\(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{\left(3x-2\right)\left(3x+1\right)}=\frac{670}{2011}\)

\(\Rightarrow\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{\left(3x-2\right)\left(3x+1\right)}\right)=\frac{670}{2011}\)

\(\Rightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{3x-2}-\frac{1}{3x+1}=\frac{670}{2011}:\frac{1}{3}\)

\(\Rightarrow1-\frac{1}{3x+1}=\frac{2010}{2011}\)

\(\Rightarrow\frac{1}{3x+1}=1-\frac{2010}{2011}\)

\(\Rightarrow\frac{1}{3x+1}=\frac{1}{2011}\)

=>3x+1=2011

=>3x=2011-1

=>x=2010:3

=>x=670

vậy x=670

Dặt \(A=\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{\left(3x-2\right).\left(3x+1\right)}\)

\(3A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{\left(3x-2\right)\left(3x+1\right)}\)

\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{\left(3x-2\right)}-\frac{1}{\left(3x+1\right)}\)

\(3A=1-\frac{1}{3x+1}\)

\(A=\left(1-\frac{1}{3x+1}\right):3=\frac{670}{2011}\)

\(1-\frac{1}{3x+1}=\frac{670}{2011}.3\)

\(1-\frac{1}{3x+1}=\frac{2010}{2011}\)

\(\frac{1}{3x+1}=1-\frac{2010}{2011}\)suy ra \(\frac{1}{3x+1}=\frac{1}{2011}\)

suy ra 3x+1=2011

3x=2000

x=2000/3

đầu bài trên tớ làm luôn nhá !!!

a,  / 3x+1/= 5-3

    / 3x+1/= 2

   3x+1=2

  x+1 = 2:3 

 x+1 = 2 phần 3

x= 2/3 -1 

x= -1/3 

còn phần b.c.d lần lượt nha bạn 

^x2-3.(x-1)

(x-1)²

^=>x2-3

x-1

Cách 1:

(x³ – 2x² + x – 1) (3x – 2)

= x³ . (3x – 2) + (-2x²) .(3x – 2) + x .(3x – 2) + (-1) . (3x – 2)

= x³ . 3x + x³ . (-2) + (-2x²). 3x + (-2x²) . (-2) + x . 3x + x. (-2) + (-1). 3x + (-1). (-2)

= 3x⁴ – 2x³ – 6x³ + 4x² + 3x² – 2x – 3x + 2

= 3x⁴ + (-2x³ -6x³) + (4x² + 3x² ) + (-2x – 3x) + 2

= x⁴ + (-8x³) + 7x² + (-5x) + 2

= x⁴ – 8x³ +7x² – 5x + 2

Cách 2:

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{125}{376}\)

\(\Leftrightarrow\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{125}{376}\)

\(\Leftrightarrow\dfrac{1}{3}\left(1-\dfrac{1}{x+3}\right)=\dfrac{125}{376}\left(x\ne0;x\ne-3\right)\)

\(\Leftrightarrow\dfrac{x+3-1}{x+3}=\dfrac{3.125}{376}\Leftrightarrow\dfrac{x+2}{x+3}=\dfrac{3.125.}{376}.\dfrac{\left(x+3\right)}{x+3}\)

\(\Leftrightarrow376\left(x+2\right)=3.125.\left(x+3\right)\)

\(\Leftrightarrow376x+752=375x+1125\)

\(\Leftrightarrow376x-375x=1125-752\Leftrightarrow x=373\left(x\in N^{\cdot}\right)\)

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

\(\Rightarrow\frac{3}{x+3}\Rightarrow x+3\in\text{Ư}\left(3\right)=\left\{1;3\right\}\)

\(\Rightarrow x+3=1\Rightarrow x=-2\)(loại vì x < 0)

\(\Rightarrow x+3=3\Rightarrow x=0\)

Vậy x = 0

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

\(x\times\dfrac{3}{5}=\dfrac{3}{3}\)

\(x\times\dfrac{3}{5}=1\)

\(x=1:\dfrac{3}{5}\)

\(x=1\times\dfrac{5}{3}\)

\(x=\dfrac{5}{3}\)

\(x-\dfrac{4}{9}=\dfrac{3}{7}:\dfrac{9}{14}\)

\(x-\dfrac{4}{9}=\dfrac{3}{7}\times\dfrac{14}{9}\)

\(x-\dfrac{4}{9}=\dfrac{2}{3}\)

\(x=\dfrac{2}{3}+\dfrac{4}{9}\)

`x`\(=\dfrac{18}{27}+\dfrac{12}{27}\)

\(x=\dfrac{30}{27}=\dfrac{10}{9}\)

Câu 1: Tác phẩm Quốc âm thi tập là của ai ? (1đ)

a. Nguyễn Trãi.

b. Lê Thánh Tông.

c. Lý Tử Tấn.

d .Lương Thế Vinh.

[3x-1]⁵=[3x-1]⁸

=>[3x-1]⁸-[3x-1]⁵⁼⁰

   [3x-1]⁵.[[3x-1]³-1]=0

=>[3x-1]⁵ =0​ hoặc [3x-1]³-1=0

=>3x-1=0 hoặc3x-1 =1

=>x=1/3 hoac x=2/3

Vay....

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