\(=\left(x-\dfrac{1}{3}\right)\left(\dfrac{4}{3}x+\dfrac{1}{9}-x+\dfrac{1}{3}\right)\\ =\left(x-\dfrac{1}{3}\right)\left(\dfrac{1}{3}x+\dfrac{4}{9}\right)\\ =\dfrac{1}{3}x^2+\dfrac{4}{9}x-\dfrac{1}{9}x-\dfrac{4}{27}\\ =\dfrac{1}{3}x^2+\dfrac{1}{3}x-\dfrac{4}{27}\)

a) \(=x^3-\dfrac{1}{27}-x^2+\dfrac{2}{3}x-\dfrac{1}{9}=x^3-x^2+\dfrac{2}{3}x-\dfrac{2}{27}\)

b) \(=x^6-6x^4+12x^2-8-x^3+x+x^2-3x=x^6-6x^4-x^3+13x^2-2x-8\)

a.\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{8}{x}+\dfrac{5}{y}=1\end{matrix}\right.\)

\(ĐK:x;y\ne0\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\) 

hpt trở thành:

\(\left\{{}\begin{matrix}a+b=\dfrac{1}{6}\\8a+5b=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{18}\\b=\dfrac{1}{9}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{18}\\\dfrac{1}{y}=\dfrac{1}{9}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=18\\y=9\end{matrix}\right.\) ( tm )

Vậy nghiệm hpt: \(\left\{{}\begin{matrix}x=18\\y=9\end{matrix}\right.\)

b.\(\left\{{}\begin{matrix}\dfrac{x-1}{2}-y=1\\2x+y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}+2x=2\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1+4x}{2}=\dfrac{4}{2}\\2x+y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x=5\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\2.1+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

a: ĐKXĐ: x<>0; x<>-1

PT =>x+1-2x=3

=>1-x=3

=>x=-2(nhận)

b: Sửa đề: \(\dfrac{1}{2x-3}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5}{x}\)

=>x-3=5(2x-3)

=>10x-15=x-3

=>9x=12

=>x=4/3(nhận)

c: ĐKXĐ: x<>0; x<>2

PT =>x(x+2)-x+2=2

=>x^2+2x-x=0

=>x(x+1)=0

=>x=-1

a: =>4x^2-2x-x-4=0

=>4x^2-3x-4=0

=>\(x=\dfrac{3\pm\sqrt{73}}{8}\)

b: =>x^2+x-x+2=5x-4

=>x^2+2=5x-4

=>x^2-5x+6=0

=>x=2(loại) hoặc x=3(nhận)

Bài 1:

a.

$|x+\frac{7}{4}|=\frac{1}{2}$

\(\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.\)

b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$

$|2x+1|=\frac{11}{15}$

\(\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.\)

c.

$3x(x+\frac{2}{3})=0$

\(\Leftrightarrow \left[\begin{matrix} 3x=0\\ x+\frac{2}{3}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\frac{-3}{2}\end{matrix}\right.\)

d.

$x+\frac{1}{3}=\frac{2}{5}-(\frac{-1}{3})=\frac{2}{5}+\frac{1}{3}$

$\Leftrightarrow x=\frac{2}{5}$

Bài 2:

$\frac{1}{100}-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}$

$=\frac{99}{100}$

$\Rightarrow A=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}=\frac{-49}{50}$

\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)

x = nhân ạ

Bài 1:

a: ĐKXĐ: x∉{-1/2;1/2}

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

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

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

\(=\frac{4x-2+4x^2+2x+2}{2}=\frac{4x^2+6x}{2}=2x^2+3x\)

b: Thay x=0 vào A, ta được:

\(A=2\cdot0^2+3\cdot0=0\)

THay x=-3 vào A, ta được:

\(A=2\cdot\left(-3\right)^2+3\cdot\left(-3\right)\)

=18-9

=9

Vì x=1/2 không thỏa mãn ĐKXĐ
nên khi x=1/2 thì A không có giá trị

c: A=2

=>\(2x^2+3x=2\)

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

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

=>(x+2)(2x-1)=0

=>x=1/2(loại) hoặc x=-2(nhận)

Bài 2: Để B là số nguyên thì \(2n^2+5n-1\) ⋮2n-1

=>\(2n^2-n+6n-3+2\) ⋮2n-1

=>2⋮2n-1

mà 2n-1 lẻ

nên 2n-1∈{1;-1}

=>2n∈{2;0}

=>n∈{1;0}

a: Ta có: \(\frac{x}{x^2-2x+1}+\frac{x}{1-x^2}\)

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

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

Ta có: \(A=\frac{x^2+x+1}{1-x^3}+\left(\frac{x}{x^2-2x+1}+\frac{x}{1-x^2}\right)\cdot\frac{x^2-1}{x^2+1}\)

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

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