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a) \(7⋮x+1\Rightarrow x+1\inƯ\left(7\right)\)
\(\Rightarrow\)X+1 \(\in\)\(\left\{\pm1;\pm7\right\}\)\(\Rightarrow x\in\left\{0;-2;6;-8\right\}\)
các câu b và c làm tương tự
a) \(\Rightarrow x+1\inƯ\left(7\right)\)
Mà Ư(7) = \([\)\(\pm1;\pm7\)\(]\)
Ta có bảng
| x+1 | x | kết luận |
| 1 | 0 | thoã mãn |
| -1 | -2 | thỏa mãn |
| 7 | 6 | thỏa mãn |
| -7 | -8 | thỏa mãn |
Ta có :
\(VT=\frac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}=\frac{\left(y-x\right)\left(y+x\right)}{\left(x-y\right)^3}\)
\(VT=\frac{-\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^3}=\frac{-\left(x+y\right)}{\left(x-y\right)^2}=\frac{-x-y}{\left(x-y\right)^2}=VP\)
Vậy .......................
Áp dụng định lý Bezout ta có:
\(A\left(x\right)⋮\left(x+1\right)\Rightarrow A\left(-1\right)=0\)
\(\Leftrightarrow a^2\left(-1\right)^3+3a\left(-1\right)^2-6.\left(-1\right)-2a=0\)
\(\Leftrightarrow-a+3a+6+2a=0\)
\(\Leftrightarrow4a+6=0\)
\(\Leftrightarrow a=\frac{-3}{2}\)
Vậy \(a=\frac{-3}{2}\)để \(A\left(x\right)⋮\left(x+1\right)\)
ĐKXĐ: \(x\ne1;x\ne-1\)
\(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}\) \(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}-\dfrac{4x}{2\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{x^2+2x+1-4x}{2\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{x^2-2x+1}{2\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{x-1}{2\left(x+1\right)}\)
a)khong ton tai x thoa man (x+1)^2=-25
b) x^5=x^3=> x=0 hoac x=1 hoac x=-1
nho k de ung ho mik nhe
a) -12(x - 5) + 7(3 - x) = 5
=> -12x + 60 + 21 - 7x = 5
=> -19x + 81 = 5
=> -19x = 5 - 81
=> -19x = -76
=> x = 4
b) x + {(x + 3) - [(x + 3) - (-x - 2)]} = x
=> (x + 3) - (x + 3 + x + 2) = 0
=> x + 3 - x - 3 - x - 2 = 0
=> -x - 2 = 0
=> -x = 2
=> x = -2
a. -12. ( x- 5 ) + 7. ( 3 - x ) = 5
= 12x + 61 + 21 - 7x = 5
= 12x - 7x = 5 - 61 - 21
= -19x = -76
= x = 76 : ( -19)
= x = 4
Đề: \(\left(x^2-1\right)^{2018}+\)\(\left|y+1\right|=0\)
Bài giải:
Dễ thấy:\(\left(x^2-1\right)^{2018}\ge0\)và \(\left|y+1\right|\ge0\)
Mà tổng của chúng lại bằng 0 \(\Rightarrow\hept{\begin{cases}\left|y+1\right|=0\Rightarrow y+1=0\Rightarrow y=-1\\\left(x^2-1\right)^{2018}=0\Rightarrow x^2-1=0\Rightarrow x^2=1\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\end{cases}}\)
+) \(\left(x^2-1\right)^{2018}=0\)
\(\Rightarrow x^2-1=0\Rightarrow x^2=1\)
\(\Rightarrow x=\pm1\)
+) \(\left|y+1\right|=0\)
\(\Rightarrow y+1=0\)
\(\Rightarrow y=-1\)
Vậy cặp số (x,y) cần tìm là (-1; -1) hoặc (1;-1)
\(\left(2x+1\right)\left(2x+3\right)\left(x+1\right)^2-18=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)\left(4x^2+8x+3\right)-18=0\)
Đặt \(x^2+2x+1=a\ge0\)
\(\Rightarrow a\left(4a-1\right)-18=0\)
\(\Leftrightarrow4a^2-a-18=0\)
\(\Leftrightarrow\left(4a^2+8a\right)+\left(-9a-18\right)=0\)
\(\Leftrightarrow\left(a+2\right)\left(4a-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-2\left(l\right)\\a=\frac{9}{4}\end{cases}}\)
\(\Rightarrow x^2+2x+1=\frac{9}{4}\)
\(\Leftrightarrow4x^2+8x-5=0\)
\(\Leftrightarrow\left(4x^2-2x\right)+\left(10x-5\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{5}{2}\end{cases}}\)
a, (x+1)+(x+2)+(x+3)+......+(x+2000) = 2005000
(x+x+....+x) + (1 + 2 + 3 + ....+2000) = 2005000
x.2000 + (1+2000)x2000:2 = 2005000
2000x + 2001000 = 2005000
2000x = 20005000 - 2001000
2000x = 4000
x = 4000:2000
x = 2
b, 1 + 2 + 3 +....+ x = 210
(x+1)x:2 = 210
(x+1)x = 210.2
(x+1)x = 420
(x+1)x = 21.20
x = 20
c, x + 2x+ 3x + 4x+.....+1000x = 1001000
(x+1000x)1000:2 = 1001000
1001x .1000 = 2.1001000
x = 2.1001000: 1000:1001
X = 2
a)
x+(x+1)+(x+2)+....+(x+30)=1240
31 . x + (1 + 2 + 3 + 4 +...+ 29 + 30) = 1240
31 . x + 31.15 = 1240
31 . x = 1240 - 31.15
31 . x = 775
x = 775 : 31
x = 25
b)
1 + 2 + 3 +...+ x = 210
(x + 1) . x : 2=210
x . (x + 1) = 420
x . x + x = 420
x . x = 420 - x
Ta thấy:
20 . 20 = 420 - 20
Vậy x = 20