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a)\(\frac{x+11}{x-6}=\frac{x-6+17}{x-6}=\frac{x-6}{x-6}+\frac{17}{x-6}\)
=>x-6\(\in\) Ư(17)
| x-6 | 1 | -1 | 17 | -17 |
| x | 7 | 5 | 23 | -11 |
\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{\left(2x-2\right)\cdot2x}=\frac{1}{8}\left(x\inℕ;x\ge2\right)\)
Đặt \(A=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{\left(2x-2\right)2x}\)
\(2A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{\left(2x-2\right)2x}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2x-2}-\frac{1}{2x}\)
\(2A=\frac{1}{2}-\frac{1}{2x}=\frac{x-1}{2x}\)
\(\Rightarrow A=\frac{x-1}{2x}:2=\frac{x-1}{2x}\cdot\frac{1}{2}=\frac{x-1}{4x}\)
Mà \(A=\frac{1}{8}\Rightarrow\frac{x-1}{4}=\frac{1}{8}\)
\(\Leftrightarrow8x-8=4\)
\(\Leftrightarrow8x=12\)
\(\Leftrightarrow x=\frac{12}{8}=\frac{3}{2}\left(ktm\right)\)
Vậy không có x thỏa mãn yêu cầu đề bài
a. vì x+3 chia hết cho(chc) x+3 => 5(x+3) chc x+3 => 5x+15 chc x+3 (1)
ta có 12+5x= 5x+12 (2)
từ (1) và (2) => (5x+15)-(5x+12) chc x+3
=> (5x+15-5x-12) chc x+3
=> 3 chc x+3
=> x+3 thuộc Ư(3)= {1; -1; 3; -3}
bảng xét dấu:
| x+3 | 1 | -1 | 3 | -3 |
| x | -2 | -4 | 0 | -6 |
vậy x thuộc {-2;-4;0;-6} để 12+5x chc x+3
các câu sau làm tương tự nhé :)))))
Số số hạng là :
(2x - 2) : 2 + 1 = x - 1 + 1 = x (số)
Tổng là :
(2x + 2).x : 2 = 210
=> (2x2 + 2x) : 2 = 210
=> x2 + x = 210
=> x(x + 1) = 210
=> x(x + 1) = 20.21
=> x = 20
Vậy x = 20
Ta có : \(\frac{x}{2}=\frac{10}{x+1}\)
=> x(x + 1) = 10.2
=> x(x + 1) = 20
=> sai đề
\(\frac{x+2}{2x+1}=\frac{x}{2x-1}\) (ĐK:\(x\ne\pm\frac{1}{2}\))
\(\Leftrightarrow\)\(\frac{\left(x+2\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\frac{x\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\)
\(\Leftrightarrow\)\(\left(x+2\right)\left(2x-1\right)=x\left(2x+1\right)\)
\(\Leftrightarrow2x^2-x+4x-2=2x^2+x\)
\(\Leftrightarrow2x^2-x+4x-2-2x^2-x=0\)
\(\Leftrightarrow2x-2=0\)
\(\Leftrightarrow2\left(x-1\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\left(TM\right)\)
Vậy x=1
\(\frac{x+2}{2x+1}=\frac{x}{2x-1}\)
Ta có:(x+2)(2x-1)=x.(2x+1)
2x2+3x-2=2x2+x
Sau đó tự làm
f) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)( ĐKXĐ : \(x\ne-\frac{1}{2}\))
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=21\cdot3\)
\(\Leftrightarrow4x^2-1=63\)
\(\Leftrightarrow4x^2=64\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x^2=\left(\pm4\right)^2\)
\(\Leftrightarrow x=\pm4\)(tmđk)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\)( ĐKXĐ : \(x\ne-1\))
\(\Leftrightarrow\left(10x+5\right)\left(x+1\right)=6\cdot5\)
\(\Leftrightarrow10x^2+15x+5=30\)
\(\Leftrightarrow10x^2+15x+5-30=0\)
\(\Leftrightarrow10x^2+15x-25=0\)
\(\Leftrightarrow5\left(2x^2+3x-5\right)=0\)
\(\Leftrightarrow2x^2+3x-5=0\)
\(\Leftrightarrow2x^2-2x+5x-5=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)(tmđk)
f) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=21.3\)
\(\Leftrightarrow4x^2-1=63\)
\(\Leftrightarrow4x^2=64\)
\(\Leftrightarrow x^2=16\)\(\Leftrightarrow x^2=4^2\)\(\Leftrightarrow x=4\)
Vậy \(x=4\)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\)
\(\Leftrightarrow\left(10x+5\right)\left(x+1\right)=5.6\)
\(\Leftrightarrow5\left(2x+1\right)\left(x+1\right)=30\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=6\)
\(\Leftrightarrow2x^2+3x+1=6\)
\(\Leftrightarrow2x^2+3x-5=0\)
\(\Leftrightarrow\left(2x^2-2x\right)+\left(5x-5\right)=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\2x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-5}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{-5}{2};1\right\}\)
Bài 1 :
1. a, 5\(^{2x-3}\)-2.5\(^2\)=5\(^2\).3
5\(^{2x}\) : 5\(^3\) -2.25 = 25.3
5\(^{2x}\): 5\(^3\) - 50 = 75
5\(^{2x}\): 5\(^3\) = 75+50
5\(^{2x}\): 5\(^3\) = 125
5\(^{2x}\) = 125.5\(^3\)
5\(^{2x}\) = 5\(^3\). 5\(^3\)
5 \(^{2x}\) = 5\(^{3+3}\)
5 \(^{2x}\) = 5\(^6\)
Có 5=5 => 2x = 6
x = 6 : 2
x = 3
Vậy x = 3.
b. / 2x -1 / = 5
=> 2x-1 = 5 hoặc 2x-1 = -5
* Với 2x - 1 = 5
thì 2x = 5+1
2x = 6
x = 6:2
x = 3
* Với 2x - 1 = - 5
thì 2x = -5 + 1
2x = -4
x = -4 : 2
x = -2