Question

[ @Phan Đức Gia Linh _ Xin cảm ơn những bạn đã quan tâm tới câu hỏi của mình! ]

Tìm x, biết:

1) x\(^3\) - \(\dfrac{4}{25}\)x = 0

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

3) 3 . \(\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{97.100}\right)\) = \(\dfrac{319}{100}\)

4) (6 - x) \(^{2014}\) = (6 - x) \(^{2015}\)

5) 2 + 4 + 6 + ... + 2x = 210

6) \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\) + ... + \(\dfrac{2}{x.\left(x+1\right)}\) = \(\dfrac{2}{9}\)

Các bạn cố gắng giải đầy đủ giúp mình!

Answer 1

1. x³ - \(\dfrac{4}{25}\)x = 0
<=> x(x² - \(\dfrac{4}{25}\)) = 0
<=> \(\left[{}\begin{matrix}x=0\\x^2-\dfrac{4}{25}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{4}{25}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\end{matrix}\right.\) (thỏa mãn)
Vậy x = 0; 2/5
@Phan Đức Gia Linh

Answer 2

1 ) \(x^3-\dfrac{4}{25}x=0\)

\(\Leftrightarrow x\left(x^2-\dfrac{4}{25}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{4}{25}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x-\dfrac{2}{5}=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{2}{5}\end{matrix}\right.\)

Vậy .............

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

\(\Leftrightarrow3^{4x+4}=\left(3^2\right)^{x+2}\)

\(\Leftrightarrow4x+4=2x+4\)

\(\Leftrightarrow2x=0\Leftrightarrow x=0.\)

3 ) \(3\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{97.100}\right)=\dfrac{319}{100}\) ( thiếu đề hay sao )

4 ) \(\left(6-x\right)^{2014}=\left(6-x\right)^{2015}\)

\(\Leftrightarrow\left(6-x\right)^{2014}-\left(6-x\right)^{2015}=0\)

\(\Leftrightarrow\left(6-x\right)^{2014}\left(1-6+x\right)=0\)

\(\Leftrightarrow\left(6-x\right)^{2014}\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(6-x\right)^{2014}=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=5\end{matrix}\right.\)

Vậy ......

5) \(2+4+6+...+2x=210\)

\(\Leftrightarrow2.1+2.2+2.3+...+2.x=210\)

\(\Leftrightarrow2\left(1+2+3+...+x\right)=210\)

\(\Leftrightarrow1+2+3+...+x=105\)

\(\Leftrightarrow\dfrac{\left(x+1\right).x}{2}=105\)

\(\Leftrightarrow x\left(x+1\right)=210\)

Ta lại có : \(x\left(x+1\right)=14\left(14+1\right)\)

\(\Leftrightarrow x=14\)

Vậy ......

6 ) \(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+..+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.7}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{2}{2.3.7}+\dfrac{2}{2.4.7}+\dfrac{2}{2.4.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{2}{6.7}+\dfrac{2}{8.7}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow2\left(\dfrac{1}{6.7}+\dfrac{1}{8.7}+\dfrac{1}{8.9}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\)

\(\Leftrightarrow2.\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{\dfrac{x-1}{x+1}}\right)=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{x+1}=\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)

\(\Leftrightarrow x=17.\)

Vậy ...........

\(\)

Answer 3

2. 3^4x+4 = 9^x+2
<=> 3^2(2x+2) = 9^x+2
<=> (3²)^2x+2 = 9^x+2
<=> 9^2x+2 = 9^x+2
<=> 2x + 2 = x + 2
<=> 2x - x = 2 - 2
<=> x = 0
@Phan Đức Gia Linh

Answer 4

4. (6 - x)²⁰¹⁴ = (6 - x)²⁰¹⁵
<=> \(\left[{}\begin{matrix}6-x=0\\6-x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=5\end{matrix}\right.\) (thỏa mãn)
Vậy x = 6; 5
@Phan Đức Gia Linh

Answer 5

5. 2 + 4 + 6 + ... + 2x = 210
<=> 2(1 + 2 + 3 + ... + x) = 210
<=> 1 + 2 + 3 + ... + x = 105
<=> (1 + x)[(x - 1) : 1 + 1] : 2 = 105
<=> (1 + x)x = 210
Mà x; x + 1 là 2 số tự nhiên liên tiếp => x = 14
@Phan Đức Gia Linh

Answer 6

\(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{1}{72}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow2\left(\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\)

\(\Leftrightarrow2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)

<=> x + 1 = 18
<=> x = 17
@Phan Đức Gia Linh

Answer 7

Câu 3) trong ngoặc đơn đằng sau cái \(\dfrac{1}{97.100}\) thì cộng x