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làm bài & thôi :
(x2 - 2x + 3) \(⋮\)(x - 1)
= x2 - 2x + 3
=) x2 - 2x + 3 - ( x - 1 )
=) x2 - 1
=) x2 - 1 - x( x - 1 )
=) 2 \(⋮\)x - 1
tự làm
a) Ta có: (x2 - 2x + 3) \(⋮\)(x - 1)
<=> [x(x - 1) - (x - 1) + 2] \(⋮\)(x - 1)
<=> [(x - 1)2 + 2] \(⋮\)(x - 1)
Do (x - 1)2 \(⋮\)(x - 1) => 2 \(⋮\)(x - 1)
=> (x - 1) \(\in\)Ư(2) = {1; -1; 2; -2}
Lập bảng :
| x - 1 | 1 | -1 | 2 | -2 |
| x | 2 | 0 | 3 | -1 |
Vậy ...
b) (3x - 1) \(⋮\)(x - 4)
<=> [3(x - 4) + 11] \(⋮\)(x - 4)
Do 3(x - 4) \(⋮\)(x - 4) => 11 \(⋮\)(x - 4)
=> (x - 4) \(\in\)Ư(11) = {1; -1; 11; -11}
Lập bảng:
| x - 4 | 1 | -1 | 11 | -11 |
| x | 5 | 3 | 15 | -7 |
vậy ...
c;d tương tự trên
Do \(\frac{3+x}{7+y}=\frac{3}{7}\)=> x=3p, y=7q (p, q\(\in\)Z)
Ta có: x+y=3p+7q=20 hay 3(p+q)+4q=20 => 0<p+q<6
Do 20\(⋮\)4, 4q\(⋮\)4 => 3(p+q)\(⋮\)4 mà (3,4)=1 => p+q\(⋮\)4.
=> p+q=4 => q=(20-3.4):4=2 => y=2.7=14
=> p=4-2=2 => x=2.3=6
=>\(\frac{3+x}{7+y}=\)một phân số có thể rút gọn thành\(\frac{3}{7}\)
Giả sử x=3; y=7. Vì \(\frac{3+3}{7+7}=\frac{6}{14}=\frac{3}{7}\)Nhưng 3+7=10 (loại)
x=6; y=14. Vì\(\frac{3+6}{7+14}=\frac{9}{21}=\frac{3}{7}\)Và 6+14=20 (thỏa mãn)
Vậy x=6; y=14
1.Vì \(\frac{x}{-2}=\frac{-8}{x}\Rightarrow-2.\left(-8\right)=x.x\)
\(16=x.x\)hay \(4^2=x^2\Rightarrow x=4\)
2. Rút gọn : \(\frac{20}{28}=\frac{5}{7}=\frac{-5}{-7}\)
\(\Rightarrow x=-7\)
3. \(\frac{x}{2}-\frac{11}{5}=\frac{7}{8}\times\frac{64}{49}\)
\(\frac{x}{2}-\frac{11}{5}=\frac{8}{7}\)
Mà \(\frac{8}{7}+\frac{11}{5}=\frac{502}{35}\)
\(\Rightarrow x=\frac{234}{35}\)
1) \(\frac{x}{-2}=\frac{-8}{x}\)
\(\Rightarrow x\times x=\left(-2\right)\times\left(-8\right)\)
\(\Rightarrow x^2=16\)
\(\Rightarrow\orbr{\begin{cases}x^2=4^2\\x^2=\left(-4\right)^2\end{cases}\Rightarrow}\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
Vậy x = 4 hoặc x = -4
2) \(\frac{-5}{x}=\frac{20}{28}\)
\(\Rightarrow\frac{-5}{x}=\frac{5}{7}\)
\(\Rightarrow5\times x=\left(-5\right)\times7\)
\(\Rightarrow5\times x=-35\)
\(\Rightarrow x=\left(-35\right):5\)
\(\Rightarrow x=-7\)
Vậy x = -7
