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TM
17 tháng 12 2016
Câu 1:
\(x^4+5x^3-12x^2+5x+1=x^4+7x^3+x^2-2x^3-14x^2-x+x^2+7x+1\)
\(=\left(x^4+7x^3+x^2\right)-\left(2x^3+14x^2+x\right)+\left(x^2+7x+1\right)\)
\(=x^2\left(x^2+7x+1\right)-2x\left(x^2+7x+1\right)+\left(x^2+7x+1\right)\)
\(=\left(x^2-2x+1\right)\left(x^2+7x+1\right)\)
\(=\left(x-1\right)^2\left(x^2+7x+1\right)\)
Câu 2:
\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2=x^4-24x^3+203x^2-720x+900-24x^2\)
\(=x^4-24x^3+179x^2-720x+900\)
\(=\left(x^4-7x^3+30x^2\right)-\left(17x^3-119x^2+510x\right)+\left(30x^2-210x+900\right)\)
\(=x^2\left(x^2-7x+30\right)-17x\left(x^2-7x+30\right)+30\left(x^2-7x+30\right)\)
\(=\left(x^2-17x+30\right)\left(x^2-7x+30\right)\)
\(=\left(x^2-2x-15x+30\right)\left(x^2-7x+30\right)\)
\(=\left[x\left(x-2\right)-15\left(x-2\right)\right]\left(x^2-7x+30\right)\)
\(=\left(x-15\right)\left(x-2\right)\left(x^2-7x+30\right)\)
Câu 3:
\(2x^3+11x^2+3x-36=\left(2x^3+14x^2+24x\right)-\left(3x^2+21x+36\right)\)
\(=2x\left(x^2+7x+12\right)-3\left(x^2+7x+12\right)\)
\(=\left(2x-3\right)\left(x^2+7x+12\right)\)
\(=\left(2x-3\right)\left(x^2+3x+4x+12\right)\)
\(=\left(2x-3\right)\left[x\left(x+3\right)+4\left(x+3\right)\right]\)
\(=\left(2x-3\right)\left(x+3\right)\left(x+4\right)\)
19 tháng 6 2021
a, 4x2 - 49 = 0
⇔⇔ (2x)2 - 72 = 0
⇔⇔ (2x - 7)(2x + 7) = 0
⇔{2x−7=02x+7=0⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=72x=−72⇔{2x−7=02x+7=0⇔{x=72x=−72
b, x2 + 36 = 12x
⇔⇔ x2 + 36 - 12x = 0
⇔⇔ x2 - 2.x.6 + 62 = 0
⇔⇔ (x - 6)2 = 0
⇔⇔ x = 6
19 tháng 6 2021
e, (x - 2)2 - 16 = 0
⇔⇔ (x - 2)2 - 42 = 0
⇔⇔ (x - 2 - 4)(x - 2 + 4) = 0
⇔⇔ (x - 6)(x + 2) = 0
⇔{x−6=0x+2=0⇔{x=6x=−2⇔{x−6=0x+2=0⇔{x=6x=−2
f, x2 - 5x -14 = 0
⇔⇔ x2 + 2x - 7x -14 = 0
⇔⇔ x(x + 2) - 7(x + 2) = 0
⇔⇔ (x + 2)(x - 7) = 0
⇔{x+2=0x−7=0⇔{x=−2x=7
TD
1
TD
18 tháng 10 2020
\(4\left(6-x\right)+x^2-12x+36=0\)
\(24-4x+x^2-12x+36=0\)
\(x^2-16x+60=0\)
\(x^2-2x8+8^2-8^2+60=0\)
\(\left(x-8\right)^2-4=0\)
\(\left(x-8\right)^2=4\)
\(\left(x-8\right)^2=\left(\pm2\right)^2\)
\(\orbr{\begin{cases}x-8=2\Rightarrow x=10\\x-8=-2\Rightarrow x=6\end{cases}}\)
NK
1
4 tháng 12 2015
a/ => 6x3 + x2 - 2x = 0
=> x (6x2 + x - 2) = 0
=> x (6x2 + 4x - 3x - 2) = 0
=> x [ 2x (3x + 2) - (3x + 2) ] =0
=> x (3x + 2) (2x - 1) = 0
=> x = 0
hoặc 3x + 2 = 0 => 3x = -2 => x = -2/3
hoặc 2x - 1 = 0 => 2x = 1 => x = 1/2
Vậy x = 0; x = -2/3 ; x = 1/2
Câu b,c,d tương tự
CN
31 tháng 12 2017
Bài 1:
a) x2 + 5x = 0
⇔ x(x + 5) = 0
