K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
13 tháng 1 2018
a. ( x + 3 )( x - 3 ) = 16
⇔x2-9=16
⇔x2-16-9=0
⇔x2-25=0
⇔(x-5)(x+5)=0
⇔\(\left[{}\begin{matrix}x+5=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=5\end{matrix}\right.\)
20 tháng 1 2016
1/
-x^3 -5x^2 + 4x +4
=> x1 =-5.5877............
x2=1.1895.............
x3=-0.6018............
ST
26 tháng 11 2017
a) \(\dfrac{x^2-1}{120}+\dfrac{x^2-2}{119}+\dfrac{x^2-3}{118}=3\)
\(=\dfrac{x^2-1}{120}-1+\dfrac{x^2-2}{119}-1+\dfrac{x^2-3}{118}-1=0\)\(=\dfrac{x^2-121}{120}+\dfrac{x^2-121}{119}+\dfrac{x^2-121}{118}=0\)
\(=\left(x^2-121\right).\left(\dfrac{1}{120}+\dfrac{1}{119}+\dfrac{1}{118}\right)=0\)
\(=\left(x+11\right)\left(x-11\right)\left(\dfrac{1}{120}+\dfrac{1}{119}+\dfrac{1}{118}\right)=0\)
⇒\(\left[{}\begin{matrix}x+11=0\\x-11=0\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}x=-11\\x=11\end{matrix}\right.\)
10 tháng 1 2018
1 ) \(\left(x-4\right)^2-25=0\)
\(\Leftrightarrow\left(x-4-5\right)\left(x-4+5\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
2 ) \(\left(x-3\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-3+x-1\right)\left(x-3-x+1\right)=0\)
\(\Leftrightarrow-2\left(2x-4\right)=0\)
\(\Leftrightarrow x=2.\)
3 ) \(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x+3-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=-4\end{matrix}\right.\)
4 ) \(\left(x^2-1\right)-\left(x+1\right)\left(2-3x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1-2+3x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{3}{4}\end{matrix}\right.\)
5 ) \(x^3+x^2+x+1=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=-1.\end{matrix}\right.\)
6 ) \(x^3+x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
7 ) \(2x^3+3x^2+6x+5=0\)
\(\Leftrightarrow2x^3+2x^2+x^2+x+5x+5=0\)
\(\Leftrightarrow2x^2\left(x+1\right)+x\left(x+1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow x=-1.\)
8 ) \(x^4-4x^3-19x^2+106x-120=0\)
\(\Leftrightarrow x^4-4x^3-19x^2+76x+30x-120=0\)
\(\Leftrightarrow x^3\left(x-4\right)-19x\left(x-4\right)+30\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-19x+30\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-8-19x+38\right)\left(x-4\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+23\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
9 ) \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+7\right)\left(x+8\right)+8=0\)
\(\Leftrightarrow\left(x^2+7x-x-7\right)\left(x^2+8x-2x-16\right)+8=0\)
\(\Leftrightarrow\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8=0\)
Đặt \(x^2+6x-7=t\)
\(\Leftrightarrow t\left(t-9\right)+8=0\)
\(\Leftrightarrow t^2-9t+8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=8\\t=1\end{matrix}\right.\)
Khi t = 8 \(\Leftrightarrow x^2+6x-7=8\Leftrightarrow x^2+6x-15\Leftrightarrow\left[{}\begin{matrix}x=-3+2\sqrt{6}\\x=-3-2\sqrt{6}\end{matrix}\right.\)
Khi t = 1 \(\Leftrightarrow x^2+6x-7=1\Leftrightarrow x^2+6x-8=0\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{17}\\x=-3-\sqrt{17}\end{matrix}\right.\)
Vậy ........
S
11 tháng 12 2018
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-120=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-120\)
Đặt: x2+5x+4=t
Ta có:
\(t\left(t+2\right)-120=t^2+2t-120=t^2+12t-10t-120=t\left(t+12\right)-10\left(t+12\right)\)
\(=\left(t+12\right)\left(t-10\right)=\left(x^2+5x+16\right)\left(x^2+5x-6\right)\)
9 tháng 1 2019
PTĐTTNT?