3) \(\frac{x}{2}-\frac{11}{5}=\frac{7}{8}\times\frac{64}{49}\)
\(\Rightarrow\frac{x}{2}-\frac{11}{5}=\frac{8}{7}\)
\(\Rightarrow\frac{x}{2}=\frac{8}{7}+\frac{11}{5}\)
\(\Rightarrow\frac{x}{2}=\frac{117}{35}\)
\(\Rightarrow35x=117\times2\)
\(\Rightarrow35x=234\)
\(\Rightarrow x=234:35\)
\(\Rightarrow x=\frac{234}{35}\)
Vậy \(x=\frac{234}{35}\)
4) \(\frac{x}{5}+\frac{9}{2}=\frac{6}{7}\times\frac{36}{48}\)
\(\Rightarrow\frac{x}{5}+\frac{9}{2}=\frac{9}{14}\)
\(\Rightarrow\frac{x}{5}=\frac{9}{14}-\frac{9}{2}\)
\(\Rightarrow\frac{x}{5}=\frac{-27}{7}\)
\(\Rightarrow7x=\left(-27\right)\times5\)
\(\Rightarrow7x=-135\)
\(\Rightarrow x=\left(-135\right):7\)
\(\Rightarrow x=\frac{-135}{7}\)
Vậy \(x=\frac{-135}{7}\)
5) \(\frac{3}{x-5}=\frac{-4}{x+2}\)
\(\Rightarrow\frac{3}{x-5}+\frac{4}{x+2}=0\)
\(\Rightarrow3\left(x+2\right)+4\left(x-5\right)=0\)
\(\Rightarrow3x+6+4x-20=0\)
\(\Rightarrow\left(3x+4x\right)+\left(6-20\right)=0\)
\(\Rightarrow7x-14=0\)
\(\Rightarrow7x=14\)
\(\Rightarrow x=14:7\)
\(\Rightarrow x=2\)
Vậy x = 2
_Chúc bạn học tốt_
\(\frac{1}{2}\cdot2^x+2^x\cdot2^2=2^8+2^5\)
\(2^x\left(\frac{1}{2}+4\right)=2^8+2^5\)
\(2^x\cdot\frac{9}{2}=288\)
\(2^x=64\)
\(2^x=2^6\)
\(x=6\)
\(9^x:3^x=3^7\)
\(3^{2x}:3^x=3^7\)
\(3^x=3^7\)
\(x=7\)
\(7^{x+2}+2\cdot7^{x-1}=345\)
\(7^x\cdot7^2+2\cdot7^x:7=345\)
\(7^x\left(7^2+\frac{2}{7}\right)=345\)
\(7^x\cdot\frac{345}{7}=345\)
\(7^x=7\)
\(x=1\)
a) 1/2.2^x + 2^x+2 = 256 + 32
1/2.2^x + 2^2.2^x=288
2^x(1/2+4)= 288
2^x.4,5=288
2^x= 288:4,5
2^x=64=2^6
x=6
Đặt \(A=\frac{x^2+2x-1}{x-1}\)
Ta có:\(A=\frac{x^2+2x-1}{x-1}=\frac{\left(x-1\right)^2}{x-1}=x-1\)
Vậy để A nguyên thì x thỏa mãn mõi số nguyên
a,\(\left(x-3\right).\left(2y+1\right)=7\)
Vì \(x;y\inℤ=>x-3;2y+1\inℤ\)
\(=>x-3;2y+1\inƯ\left(7\right)\)
Nên ta có bảng sau
| x-3 | 1 | 7 | -7 | -1 |
| 2y+1 | 7 | 1 | -1 | -7 |
| x | 4 | 10 | -4 | 2 |
| y | 3 | 0 | -1 | -4 |
Vậy ...
b,\(A=-126-\left(4^2-5\right)^2+870:29\)
\(=-126-\left(16-5\right)^2+30\)
\(=-126-11^2+30\)
\(=-247+30=-217\)
=>7 chia hết cho x-3
=>\(x-3\in\left\{1;-1;7;-7\right\}\)
=>\(x\in\left\{4;2;10;-4\right\}\)