Giúp m vs: x^3-4x^2-12x+27
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13 tháng 12 2020
c \(\frac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(3x^2-27\right)}=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x^2-9\right)}\)
\(=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{2}{3\left(x+3\right)}\)
d, \(\frac{x^2+5x+6}{x^2+4x+4}=\frac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\frac{x+3}{x+2}\)
Tương tự với a ; b
18 tháng 6 2016
x^3+3x^2-7x^2-21x+9x+27=0
x^2(x+3)-7x(x+3)+9(x+3)=0
(x+3)(x^2-7x+9)=0
x = -3 hoặc x^2 - 7x + 9 = 0 (chuyển về pt dạng kx^2 + m)
Bạn chuyển 7x = 2 . x . 7/2 + 49/4 - 49/4
18 tháng 6 2016
x^3+3x^2-7x^2-21x+9x+27=0
x^2(x+3)-7x(x+3)+9(x+3)=0
(x+3)(x^2-7x+9)=0
x = -3 hoặc x^2 - 7x + 9 = 0 (chuyển về pt dạng kx^2 + m)
28 tháng 8 2021
3: Ta có: \(\sqrt{4x+1}=x+1\)
\(\Leftrightarrow x^2+2x+1=4x+1\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)
\(\Leftrightarrow3\sqrt{x-1}=15\)
\(\Leftrightarrow x-1=25\)
hay x=26
5: Ta có: \(\sqrt{4x^2-12x+9}=7\)
\(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
HT
0
23 tháng 7 2023
a: =(x-y)^2+2(x-y)
=(x-y)(x-y+2)
c: =(x-3)(x+3)+(x-3)^2
=(x-3)(x+3+x-3)
=2x(x-3)
d: =(x+3)(x^2-3x+9)-4x(x+3)
=(x+3)(x^2-7x+9)
e: =(x^2-8x+7)(x^2-8x+15)-20
=(x^2-8x)^2+22(x^2-8x)+85
=(x^2-8x+17)(x^2-8x+5)
TG
20 tháng 8 2020
a) \(x^2-xz-9y^2+3yz\)
\(=\left(x^2-9y^2\right)-\left(xz-3yz\right)\)
\(=\left[x^2-\left(3y\right)^2\right]-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
b) \(x^3-x^2-5x+125\)
\(=\left(x^3+125\right)-\left(x^2+5x\right)\)
\(=\left(x^3+5^3\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+5^2\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+5^2-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
c) \(x^3+2x^2-6x-27\)
\(=\left(x^3-27\right)-\left(2x^2-6x\right)\)
\(=\left(x^3-3^3\right)-2x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+3^2\right)-2x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+3^2-2x\right)\)
\(=\left(x-3\right)\left(x^2+x+9\right)\)
e) \(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=\left(x+1\right)\left(4x^3-x\right)\)
f) \(x^6-x^4-9x^3+9x^2\)
\(=x^4\left(x^2-1\right)-9x^2\left(x-1\right)\)
\(=x^4\left(x-1\right)\left(x+1\right)-9x^2\left(x-1\right)\)
\(=\left(x-1\right)\left[x^4\left(x+1\right)-9x^2\right]\)
\(=\left(x-1\right)\left(x^5+x^4-9x^2\right)\)
TT
2
5 tháng 3 2020
\(a,9\left(2x+1\right)=4\left(x-5\right)^2\)
\(4x^2-40x+100=18x+9\)
\(4x^2-58x+91=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{29+3\sqrt{53}}{4}\\x=\frac{29-3\sqrt{53}}{4}\end{cases}}\)
\(b,x^3-4x^2-12x+27=0\)
\(\left(x+3\right)\left(x^2-7x+9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}}\)
\(c,x^3+3x^2-6x-8=0\)
\(\left(x+4\right)\left(x-2\right)\left(x+1\right)=0\)
\(Th1:x+4=0\Leftrightarrow x=-4\)
\(Th2:x-2=0\Leftrightarrow x=2\)
\(Th3:x+1=0\Leftrightarrow x=-1\)
5 tháng 3 2020
\(a,9.\left(2x+1\right)=4.\left(x-5\right)^2\)
\(< =>4x^2-40x+100=18x+9\)
\(< =>4x^2+58x+91=0\)
\(< =>\orbr{\begin{cases}x=\frac{29-3\sqrt{53}}{4}\\x=\frac{29+3\sqrt{53}}{4}\end{cases}}\)
\(b,x^3-4x^2-12x+27=0\)
\(< =>\left(x+3\right)\left(x^2-7x+9\right)=0\)
\(< =>\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}}\)
\(< =>\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}\)
(𝑥+3)(𝑥2−7𝑥+9)