\(1\text{)} \left(x-3\right)-9.\left(2x-7\right)=16:2\\ x-3-18x+63=8\\ -17x=-52\\ x=\dfrac{52}{17}\)

\(2\text{)} x-\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\\ -\dfrac{1}{6}x=\dfrac{5}{12}\\ x=-2,5\)

\(3\text{)} -2.\left(x-1\right)+5.\left(x-2\right)=x-14\\ -2x+2+5x-10=x-14\\ 2x=-6\\ x=-3\)

\(4\text{)} x.\left(x+1\right)-x-1=0\\ \left(x+1\right)\left(x-1\right)=0\\ x^2-1=0\\ x^2=1\\ x=\pm1\)

1) \(\left|5x-4\right|=\left|x+2\right|\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(x=\dfrac{1}{3}\) hoặc \(x=\dfrac{3}{2}\)

2) \(\left|x+15\right|=\left|3x-4\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}x+15=3x-4\\x+15=-\left(3x-4\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+15=3x-4\\x+15=-3x+4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3x=-4-15\\x+3x=4-15\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=-19\\4x=-11\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{2}\\x=-\dfrac{11}{4}\end{matrix}\right.\)

Vậy \(x=-\dfrac{11}{4}\) hoặc \(x=\dfrac{19}{2}\)

3) \(\left|\dfrac{5}{4}x-\dfrac{7}{2}\right|-\left|\dfrac{5}{8}x+\dfrac{3}{5}\right|=0\)

\(\Leftrightarrow\left|\dfrac{5}{4}x-\dfrac{7}{2}\right|=\left|\dfrac{5}{8}x+\dfrac{3}{5}\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{7}{2}=\dfrac{5}{8}x+\dfrac{3}{5}\\\dfrac{5}{4}x-\dfrac{7}{2}=-\left(\dfrac{5}{8}x+\dfrac{3}{5}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{7}{2}=\dfrac{5}{8}x+\dfrac{3}{5}\\\dfrac{5}{4}x-\dfrac{7}{2}=-\dfrac{5}{8}x-\dfrac{3}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{5}{8}x=\dfrac{3}{5}+\dfrac{7}{2}\\\dfrac{5}{4}x+\dfrac{5}{8}x=-\dfrac{3}{5}+\dfrac{7}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{8}x=\dfrac{41}{10}\\\dfrac{15}{8}x=\dfrac{29}{10}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{164}{25}\\x=\dfrac{116}{75}\end{matrix}\right.\)

Vậy \(x=\dfrac{116}{75}\) hoặc \(x=\dfrac{164}{25}\)

4) \(\left|2x-6\right|-\left|x+3\right|=0\)

\(\Leftrightarrow\left|2x-6\right|=\left|x+3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=x+3\\2x-6=-\left(x+3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=x+3\\2x-6=-x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3+6\\2x+x=-3+6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\3x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)

Vậy \(x=1\) hoặc \(x=9\)

@@ nhiều z @@

Xàm như đọc bản tuyên ngôn không bằng

Ace Legona, Hoàng Thị Ngọc Anh, ... giúp mình câu này với!

Đoàn Đức Hiếu, ... giúp với!

1. |x| - x = 0
<=> |x| = x
<=> \(\left[{}\begin{matrix}x=x\\x=-x\end{matrix}\right.\) (thỏa mãn)
@Phan Đức Gia Linh

2. |x| + x = 0
<=> |x| = -x
Do |x| \(\ge\) 0, mà -x < 0 => không tồn tại x thỏa mãn
@Phan Đức Gia Linh

3. |x - 1| - x = - 1
<=> |x - 1| = -1 + x
Do |x - 1| \(\ge\) 0 => -1 + x \(\ge\) 0 => x \(\ge\) 1
Vậy x \(\ge\) 1 thì |x - 1| - x = -1
@Phan Đức Gia Linh

1) \(\dfrac{2}{x+1}=\dfrac{x+1}{8}\Leftrightarrow\left(x+1\right)\left(x+1\right)=2.8\Leftrightarrow\left(x+1\right)^2=16\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\) vậy \(x=3;x=-5\)

2) thiếu quế phải nha

3) \(\dfrac{x-4}{x-7}=\left(\dfrac{-3}{5}\right)^2\Leftrightarrow\dfrac{x-4}{x-7}=\dfrac{9}{25}\Leftrightarrow9.\left(x-7\right)=25.\left(x-4\right)\)

\(\Leftrightarrow9x-63=25x-100\Leftrightarrow25x-9x=-63+100\)

