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AT
3
2T
17 tháng 8 2019
\(36x^2-\left(3x-2\right)^2\)
\(=\left(6x\right)^2-\left(3x-2\right)^2\)
\(=\left(6x+3x-2\right)\left(6x-3x+2\right)\)
\(=\left(9x-2\right)\left(3x+2\right)\)
2T
17 tháng 8 2019
\(-49x^2+9\)
\(=3^2-\left(7x\right)^2\)
\(=\left(3-7x\right)\left(3+7x\right)\)
TT
5
2T
17 tháng 8 2019
\(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
2T
17 tháng 8 2019
\(x^5-9x=0\)
\(\Leftrightarrow x\left(x^4-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^4-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt[4]{9}\end{cases}}\)
23 tháng 8 2017
Câu 1:
Ta có:\(x\left(x^2-y\right)+x\left(y^2-y\right)-x\left(x^2+y^2\right)\)
\(=x\left(x^2-y+y^2-y-x^2-y^2\right)\)
\(=-2xy\)
Tại \(x=\frac{1}{2};y=-100\) PT có dạng:
\(=-2.\frac{1}{2}.\left(-100\right)=100\)
13 tháng 10 2020
a, 3x(x - 1) + x - 1 = 0
\(\Leftrightarrow\) (x - 1)(3x + 1) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=1\\x=\frac{-1}{3}\end{matrix}\right.\)
Vậy S = {1; \(\frac{-1}{3}\)}
b, (x - 2)(x2 + 2x + 7) + 2(x2 - 4) - 5(x - 2) = 0
\(\Leftrightarrow\) (x - 2)(x2 + 2x + 7) + 2(x - 2)(x + 2) - 5(x - 2) = 0
\(\Leftrightarrow\) (x - 2)[(x2 + 2x + 7 + 2(x + 2) - 5] = 0
\(\Leftrightarrow\) (x - 2)(x2 + 2x + 7 + 2x + 4 - 5) = 0
\(\Leftrightarrow\) (x - 2)(x2 + 4x + 6) = 0
\(\Leftrightarrow\) (x - 2)[(x + 2)2 + 2] = 0
Vì [(x + 2)2 + 2] > 0 với mọi x nên
\(\Rightarrow\) x - 2 = 0
\(\Leftrightarrow\) x = 2
Vậy S = {2}
c, (2x - 1)2 - 25 = 0
\(\Leftrightarrow\) (2x - 1 - 5)(2x - 1 + 5) = 0
\(\Leftrightarrow\) (2x - 6)(2x + 4) = 0
\(\Leftrightarrow\) (x - 3)(x + 2) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy S = {3; -2}
d, x3 + 27 + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) (x + 3)(x2 - 3x + 9) + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) (x + 3)(x2 - 3x + 9 + x - 9) = 0
\(\Leftrightarrow\) (x + 3)(x2 - 2x) = 0
\(\Leftrightarrow\) x(x + 3)(x - 2) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x+3=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
Vậy S = {0; -3; 2}
Chúc bn học tốt! (Dễ mà :v)
11 tháng 8 2017
a. \(\left(3x-5\right)^2-\left(x+1\right)^2=0\Leftrightarrow\left(3x-5+x+1\right)\left(3x-5-x-1\right)=0\Leftrightarrow\left(4x-4\right)\left(2x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}4x-4=0\\2x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy ...
b. \(\left(5x-4\right)^2-49x^2=0\Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ...
c. \(4x^3-36x=0\Leftrightarrow4x\left(x^2-9\right)=0\Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy ...
d. \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\Leftrightarrow\left(2x+3\right)\left(x-1\right)-\left(2x-3\right)\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x+3-2x+3\right)=0\Leftrightarrow6\left(x-1\right)=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy ...
