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

\(=\dfrac{4-x}{x}:\dfrac{x^2+x+1-x+3}{x^2+x+1}\)

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

b: x^4-7x^2-4x+20=0

=>(x-2)^2(x^2+4x+5)=0

=>x=2

Khi x=2 thì \(A=\dfrac{\left(4-2\right)\left(4+2+1\right)}{2\left(4+4\right)}=\dfrac{7}{8}\)

Bài 1: 

a: x+1/2=5/6

nên x=5/6-1/2=1/3

b: x+1/4=3/4

nên x=3/4-1/4=2/4=1/2

c: x+3/10=1/2

nên x=1/2-3/10=5/10-3/10=1/5

d: x+1/4=3/8

nên x=3/8-1/4=3/8-2/8=1/8

TOT LAM

1. ĐKXĐ: \(x\ne\pm1\)

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

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

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

\(=\dfrac{6x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

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

\(=\dfrac{x-3}{x-1}\)

3. Tại x = 5, A có giá trị là:

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

4. \(A=\dfrac{x-3}{x-1}\) \(=\dfrac{x-1-3}{x-1}=1-\dfrac{3}{x-1}\)

Để A nguyên => \(3⋮\left(x-1\right)\) hay \(\left(x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(tmđk\right)\\x=0\left(tmđk\right)\\x=4\left(tmđk\right)\\x=-2\left(tmđk\right)\end{matrix}\right.\)

Vậy: A nguyên khi \(x=\left\{2;0;4;-2\right\}\)

a: Ta có: |x+4|=1

=>x+4=1 hoặc x+4=-1

=>x=-3(loại) hoặc x=-5

Khi x=-5 thì \(A=\dfrac{\left(-5\right)^2-5}{3\left(-5+3\right)}=\dfrac{20}{3\cdot\left(-2\right)}=\dfrac{-10}{3}\)

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

Bài 5: 

a: Thay \(x=4+2\sqrt{3}\) vào E, ta được:

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

b: Để E<1 thì E-1<0

\(\Leftrightarrow\dfrac{\sqrt{x}-1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)

\(\Leftrightarrow\sqrt{x}-3< 0\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)

c: Để E nguyên thì \(4⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{-2;1;2;4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;5;7\right\}\)

hay \(x\in\left\{16;25;49\right\}\)

Câu 2:
a) Ta có \(x=4-2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-2\right)^2}=\sqrt{3}-2\)

Thay \(x=\sqrt{3}-1\) vào \(B\), ta được

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

b) Để \(B\) âm thì \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\) mà \(\sqrt{x}+1\ge1>0\forall x\) \(\Rightarrow\sqrt{x}-2< 0\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)

c) Ta có \(B=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=1-\dfrac{3}{\sqrt{x}+1}\)

Với mọi \(x\ge0\) thì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{3}{\sqrt{x}+1}\le3\Rightarrow B=1-\dfrac{3}{\sqrt{x}+1}\ge-2\)

Dấu "=" xảy ra khi \(\sqrt{x}+1=1\Leftrightarrow x=0\)

Vậy \(B_{min}=-2\) khi \(x=0\)

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

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

b: Khi x=9 thì P=9-3+1=7

c: P=3

=>x-căn x-2=0

=>(căn x-2)(căn x+1)=0

=>x=4

1,

\(A=\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)

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

\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)

2.

\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)

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

3.

Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)

4.

\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)

\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)

5.

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

\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)

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

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

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

a) Ta có: \(P=\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\cdot\dfrac{1-x^2}{2}\)

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

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

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

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

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

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

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

\(=\dfrac{-x^3+5x-6}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-\left(x^3-5x+6\right)}{2\left(x-1\right)\left(x+1\right)}\)