(x-3)^2 =16
(1-3x )^3 =-64
x^13 = 27. x^10
(4x-1)^2 = (1 -4x )^4
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
25 tháng 9 2017
Ta có : x3 - 7x + 6
= x3 - x - 6x + 6
= x(x2 - 1) - 6(x - 1)
= x(x + 1)(x - 1) - 6(x - 1)
= (x - 1) [x(x + 1) - 6]
= (x - 1) (x2 + x - 6) .
CÁC Ý SAU TƯƠNG TỰ
KK
25 tháng 9 2017
1
x3-7x+6
=x3+0x2-7x +6
= x3-x2+x2-x-6x+6
=(x3-x2)+(x2-x)-(6x-6)
=x2(x-1)+x(x-1)-6(x-1)
=(x-1)(x2+x-6)
=(x-1)(x2+3x-2x-6)
=(x-1)[x(x+3)-2(x+3)]
=(x-1)(x-2)(x+3)
25 tháng 9 2017
7) (x+2)(x+3)(x+4)(x+5)-24
=(x+2)(x+5) (x+3)(x+4)-24
=[x(x+5)+2(x+5)][x(x+4)+3(x+4)]-24
=[x2+5x+2x+10][x2+4x+3x+12]-24
=[x2+7x+10][x2+7x+12]-24
đặt a=x2+7x+10
=>x2+7x+12=a+2
=a(a+2)-24
=a2+2a-24
=a2+6a-4a-24
=(a2+6a)-(4a+24)
=a(a+6)-4(a+6)
=(a+6)(a-4)
thay a= x2+7x+10 vào ta được
(x2+7x+10+6)(x2+7x+10-4)
=(x2+7x+16)(x2+7x+6)
11 tháng 7 2021
9) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)
\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)
\(\Leftrightarrow-4x=9\)
hay \(x=-\dfrac{9}{4}\)
11 tháng 7 2021
10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}
11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{9x+5}{\left(x-4\right)\left(x+5\right)}\)
Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)
\(\Leftrightarrow5x^2-7x=0\)
\(\Leftrightarrow x\left(5x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)
12) Ta có: \(\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{\left(2x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow\dfrac{2x\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{4x^2-1+4}{\left(2x-1\right)\left(2x+1\right)}\)
Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)
\(\Leftrightarrow2x^2+x-3=0\)
\(\Leftrightarrow2x^2+3x-2x-3=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)
12 tháng 6 2023
2:
=>x^3-1-2x^3-4x^6+4x^6+4x=6
=>-x^3+4x-7=0
=>x=-2,59
4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50
=>-62x+12=-50
=>x=1
VK
28 tháng 7 2019
a) x3 - 9x2 + 27x - 27 = -8
<=> x3 - 3x2.3 + 3x.32 - 33 = -8
<=> (x - 3)3 = -23
<=> x - 3 = -2
<=> x = 1 (T/m)
Vậy x = 1.
VK
28 tháng 7 2019
b) 64x3 + 48x2 + 12x + 1 = 27
<=> (4x)3 + 3.(4x)2.1 + 3.4x.12 + 13 = 27
<=> (4x + 1)3 = 33
<=> 4x + 1 = 3
<=> 4x = 2
<=> x = \(\frac{1}{2}\)(T/m)
Vậy x = \(\frac{1}{2}\).
S
28 tháng 7 2019
\(x^3-9x^2+27x-27=-8\Leftrightarrow\left(x^3-27\right)-\left(9x^2-27x\right)=\left(x-3\right)\left(x^2+3x+9\right)-9x\left(x-3\right)=\left(x-3\right)\left(x^2-6x+9\right)=\left(x-3\right)^3=-8=\left(-2\right)^3\Rightarrow x=\left(-2\right)+3=1\)
\(64x^3+48x^2+12x+1=\left(64x^3+1\right)+\left(48x^2+12x\right)=\left(4x+1\right)\left(16x^2-4x+1\right)+12x\left(4x+1\right)=\left(4x+1\right)\left(16x^2+8x+1\right)=\left(4x+1\right)^3=27\Rightarrow4x=2\Leftrightarrow x=\frac{1}{2}\)
29 tháng 7 2019
c) \(\left(2x-1\right)^3-4x^2.\left(2x-3\right)=5\)
\(\Leftrightarrow\left(8x^3-12x^2+6x-1\right)-\left(8x^3-12x^2\right)=5\)
\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\)
\(\Leftrightarrow6x-1=5\)
\(\Leftrightarrow6x=6\)
\(\Leftrightarrow x=1\)
d) \(\left(x+4\right)^3-x^2.\left(x+12\right)=16\)
\(\Leftrightarrow\left(x^3+12x^2+48x+64\right)-\left(x^3+12x^2\right)=16\)
\(\Leftrightarrow x^3+12x^2+48x+64-x^3-12x^2=16\)
\(\Leftrightarrow48x+64=16\)
\(\Leftrightarrow48x=-48\)
\(\Leftrightarrow x=-1\)
#vì câu a,b có người làm rồi nên mình chỉ làm c,d thôi nhé ! :)
Học Tốt !!
+) \(\left(x-3\right)^2=16\)
\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^2=4^2\\\left(x-3\right)^2=\left(-4\right)^2\end{cases}\Rightarrow}\orbr{\begin{cases}x-3=4\\x-3=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=-1\end{cases}}\)
Vậy x = 7 hoặc x = -1
+) \(\left(1-3x\right)^3=-64\)
\(\Rightarrow\left(1-3x\right)^3=\left(-4\right)^3\)
\(\Rightarrow1-3x=-4\)
\(\Rightarrow3x=1+4\)
\(\Rightarrow3x=5\)
\(\Rightarrow x=5:3\)
\(\Rightarrow x=\frac{5}{3}\)
Vậy \(x=\frac{5}{3}\)
+) \(x^{13}=27.x^{10}\)
\(\Rightarrow x^{13}:x^{10}=27\)
\(\Rightarrow x^3=27\)
\(\Rightarrow x^3=3^3\)
\(\Rightarrow x=3\)
Vậy x = 3
+) \(\left(4x-1\right)^2=\left(1-4x\right)^4\)
\(\Rightarrow\left(4x-1\right)^2=\left(4x-1\right)^4\)
\(\Rightarrow\left(4x-1\right)^2-\left(4x-1\right)^4=0\)
\(\Rightarrow\left(4x-1\right)^2\left[1-\left(4x-1\right)^2\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(4x-1\right)^2=0\\1-\left(4x-1\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(4x-1\right)^2=0\\\left(4x-1\right)^2=1\end{cases}}\)
TH 1 : \(\left(4x-1\right)^2=0\Rightarrow4x-1=0\Rightarrow4x=1\Rightarrow x=\frac{1}{4}\)
TH 2 : \(\left(4x-1\right)^2=1\Rightarrow\orbr{\begin{cases}4x-1=1\\4x-1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}4x=2\\4x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=0\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{4};\frac{1}{2};0\right\}\)
_Chúc bạn học tốt_
a, (x-3)^2 = 16
=> (x-3)^2=4^2
=> x-3=4
=> x= 4+3
=> x = 7 .Vậy x =7
b,(1-3x)^3 = 64
=> ( 1-3x)^3 = 4^3
=> 1-3x = 4
=> 3x = 1-4
=> 3x = -3
=> x = -1 . Vậy x = -1
c, x^13 = 27.x^10
=> x^13 : x^10 = 27
=> x^3 = 3^3
=> x = 3 . Vậy x = 3