(6x - 11)³ = (-3)² . 15 + 208

(6x - 11)³ =  9 . 15 + 208

(6x - 11)³ = 135 + 208

(6x - 11)³ = 343

(6x - 11)³ = 7³ (cùng số mũ)

⇒6x - 11 = 7

6x = 7 + 11

6x = 18

x  =3

Thanks hai bạn

:D

Bài 2 : Gọi số tự nhiên chẵn đầu tiên là a (hai số tiếp theo lần lượt là a+2 và a+4)

Theo bài ra ta có : \(\left(a+2\right)\left(a+4\right)-a\left(a+2\right)=208\)

\(\Rightarrow\left(a+2\right)\left(a+4-a\right)=208\)

\(\Rightarrow\left(a+2\right).4=208\)

\(\Rightarrow a+2=208:4=52\)

\(\Rightarrow a=50\)

Vậy các số tiếp theo là 52 ;54

Vậy....................

a) (2010 - x)(x + 2021) = 0

<=> \(\left[{}\begin{matrix}2010-x=0\\x+2021=0\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=2010\\x=-2021\end{matrix}\right.\)

b) (x - 2)^2 + 5(x - 2) = 0

<=> x^2 - 4x + 4 + 5x - 10 = 0

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

<=> x^2 - 2x + 3x - 6 = 0

<=> x(x - 2) + 3(x - 2) = 0

<=> (x + 3)(x - 2) = 0

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

Câu c hình như sai đề hay sao ý?

Tìm x

(7x - 1)³ = (-3)² . 15 + 208

\(<=> (7x-1)^3 = 135 +208\)

\(<=> (7x-1)^3 = 343\)

\(<=> \)\((7x-1)^3 = 7^3\)

\(<=> 7x-1= 7\)

\(<=> 7x=8\)

\(<=> x=\dfrac{7}{8}\)

Vậy \( x=\dfrac{7}{8} \)

a)

 \(\left(2x-15\right)^5=\left(2x-15\right)^3\\ \Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\\ \Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15-1\right).\left(2d-15+1\right)=0\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right. \)

b) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)

a) Đề thiếu nhé. sửa đề:

 \(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{195}\)

\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)

\(=1-\frac{1}{15}\)

\(=\frac{14}{15}\)

mình sửa lại nha, nãy mình nhìn nhầm dấu:

343x³ - 1617x² + 2541x - 1674 = 0

x = 18/7 là nghiệm duy nhất của phương trình đã cho

bài 5:

1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)

2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)

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

3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)

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

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

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

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

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

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

5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)

\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)

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

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

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

Bài 3:

1: \(9x^3-xy^2\)

\(=x\cdot9x^2-x\cdot y^2\)

\(=x\left(9x^2-y^2\right)\)

\(=x\left(3x-y\right)\left(3x+y\right)\)

2: \(x^2-3xy-6x+18y\)

\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)

\(=x\left(x-3y\right)-6\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x-6\right)\)

3: \(x^2-3xy-6x+18y\)

\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)

\(=x\left(x-3y\right)-6\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x-6\right)\)

4: \(6xy-x^2+36-9y^2\)

\(=36-\left(x^2-6xy+9y^2\right)\)

\(=36-\left(x-3y\right)^2\)

\(=\left(6-x+3y\right)\left(6+x-3y\right)\)

5: \(x^4-6x^2+5\)

\(=x^4-x^2-5x^2+5\)

\(=x^2\left(x^2-1\right)-5\left(x^2-1\right)\)

\(=\left(x^2-5\right)\left(x^2-1\right)\)

\(=\left(x^2-5\right)\left(x-1\right)\left(x+1\right)\)

6: \(9x^2-6x-y^2+2y\)

\(=\left(9x^2-y^2\right)-\left(6x-2y\right)\)

\(=\left(3x-y\right)\left(3x+y\right)-2\left(3x-y\right)\)

\(=\left(3x-y\right)\left(3x+y-2\right)\)

bài này dễ mà

bai nay hay nhung de

1) \(x=\frac{8}{-3}-\frac{1}{2}=\frac{-19}{6}\)

2) \(\left(\frac{-5}{3}\right)^3=\frac{-125}{27}\)

     \(\frac{-24}{35}.\frac{-5}{6}=\frac{4}{7}\)

Mà \(\frac{-125}{27}<0<\frac{4}{7}\)

=> \(x=0\)

3) \(\left(7.x-11\right)^3=343\)

=> \(\left(7.x-11\right)^3=7^3\)

=> \(7.x-11=7\)

=> \(x=\frac{7+11}{7}=\frac{18}{7}\)