x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2+2x^2-4x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x^2-3x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x^2-2x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x\left(x-2\right)-\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}=\dfrac{x-1}{x-3}\)

c)

Để P\(\ge1\) thì:

\(\dfrac{x-1}{x-3}\ge1\\ \Leftrightarrow x-3-x+1-1\ge0\\ \Leftrightarrow-3\ge0\left(vô.lý\right)\)

Vậy không tồn tại giá trị x để \(P\ge1\)

`HaNa☘D`

b)

\(P=A-B=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x^2-9}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-9-x^2+9}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x\left(2-x\right)}{\left(x-3\right)\left(x-2\right)}\\ =-\dfrac{x\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =-\dfrac{x}{x-3}\)

c)

Để \(P\le1\) thì:

\(-\dfrac{x}{x-3}\le1\)

\(\Leftrightarrow\dfrac{x}{x-3}\ge1\\ \Leftrightarrow x-3-x\ge1\\ \Leftrightarrow-3\ge1\left(vô.lý\right)\)

Vậy không tồn tại giá trị x để \(P\le1\)

`HaNa♬D`

\(a,ĐK:x\ne1;x\ne-1\\ b,C=\dfrac{x^2+x+x^2+1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{2x^2+2x+1}{2x^2-2}\\ c,C=-\dfrac{1}{2}\Leftrightarrow2-2x^2=2x^2+2x+1\\ \Leftrightarrow4x^2+2x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}-1}{4}\\x=\dfrac{-\sqrt{5}-1}{4}\end{matrix}\right.\\ d,C>0\Leftrightarrow2x^2-2>0\left(2x^2+2x+1>0\right)\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)>0\\ \Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)

a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b) Ta có: \(B=\left(\dfrac{x-2}{2x-2}+\dfrac{3}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(1-\dfrac{x-3}{x+1}\right)\)

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

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

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

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

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

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

Vậy: Khi x=2005 thì \(B=\dfrac{1}{2}\)

a/

Để biểu thức được xác định

\(=>\left\{{}\begin{matrix}2x-2\ne0\\2x+2\ne0\\x+1\ne0\end{matrix}\right.\)

\(\odot2x-2\ne0\)

\(2x\ne2\)

\(x\ne1\)

\(\odot2x+2\ne0\)

\(2x\ne-2\)

\(x\ne-1\)

\(\odot x+1\ne0\)

\(x\ne-1\)

Vậy điều kiện xác định của bt là: \(x\ne-1;x\ne\pm2\)

a: \(A=\sqrt{x}+\dfrac{\sqrt{x}\left(1+2\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\sqrt{x}+\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)

Khi x=4 thì \(A=2+\dfrac{2\cdot2+1}{2+1}=2+\dfrac{5}{3}=\dfrac{11}{3}\)

b: Khi x=(2-căn 3)^2 thì \(A=2-\sqrt{3}+\dfrac{2\left(2-\sqrt{3}\right)+1}{2-\sqrt{3}+1}\)

\(=2-\sqrt{3}+\dfrac{4-2\sqrt{3}+1}{3-\sqrt{3}}\)

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

\(=\dfrac{\left(2-\sqrt{3}\right)\left(3-\sqrt{3}\right)+5-2\sqrt{3}}{3-\sqrt{3}}\)

\(=\dfrac{6-2\sqrt{3}-3\sqrt{3}+3+5-2\sqrt{3}}{3-\sqrt{3}}\)

\(=\dfrac{14-7\sqrt{3}}{3-\sqrt{3}}\)

d: A=2

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

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

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

=>\(\left[{}\begin{matrix}\sqrt{x}=\dfrac{-1+\sqrt{5}}{2}\left(nhận\right)\\\sqrt{x}=\dfrac{-1-\sqrt{5}}{2}\left(loại\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{6-2\sqrt{5}}{4}=\dfrac{3-\sqrt{5}}{2}\)

a. \(A=\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\left(ĐKXĐ:x\ne1;x\ne-3\right)\)

\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{x+3}{x-1}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{\left(x+3\right)^2}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{2-3x+x^2+6x+9-x^2+1}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}.\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{3x+12}=\dfrac{x^2+x+1}{x+3}\)

\(M=A.B=\dfrac{x^2+x+1}{x+3}.\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x-2}{x+3}\)

b. -Để M thuộc Z thì:

\(\left(x^2+x-2\right)⋮\left(x+3\right)\)

\(\Rightarrow\left(x^2+3x-2x-6+4\right)⋮\left(x+3\right)\)

\(\Rightarrow\left[x\left(x+3\right)-2\left(x+3\right)+4\right]⋮\left(x+3\right)\)

\(\Rightarrow4⋮\left(x+3\right)\)

\(\Rightarrow x+3\in\left\{1;2;4;-1;-2;-4\right\}\)

\(\Rightarrow x\in\left\{-2;-1;1;-4;-5;-7\right\}\)

c. \(A^{-1}-B=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{x^3-1}\)

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

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

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

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

\(=\dfrac{1}{x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)

\(Max=\dfrac{4}{3}\Leftrightarrow x=\dfrac{-1}{2}\)

a) ĐKXĐ: \(x\ne-10;x\ne0;x\ne-5\)

b) \(P=\dfrac{x^2+2x}{2x+20}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+2x}{2\left(x+10\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x\left(x^2+2x\right)\left(x+5\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{2\left(x-5\right)\left(x+10\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{\left(50-5x\right)\left(x+10\right)}{2x\left(x+5\right)\left(x+10\right)}\)

\(=\dfrac{x^4+7x^3+10x^2+2x^2+10x-100+500-5x^2}{2x\left(x+10\right)\left(x+5\right)}\)

\(=\dfrac{x^4+7x^3+7x^2+10x+400}{2x\left(x+10\right)\left(x+5\right)}\)

c) \(P=0\Rightarrow x^4+7x^3+7x^2+10x+400=0\Leftrightarrow...\)

Số xấu thì câu c, d làm cũng như không. Bạn xem lại đề.

Lời giải:

a.

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

b.

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

Để $P<0\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}+3}<0$

Mà $\sqrt{x}+3>0$ nên $\sqrt{x}-2<0$

$\Leftrightarrow 0< x< 4$

Kết hợp với ĐKXĐ suy ra $0< x< 4$

Mà $x$ nguyên nên $x\in left\{1; 2; 3\right\}$