NP
2
26 tháng 5 2016
\(x^4+2x^2+1+3x^3+3x+2x^2=0\)
\(x^4+3x^3+4x^2+3x+1=0\)
\(x^4+x^3+2x^3+2x^2+2x^2+2x+x+1=0\)
\(x^3\left(x+1\right)+2x^2\left(x+1\right)+2x\left(x+1\right)+\left(x+1\right)=0\)
\(\left(x+1\right)\left(x^3+x^2+x^2+x+x+1\right)=0\)
\(\left(x+1\right)\left[x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\right]=0\)
\(\left(x+1\right)^2\left(x^2+x+1\right)=0\)
\(x=-1\)
26 tháng 5 2016
\(x^4+2x^2+1+3x^3+3x+2x^2=0\) 0
\(x^4+3x^3+4x^2+3x+1=0\)
\(x^4+x^3+2x^3+2x^2+2x^2+2x+x+1=0\)
\(x^3\left(x+1\right)+2x^2\left(x+1\right)+2x\left(x+1\right)+\left(x+1\right)=0\)
\(\left(x+1\right)\left(x^3+x^2+x^2+x+x+1\right)=0\)
\(\left(x+1\right)\left[x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\right]=0\)
\(\left(x+1\right)^2\left(x^2+x+1\right)=0\)
\(x=-1\)
NL
1
4 tháng 8 2017
x2 - 25 + y2 + 2xy
x2 - 25 + y2 + 2xy
Chưa trả lờiNhật Linh Đặng
(2x + 1)2 - 49x2 + 56x - 1
x2 - 25 + y2 + 2xy
(2x + 1)2 - 49x2 + 56x - 16
xm + 4 - xm + 3 - x - 1
a3 + b3 + c3 - 3abc
x2 - 25 + y2 + 2xy
(2x + 1)2 - 49x2 + 56x - 16
xm + 4 - xm + 3 - x - 1
a3 + b3 + c3 - 3abc
x2 - 25 + y2 + 2xy
(2x + 1)2 - 49x2 + 56x - 16
xm + 4 - xm + 3 - x - 1
a3 + b3 + c3 - 3abc
x2 - 25 + y2 + 2xy
(2x + 1)2 - 49x2 + 56x - 16
xm + 4 - xm + 3 - x - 1
a3 + b3 + c3 - 3abc
x2 - 25 + y2 + 2xy
x2 - 25 + y2 + 2xy
(2x + 1)2 - 49x2 + 56x - 16
xm + 4 - xm + 3 - x - 1
a3 + b3 + c3 - 3abc
(2x + 1)2 - 49x2 + 56x -
x2 - 25 + y2 + 2xy
(2x + 1)2 - 49x2 + 56x - 16
xm + 4 - xm + 3 - x - 1
a3 + b3 + c3 - 3abc
16
xm + 4 - xm + 3 - x - 1
a3 + b3 + c3 - 3abc
6
xm + 4 - xm + 3 - x - 1
a3 + b3 + c3 - 3abc
(2x + 1)2 - 49x2 + 56x - 16
xm + 4 - xm + 3 - x - 1
a3 + b3 + c3 - 3abc
NT
3
HT
22 tháng 8 2016
a) \(x^3\)+\(x^2\)=36
\(\Leftrightarrow\)\(x^3\)+\(x^2\)\(-36=0\)
\(\Leftrightarrow\)\(x^3\)\(-3x^2\)\(+4x^2\)\(-12x\)\(+12x-36=0\)
\(\Leftrightarrow\)\(x^2\left(x-3\right)+4x\left(x-3\right)+12\left(x-3\right)=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(x^2+4x+12\right)=0\)
Suy ra: \(x-3=0\) hoặc \(x^2+4x+12=0\)
- \(x-3=0\) \(\Leftrightarrow\) \(x=3\)
- \(x^2+4x+12=0\) (phương trình vô nghiệm)
Vậy \(x=3\)
\(49x_1^2-25\left(x_2+1\right)^2=0\)
\(\Rightarrow\left(7x_1\right)^2-25\left(x_2+1\right)^2=0\)
Xét \(\left(7x_1\right)^2\ge0\) ; \(25\left(x_2+1\right)^2\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(7x_1\right)^2=0\\25\left(x_2+1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x_1=0\\x_2=-1\end{matrix}\right.\)