⇔ \(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x-0\\x=-5\end{matrix}\right.\)
b) (12x3 - 8x) : x - 4x(3x - 1) = 0
⇔ 12x2 - 8 - 12x2 + 4x = 0
⇔ 4x - 8 = 0
⇔ 4x = 8
⇔ x = 2
Bài 2:
\(P=\dfrac{x^2-12x+36}{2x^2-72}\)
\(=\dfrac{\left(x-6\right)^2}{2\left(x^2-6^2\right)}\)
\(=\dfrac{\left(x-6\right)^2}{2\left(x-6\right)\left(x+6\right)}\)
\(=\dfrac{x-6}{2\left(x+6\right)}\)
21 tháng 12 2018
1) \(2x\left(x-3\right)+5x-15=0\)
\(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\left(x-3\right)\left(2x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-5}{2}\end{matrix}\right.\)
2) \(x\left(2x-7\right)-4x+14=0\)
\(x\left(2x-7\right)-2\left(2x-7\right)=0\)
\(\left(2x-7\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\)
3) \(x^2-12x+36=0\)
\(\left(x-6\right)^2=0\)
\(x-6=0\)
\(x=6\)
4) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)
\(\left(x^3+3^3\right)-x\left(x^2-1\right)-27=0\)
\(x^3+27-x^3+x-27=0\)
\(x=0\)
S
28 tháng 6 2018
1/
a, x2+36=12x
<=>x2-12x+36=0
<=>(x-6)2=0
<=>x-6=0
<=>x=6
b, 5x(x-3)+3-x=0
<=>5x(x-3)-(x-3)=0
<=>(5x-1)(x-3)=0
<=>\(\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}}\)
2/ Sửa đề x2z2 = y2z2
Đặt \(A=4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2=4x\left(x+y+z\right)\left(x+y\right)\left(x+z\right)+y^2z^2\)
\(=4\left(x^2+xy+xz\right)\left(x^2+xz+xy+yz\right)+y^2z^2\)
Đặt x2+xy+xz=t, ta có
\(A=4t\left(t+yz\right)+y^2z^2=4t^2+4tyz+y^2z^2=\left(2t+yz\right)^2=\left(2x^2+2xy+2xz+y^2z^2\right)^2\ge0\)
A
1 tháng 10 2020
a, \(x^2-12x-2x+24=0\Leftrightarrow x^2-14x+24=0\Leftrightarrow\left(x-12\right)\left(x-2\right)=0\)
TH1 : x = 12 ; TH2 : x = 2
b, \(x^2-5x-24=0\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
TH1 : x = 8 ; TH2 : x = -3
c, \(4x^2-12x-7=0\Leftrightarrow\left(2x+1\right)\left(2x-7\right)=0\)
TH1 : x = -1/2 ; TH2 : x = 7/2
d, \(x^3+6x^2+12x+8=0\Leftrightarrow\left(x+2\right)^3=0\Leftrightarrow x=-2\)
Tương tự HĐT thôi :)
ta có
\(x^2-12x+36+5x\left(x-6\right)=0\Leftrightarrow\left(x-6\right)^2+5x\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(6x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=6\\x=1\end{cases}}\)
Answer:
\(x^2-12x+36+5x\left(x-6\right)=0\)
\(\Rightarrow x^2-12x+36+5x^2-30x=0\)
\(\Rightarrow6x^2-42x+36=0\)
\(\Rightarrow6x^2-36x-6x+36=0\)
\(\Rightarrow6x\left(x-6\right)-6\left(x-6\right)=0\)
\(\Rightarrow\left(x-6\right)\left(6x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-6=0\\6x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=6\\x=1\end{cases}}}\)