a) Ta có:

1; 4; 7;...; 100 có (100 - 1) : 3 + 1 = 34 (số)

1 + 4 + 7+ ... + 100 = (100 + 1) × 34 : 2

= 101 × 17

(1 + 4 + 7 + ... + 100) : a = 17

101 × 17 : a = 17

a = 101 × 17 : 17

a = 100

b) (X - 1/2) × 5/3 = 7/4 - 1/2

(X - 1/2) × 5/3 = 5/4

X - 1/2 = 5/4 : 5/3

X - 1/2 = 3/4

X = 3/4 + 1/2

X = 5/4

a) (1 + 4 + 7 +...+ 100) : a = 17

1717 : a = 17

a = 101

b) \(\left(x-\dfrac{1}{2}\right)\times\dfrac{5}{3}=\dfrac{7}{4}-\dfrac{1}{2}\)

\(\left(x-\dfrac{1}{2}\right)\times\dfrac{5}{3}=\dfrac{10}{8}\)

\(\left(x-\dfrac{1}{2}\right)=\dfrac{10}{8}\div\dfrac{5}{3}\)

\(\left(x-\dfrac{1}{2}\right)=\dfrac{10}{8}\times\dfrac{3}{5}\)

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

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

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

\(x=\dfrac{5}{4}\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

b) Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)

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

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

d) Để A>0 thì \(\sqrt{x}-2>0\)

hay x>4

Ta có:

x⁴ + 1/x⁴ = x⁴ + 2.x².1/x² + 1/x⁴ - 2.x².1/x²

= (x² + 1/x²)² - 2.x².1/x²

= 4² - 2

= 14

Ta có: 

\(\dfrac{x^2+1}{x^2}=4\) (ĐK: \(x\ne0\))  

\(\Rightarrow x^2+1+4x^2\)

\(\Rightarrow4x^2-x^2=1\)

\(\Rightarrow3x^2=1\)

\(\Rightarrow x^2=\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{\sqrt{3}}{3}\left(tm\right)\)

Thay vào biểu thức ta có:

\(\dfrac{x^4+1}{x^4}\)

\(=\dfrac{\left(\dfrac{\sqrt{3}}{3}\right)^4+1}{\left(\dfrac{\sqrt{3}}{3}\right)^4}\)

\(=\dfrac{\dfrac{9}{81}+1}{\dfrac{9}{81}}\)

\(=\dfrac{\dfrac{1}{9}+1}{\dfrac{1}{9}}\)

\(=\dfrac{10}{9}:\dfrac{1}{9}\)

\(=10\)

Sửa đề: \(\left\lbrack\frac{4}{x-4}-\frac{4}{x+4}\right\rbrack\cdot\frac{x^2+8x+16}{32}\)

Đặt \(A=\left\lbrack\frac{4}{x-4}-\frac{4}{x+4}\right\rbrack\cdot\frac{x^2+8x+16}{32}\)

a: ĐKXĐ: x∉{4;-4}

b: \(A=\left\lbrack\frac{4}{x-4}-\frac{4}{x+4}\right\rbrack\cdot\frac{x^2+8x+16}{32}\)

\(=\frac{4\left(x+4\right)-4\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\cdot\frac{\left(x+4\right)^2}{32}\)

\(=\frac{4x+16-4x+16}{x-4}\cdot\frac{x+4}{32}=\frac{32}{x-4}\cdot\frac{x+4}{32}=\frac{x+4}{x-4}\)

\(A=\frac13\)

=>\(\frac{x+4}{x-4}=\frac13\)

=>3(x+4)=x-4

=>3x+12=x-4

=>2x=-16

=>x=-8(nhận)

c: A=1

=>x+4=x-4

=>4=-4(loại)

=>x∈∅

d: Để A nguyên thì x+4⋮x-4

=>x-4+8⋮x-4

=>8⋮x-4

=>x-4∈{1;-1;2;-2;4;-4;8;-8}

=>x∈{5;3;6;2;8;0;12;-4}

Kết hợp ĐKXĐ, ta được: x∈{5;3;6;2;8;0;12}

e: Để A>0 thì \(\frac{x+4}{x-4}>0\)

=>x-4>0 hoặc x+4<0

=>x>4 hoặc x<-4

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

\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)

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

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

a) đk x khác 0;2

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

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

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

= \(\dfrac{x-2}{x^2}+1\)

= \(\dfrac{x^2+x-2}{x^2}\)

b) Để \(\left|2+x\right|=1\)

<=> \(\left[{}\begin{matrix}2+x=1< =>x=-1\left(tm\right)\\2+x=-1< =>x=-3\left(tm\right)\end{matrix}\right.\)

TH1: x = -1

Thay x = -1 vào P, ta có:

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

TH2: x = -3

Thay x = -3 vào P, ta có:

\(P=\dfrac{\left(-3\right)^2-3-2}{\left(-3\right)^2}=\dfrac{4}{9}\)

c) P = \(1+\dfrac{x-2}{x^2}\)

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

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

Áp dụng bdt co-si, ta có:

\(\left(x-2\right)+\dfrac{4}{x-2}\ge2\sqrt{\left(x-2\right)\dfrac{4}{x-2}}=4\)

<=> \(\dfrac{x^2}{x-2}\ge4+4=8\)

<=> \(\dfrac{x-2}{x^2}\le\dfrac{1}{8}\)

<=> A \(\le\dfrac{9}{8}\)

Dấu "=" <=> x = 4