\(x^4-10x^3+25x^2=36\)
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NM
2
LD
29 tháng 3 2016
SUY RA \(x^4+x^3-11x^3-11x^2+36x^2-36=0\)
\(\Leftrightarrow x^3\left(x+1\right)-11x^2\left(x+1\right)+36\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3-11x^2+36x-36\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-3\right)\left(x-2\right)\left(x+1\right)=0\)
suy ra x=-1 hoặc x=6 hoặc x=3 hoặc x=2
mk làm hơi tắt nhưng vẫn dk k nha
DT
2
TD
7 tháng 1 2017
a. \(x^4-10x^3+25x^2-36=0\)
=> \(x^3\left(x-3\right)-7x^2\left(x-3\right)+4x\left(x-3\right)+12\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(x^3-7x^2+4x+12\right)=0\)
=>\(\left(x-3\right)\left[x^2\left(x-2\right)-5x\left(x-2\right)-6\left(x-2\right)\right]=0\)=> \(\left(x-3\right)\left(x-2\right)\left(x^2-5x-6\right)=0\)
=> \(\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(x-6\right)=0\)
=>\(\left[\begin{matrix}x=3\\x=2\\x=-1\\x=6\end{matrix}\right.\)
b) \(x^4\) - \(^{9x^2}\) - 24x - 16 = 0
=> \(x^3\left(x-4\right)+4x^2\left(x-4\right)+7x\left(x-4\right)+4\left(x-4\right)=0\)=>\(\left(x-4\right)\left(x^3+4x^2+7x+4\right)=0\)
=> \(\left(x-4\right)\left[x^2\left(x+1\right)+3x\left(x+1\right)+4\left(x+1\right)\right]=0\)=>\(\left(x-4\right)\left(x+1\right)\left(x^2+3x+4\right)=0\)
=> \(\left(x-4\right)\left(x+1\right)=0\) (vì x^2 + 3x + 4> 0)
=>\(\left[\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
TD
7 tháng 1 2017
a,pt\(\Leftrightarrow\left(x^4-10x^3+25x\right)-36=0\)\(\Leftrightarrow\left(x^2-5x\right)^2-36=0\)
\(\Leftrightarrow\left(x^2-5x-6\right)\left(x^2-5x+6\right)=0\)\(\Leftrightarrow\left[\begin{matrix}x^2-5x-6=0\\x^2-5x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}\left(x+1\right)\left(x-6\right)=0\\\left(x-2\right)\left(x-3\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=-1,x=6\\x=2,x=3\end{matrix}\right.\)
vậy pt có 4 nghiệm x=(-1,6,2,3)
13 tháng 7 2017
a) 2x2 - 98 = 0
2x2 = 0 + 98
2x2 = 98
x2 = 98 : 2
x2 = 49
x = \(\sqrt{49}\)
=> x = 7
13 tháng 7 2017
Ta có : 2x2 - 98 = 0
=> 2(x2 - 49) = 0
Mà : 2 > 0
Nên x2 - 49 = 0
=> x2 = 49
=> x2 = -7;7
15 tháng 7 2019
1)\(36\left(x-y\right)^2-49\left(x+y\right)^2\)
\(=\left(6x-6y+7x+7y\right)\left(6x-6y-7x-7y\right)\)
\(=\left(13x+y\right)\left(-x-13y\right)\)
\(=-\left(13x+y\right)\left(x+13y\right)\)
2)\(16x^2-9\left(x+y\right)^2=\left(4x+3x+3y\right)\left(4x-3x-3y\right)=\left(7x+3y\right)\left(x-3y\right)\)
20 tháng 8 2018
a, A = x5 - 5x4 + 5x3 - 5x2 + 5x - 1
A= x5 - ( 4+1 ) x4 + ( 4+1 ) x3 - ( 4+1) x2 + ( 4+1 ) x -1
Thay 4 = x vào biểu thức A, ta đc :
A = x5 - ( x+1 ) x4 + ( x+1 ) x3 - ( x+1 ) x2 + ( x+1 ) x - 1
A = x5 - x5 - x4 + x4 + x3 - x3 - x2 + x2 + x -1
A = x -1
Thay x = 4 vào biểu thức A, ta đc :
A = 4 -1
A = 3
b, B = x7 - 80x6 + 80x5 - 80x4 + .....+ 80x + 15
B = x7 - ( 79 +1 ) x6 + ( 79+1 )x5 - ( 79+1 ) x4 +....+( 79+1 )x + 15