\(\Leftrightarrow16x=37\Leftrightarrow x=\dfrac{37}{16}\) vậy \(x=\dfrac{37}{16}\)

4) ta có : \(x+y=20\Leftrightarrow y=20-x\)

\(\dfrac{3+x}{7+y}=\dfrac{3}{7}\Leftrightarrow7\left(3+x\right)=3\left(7+y\right)\Leftrightarrow21+7x=21+3y\)

\(\Leftrightarrow7x=3y\Leftrightarrow7x-3y=0\Leftrightarrow7x-3\left(20-x\right)=0\)

\(\Leftrightarrow7x-60+3x=0\Leftrightarrow10x=60\Leftrightarrow x=6\)

\(\Rightarrow6+y=20\Leftrightarrow y=14\) vậy \(x=6;y=14\)

a.\(\dfrac{2}{x+1}=\dfrac{x+1}{8}\)

\(\Rightarrow\left(x+1\right)\left(x+1\right)=2.8\)

\(\Rightarrow\left(x+1\right)^2=16\)

\(\Rightarrow\left(x+1\right)^2=(\pm4)^2\)

\(\rightarrow x+1=\pm4\)

\(+)x+1=-4\)

\(x=\left(-4\right)-1\)

\(x=-5\)

\(+)x+1=4\)

\(x=4-1\)

\(x=3\)

Vậy \(x\in\left\{-5;3\right\}\)

Câu còn lại tương tự nhé. Trừ câu 2 bạn xem lại nhé

Câu 2) \(\dfrac{23+x}{40-x}\) = \(\dfrac{-3}{4}\)

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

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 ...........

\(\)

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

\(\left|x-2\right|+\left|2y-5\right|=0\)

\(\left\{{}\begin{matrix}\left|x-2\right|\ge0\forall x\\\left|2y-5\right|\ge0\forall y\end{matrix}\right.\)

\(\left|x-2\right|+\left|2y-5\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x-2\right|=0\Rightarrow x=2\\\left|2y-5\right|=0\Rightarrow2y=5\Rightarrow y=\dfrac{5}{2}\end{matrix}\right.\)

\(\left|3y-2\right|+\left|xy-6\right|=0\)

\(\left\{{}\begin{matrix} \left|3y-2\right|\ge0\forall y\\\left|xy-6\right|\ge0\forall x;y\end{matrix}\right.\)

\(\Rightarrow\left|3y-2\right|+\left|xy-6\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|3y-2\right|=0\Rightarrow3y=2\Rightarrow y=\dfrac{3}{2}\\\left|xy-6\right|=0\Rightarrow\dfrac{3}{2}x=6\Rightarrow x=4\end{matrix}\right.\)

\(\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|\le0\)

\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|\ge0\forall x\\\left|2y-\dfrac{1}{3}\right|\ge0\forall y\\ \left|4z-5\right|\ge0\forall z\end{matrix}\right.\)

\(\Rightarrow\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|\ge0\\\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|\le0\end{matrix}\right.\)

\(\Rightarrow\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|=0\Rightarrow x=\dfrac{1}{2}\\\left|2y-\dfrac{1}{3}\right|=0\Rightarrow2y=\dfrac{1}{3}\Rightarrow y=\dfrac{1}{6}\\\left|4z-5\right|=0\Rightarrow4z=5\Rightarrow z=\dfrac{5}{4}\end{matrix}\right.\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2\cdot2}+\dfrac{1}{2\cdot2}-\dfrac{1}{2\cdot2\cdot2}+\dfrac{1}{2\cdot2\cdot2}-\dfrac{1}{2\cdot2\cdot2\cdot2}+.....+\dfrac{1}{2^{10}}\)

\(A=1-\dfrac{1}{2^{10}}\)

\(A+\dfrac{1}{2^{10}}=1-\dfrac{1}{2^{10}}+\dfrac{1}{2^{10}}=1\left(dpcm\right)\)

\(A=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{10^2}\)

Vì \(\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...\dfrac{1}{10^2}< \dfrac{1}{9.10}\)

\(A< \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)

Do đó \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\Rightarrow A< \dfrac{1}{2}-\dfrac{1}{10}\Rightarrow A< \dfrac{1}{2}\)

Vậy \(A=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{10^2}< \dfrac{1}{2}\)

`A = 1/3^2 + 1/4^2 + ... + 1/10^2`

Ta có:

`1/3^2 < 1/(2.3)`

`1/(4^2) < 1/(3.4)`

`...`

`1/(10^2) < 1/(9.10)`

`=> A < 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10 = 1/2 - 1/10 < 1/2`.