1.Đặt \(a^2+a=t\)
\(\Rightarrow\left(a^2+a\right)\left(a^2+a+1\right)-2\)
\(=t\left(t+1\right)-2\)
\(=t^2+t-2\)
\(=t^2+2t-\left(t+2\right)\)
\(=t\left(t+2\right)-\left(t+2\right)\)
\(=\left(t+2\right)\left(t-1\right)\)
9 tháng 1 2019
Sửa đề:
\(x^4+2011x^2+2010x+2011\)
\(=\left(x^4-x\right)+2011x^2+2011x+2011\)
\(=x\left(x^3-1\right)+2011\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2011\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2011\right)\)
3. \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-120\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-120\)
Đặt \(x^2+5x+4=t\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-120\)
\(=t\left(t+2\right)-120\)
\(=t^2+2t+1-121\)
\(=\left(t+1\right)^2-11^2\)
\(=\left(t+1-11\right)\left(t+1+11\right)\)
\(=\left(t-10\right)\left(t+12\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+16\right)\)
\(=\left[\left(x^2-x\right)+\left(6x-6\right)\right]\left(x^2+5x+16\right)\)
\(=\left[x.\left(x-1\right)+6\left(x-1\right)\right]\left(x^2+5x+16\right)\)
\(=\left(x-1\right)\left(x+6\right)\left(x^2+5x+16\right)\)
4. \(\left(x^2+x+4\right)^2+8x\left(x^2+x+1\right)+15x^2\)
\(=\left(x^2+x+4\right)^2+2.\left(x^2+x+1\right).4x+\left(4x\right)^2-x^2\)
\(=\left(x^2+x+4+4x\right)^2-x^2\)
\(=\left(x^2+4+5x-x\right)\left(x^2+5x+x+4\right)\)
\(=\left(x^2+4x+4\right)\left(x^2+6x+4\right)\)
\(=\left(x+2\right)^2\left[\left(x^2+2.x.3+3^2\right)-\left(\sqrt{5}\right)^2\right]\)
\(=\left(x+2\right)^2\left[\left(x+3\right)^2-\left(\sqrt{5}\right)^2\right]\)
\(=\left(x+2\right)^2\left(x+3-\sqrt{5}\right)\left(x+3+\sqrt{5}\right)\)
KT
6 tháng 2 2018
c) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)
\(\Leftrightarrow\)\(\left(x^2+6x+5\right)\left(x^2+6x+8\right)-40=0\)
Đặt \(x^2+6x+5=t\) ta có:
\(t\left(t+3\right)-40=0\)
\(\Leftrightarrow\)\(t^2+3t-40=0\)
\(\Leftrightarrow\)\(\left(t-5\right)\left(t+8\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}t-5=0\\t+8=0\end{cases}}\)
Thay trở lại ta có: \(\orbr{\begin{cases}x^2+6x=0\\x^2+6x+13=0\end{cases}}\)
(*) \(x^2+6x=0\)
\(\Leftrightarrow\)\(x\left(x+6\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x+6=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=-6\end{cases}}\)
(*) \(x^2+6x+13=0\)
\(\Leftrightarrow\)\(\left(x+3\right)^2+4=0\) (vô lý)
Vậy......
21 tháng 4 2020
Sorry Ngân Chu, đoạn chia hết cho 120 thì thêm cả chia hết cho 2 nữa, nên nhân vào mới ra 120 nhé!!
21 tháng 4 2020
Bài 1:
a, (n + 3)2 - (n - 1)2
= (n + 3 - n + 1)(n + 3 + n - 1)
= 4(2n - 2)
= 8(n - 1)
Vì 8 \(⋮\) 8 nên 8(n - 1) \(⋮\) 8 với n \(\in\) Z
b, n5 - 5n3 + 4n
= n(n4 - 5n2 + 4)
= n(n4 - n2 - 4n2 + 4)
= n[n2(n2 - 1) - 4(n2 - 1)]
= n(n2 - 1)(n2 - 4)
= n(n - 1)(n + 1)(n - 2)(n + 2)
= (n - 2)(n - 1)n(n + 1)(n + 2)
Vì (n - 2)(n - 1)n(n + 1)(n + 2) là tích của 5 số nguyên liên tiếp nên chia hết cho 3, 5, 8
Mà 3 x 5 x 8 = 120