Thay 79 = z vào biểu thức A, ta có :
B = x7 - ( x + 1 )x6 + ( x+1 )x5 - ( x+1 )x4 + .....+ ( x+1 )x +15
B= x7 - x7 - x6 + x6 + x5 - x5 - x4 + .....- x2 + x2 + x + 15
B= x + 15
Thay x= 79 vào biểu thức A, ta có:
A = 79 + 15
A= 94
c, C = x14 - 10x13 + 10x12 - 10x11 + ....+ 10x2 - 10x + 10
C= x14 - ( x +1 )x13 + ( x + 1 ) x12 - ( x + 1 )x11 + ..... + ( x + 1 )x2 - ( x + 1 )x - 10
C= x14 - x14 - x13 + x13 + x12 - x12 - x11 +....+ x3 - x2 + x2 - x +10
C= -x -10
Thay -x = -9 vào biểu thức C, ta có :
C = -9 + 10
C = 1
d, D = x10 - ( x+1 )x9 + (x + 1 )x8 - ( x+1 )x7 +....+( x+1 )x2 - ( x + 1 )x + 25
D = x10 - ( x + 1 ) x9 + ( x + 1 )x8 - ( x + 1 )x7 + ..... + x3 - x2 + x2 - x + 25
D = -x + 25
thay -x = -24, vào biểu thức A , ta đc ;
A = -24 + 25
A = 1
5 tháng 9 2018
\(A=x^5-5x^4+5x^3-5x^2+5x-1\)
\(=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-x+3\)
\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-x+3\)
\(=3\)
26 tháng 10 2021
1) ĐKXĐ: \(x\ge\dfrac{5}{2}\)
\(\sqrt{x^2}=2x-5\\ \Rightarrow\left|x\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x=2x-5\\x=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)
2) ĐKXĐ: \(x\ge3\)
\(\sqrt{25x^2-10x+1}=2x-6\\ \Rightarrow\left|5x-1\right|=2x-6\\ \Rightarrow\left[{}\begin{matrix}5x-1=2x-6\\5x-1=6-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
3) ĐKXĐ: \(x\ge\dfrac{5}{2}\)
\(\sqrt{25-10x+x^2}=2x-5\\ \Rightarrow\left|x-5\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x-5=2x-5\\x-5=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{10}{3}\left(tm\right)\end{matrix}\right.\)
4) ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(\sqrt{1-2x+x^2}=2x-1\\ \Rightarrow\left|x-1\right|=2x-1\\ \Rightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)
TQ
2
H
19 tháng 10 2019
TA CO: A\(=x^4-10x^3+25x^2+12\)
\(=x^2\left(x^2-10x+25\right)+12\)
\(=x^2\left(x-5\right)^2+12\)
\(Do\)\(\left(x-5\right)^2\ge0\Rightarrow x^2\left(x-5\right)^2\ge0\)
\(\Rightarrow A\ge12\)
Dau''=''xay ra khi vµ chi khi:
\(\left(x-5\right)^2=0\)
\(\Rightarrow x-5=0\)
\(\Rightarrow x=5\)
Vay MAX A=12 khi x=5
\(x^4-10x^3+25x^2=36\)
➜\(x^4-10x^3=25x^2-36=0\)
➜\(x^3\left(x-3\right)-7x^2\left(x-3\right)+4x\left(x-3\right)+12\left(x-3\right)=0\)
➜\(\left(x-3\right)\left(x^3-7x^2+x+12\right)=0\)
➜\(\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(x-6\right)=0\)
➜\(\left[{}\begin{matrix}x-3=0\\x-2=0\\x+1=0\\x-6=0\end{matrix}\right.\text{➜}\left[{}\begin{matrix}x=3\\x=2\\x=-1\\x=6\end{matrix}\right.\)
Vậy..................................................
Ta có: \(x^4-10x^3+25x^2=36\Leftrightarrow x^4-10x^3+25x^2-36=0\Leftrightarrow x^4+x^3-11x^3-11x^2+36x^2-36=0\)
\(\Leftrightarrow x^3\left(x+1\right)-11x^2\left(x+1\right)+36\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3-11x^2+36x-36\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=3\\x=6\end{matrix}\right.\)