\(\Rightarrow\) (n - 2)(n - 1)n(n + 1)(n + 2) \(⋮\) 120 hay n5 - 5n3 + 4n \(⋮\) 120 với n \(\in\) Z
Bài 2:
a, 4x(x + 1) = 8(x + 1)
\(\Leftrightarrow\) 4x(x + 1) - 8(x + 1) = 0
\(\Leftrightarrow\) (x + 1)(4x - 8) = 0
\(\Leftrightarrow\) 4(x + 1)(x - 2) = 0
\(\Leftrightarrow\) (x + 1)(x - 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy S = {-1; 2}
b, x2 - 6x + 8 = 0
\(\Leftrightarrow\) x2 - 6x + 9 - 1 = 0
\(\Leftrightarrow\) (x - 3)2 - 1 = 0
\(\Leftrightarrow\) (x - 3 - 1)(x - 3 + 1) = 0
\(\Leftrightarrow\) (x - 4)(x - 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
Vậy S = {4; 2}
c, x3 + x2 + x + 1 = 0
\(\Leftrightarrow\) x2(x + 1) + (x + 1) = 0
\(\Leftrightarrow\) (x + 1)(x2 + 1) = 0
Vì x2 + 1 > 0 với mọi x
\(\Rightarrow\) x + 1 = 0
\(\Leftrightarrow\) x = -1
Vậy S = {-1}
d, x3 - 7x - 6 = 0
\(\Leftrightarrow\) x3 - x - 6x - 6 = 0
\(\Leftrightarrow\) (x3 - x) - (6x + 6) = 0
\(\Leftrightarrow\) x(x2 - 1) - 6(x + 1) = 0
\(\Leftrightarrow\) x(x - 1)(x + 1) - 6(x + 1) = 0
\(\Leftrightarrow\) (x + 1)[x(x - 1) - 6] = 0
\(\Leftrightarrow\) (x + 1)(x2 - x - 6) = 0
\(\Leftrightarrow\) (x + 1)(x2 - 3x + 2x - 6) = 0
\(\Leftrightarrow\) (x + 1)[x(x - 3) + 2(x - 3)] = 0
\(\Leftrightarrow\) (x + 1)(x - 3)(x + 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\\x=-2\end{matrix}\right.\)
Vậy S = {-1; 3; -2}
Câu e hình như bạn viết nhầm 2 lần số 17x thì phải, mình sửa lại rồi!!
e, 3x3 - 7x2 + 17x - 5 = 0
\(\Leftrightarrow\) 3x3 - x2 - 6x2 + 2x + 15x - 5 = 0
\(\Leftrightarrow\) (3x3 - x2) + (-6x2 + 2x) + (15x - 5) = 0
\(\Leftrightarrow\) x2(3x - 1) - 2x(3x - 1) + 5(3x - 1) = 0
\(\Leftrightarrow\) (3x - 1)(x2 - 2x + 5) = 0
\(\Leftrightarrow\) (3x - 1)(x2 - 2x + \(\frac{1}{4}\) + \(\frac{19}{4}\)) = 0
\(\Leftrightarrow\) (3x - 1)[(x - \(\frac{1}{2}\))2 + \(\frac{19}{4}\)] = 0
Vì (x - \(\frac{1}{2}\))2 + \(\frac{19}{4}\) > 0 với mọi x nên
\(\Rightarrow\) 3x - 1 = 0
\(\Leftrightarrow\) x = \(\frac{1}{3}\)
Vậy S = {\(\frac{1}{3}\)}
Bài 3:
Hình như phần a thì 16(1 - x) mới đúng chứ!!
a, x2(x - 1) + 16(1 - x)
= x2(x - 1) - 16(x - 1)
= (x - 1)(x2 - 16)
= (x - 1)(x - 4)(x + 4)
Câu b, d, g mình chịu, hình như đề sai thì phải, mình ko nghĩ ra được!!
c, x3 - 3x2 - 3x + 1
= (x3 + 1) - (3x2 + 3x)
= (x + 1)(x2 + x + 1) - 3x(x + 1)
= (x + 1)(x2 + x + 1 - 3x)
= (x + 1)(x2 - 2x + 1)
= (x + 1)(x - 1)(x - 1)
e, x4 - 13x2 + 36
= x4 - 4x2 - 9x2 + 36
= x2(x2 - 4) - 9(x2 - 4)
= (x2 - 4)(x2 - 9)
= (x - 2)(x + 2)(x - 3)(x + 3)
f, (x2 + x)2 + 4x2 + 4x - 12
= (x2 + x)2 + 4x2 + 4x + 4 - 16
= (x2 + x)2 + 4(x2 + x) + 4 - 16
= (x2 + x + 2)2 - 16
= (x2 + x + 2 - 4)(x2 + x + 2 + 4)
= (x2 + x - 2)(x2 + x + 6)
2x + 2x+1 + 2x+2 + 2x+3 = 120
=> 2x ( 1 + 2 + 22 + 23) = 120
=> 2x . 15 = 120
=> 2x = 8 = 23
=> x = 3
\(2^x+2^x\cdot2^1+2^x\cdot2^2+2^x\cdot2^3=120\)
\(2^x\cdot\left(1+2+4+8\right)=120\)
\(2^x\cdot15=120\)
\(2^x=120:15\)
\(2^x=8\)
\(2^x=2^3\)
\(=>x=3\)