\(=>\left(x+2\right)\left(x-2\right)-6x+12=0\)
\(=>x^2-4x+4-6x+12=0\)
\(=>x^2-10x+16=0\)
( công thức căn bậc hai )
\(\left[{}\begin{matrix}x_1=\frac{-\left(-10\right)+\sqrt{\left(-10\right)^2-4.1.16}}{2.1}\\x_2=\frac{-\left(-10\right)-\sqrt{\left(-10\right)^2-4.1.16}}{2.1}\end{matrix}\right.=>\left[{}\begin{matrix}x_1=9\\x_2=x\end{matrix}\right.\)
Vậy.....
Tìm x, biết:
a) \(\left(2x-1\right)^2-4x^2+1=0\)
b) \(6x^3-24x=0\)
c) \(2x\left(x-3\right)-4x+12=0\)
d) \(x^3-5x^2+x-5=0\)
tìm x biết
a)5(x+3)-2x(x+3)=0
b)6x\(\left(x^2-2\right)-\left(2-x^2\right)=0\)
c)\(\left(x+1\right)^2-\left(x+1\right)\left(x-2\right)=0\)
d)\(4x\left(x-2017\right)-x+2017=0\)
e)\(\left(x+4\right)^2-16=0\)
f)\(12x-x^2-36=0\)
giải các bất phương trình và biểu diễn tập nghiệm trên trục số
\(a,\left(x+\dfrac{1}{9}\right)\left(2x-5\right)< 0\)
\(b,\left(4x-1\right)\left(x^2+12\right)\left(-x+4\right)>0\)
\(c,x^2-6x+9< 0\)
Tìm x biết:
\(\left(2x-3\right)^2-\left(3x+2\right)^2=5x\left(2-x\right)\)
\(8x^3-12x^2+6x-1=0\)
Tìm x biết: \(x^3-8-\left(x-2\right).\left(x-12\right)=0\)
Tìm giá trị nhỏ nhất : \(A=x^2-2x+9\) ; \(B=x^2+6x-3\) ; \(C=\left(x-1\right).\left(x-3\right)+9\)
Tìm giá trị lớn nhất : \(D=-x^2-4x+7\) ; \(E=5-4x^2+4\)
1.rút gọn biểu thuc P=\(\dfrac{2}{x+3}+\dfrac{1}{x-3}+\dfrac{9-x}{9-x^2}\) với x\(\ne-3vàx\ne3\)
2.thực hiện phép tính \(\left(2x^4-3x^3-3x^2+6x-1\right):\left(x^2-2\right)\)
\(\left(15x^4y^6-12^3y^4-18x^2y^3\right):\left(-6x^2y^2\right)\)
Giải các phương trình sau:
a) \(x^2+\dfrac{2x}{x-1}=8\)
b) \(\dfrac{x^2+2x+1}{x^2+2x+2}+\dfrac{x^2+2x+2}{x^2+2x+3}=\dfrac{7}{6}\)
c) \(\dfrac{x+4}{x-1}+\dfrac{x-4}{x+1}=\dfrac{x+8}{x-2}+\dfrac{x-8}{x+2}+6\)
d) \(\left(x^2+6x+8\right)\left(x^2+8x+15\right)=24\)
e) \(\left(x^2+x-2\right)\left(x^2+9x+18\right)=28\)
f) \(3\left(-x^2+2x+3\right)^4-26x^2\left(-x^2+2x+3\right)^2-9x^4=0\)
g) \(x^4+6x^3+11x^2+6x+1=0\)
h) \(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2=0\)
i) \(\left(x+2\right)^4+\left(x+8\right)^4=272\)
Cho A=\(\dfrac{\left(x+2\right)^2}{x}.\left(1-\dfrac{x^2}{x+2}\right)-\dfrac{x^2+6x+4}{x}\)
a) Rút gọn
b) Tìm Min A
1) \(\frac{3x-1}{4}+\frac{2x-3}{3}=\frac{x-1}{2}\) Mc : 12
\(\Leftrightarrow\) \(\frac{3.\left(3x-1\right)}{12}+\frac{4.\left(2x-3\right)}{12}=\frac{6.\left(x-1\right)}{12}\)
\(\Leftrightarrow\) 9x - 3 + 8x - 12 = 6x - 6
\(\Leftrightarrow\) 9x + 8x - 6x = 3 + 12 - 6
\(\Leftrightarrow\) 11x = 9
\(\Leftrightarrow\) x = 0,8
Vậy S = {0,8}
2) \(\frac{x+1}{2}-\frac{x+3}{12}=3-\frac{5-3x}{3}\) Mc : 12
\(\Leftrightarrow\) \(\frac{6.\left(x+1\right)}{12}-\frac{x+3}{12}=\frac{12.3}{12}-\frac{4.\left(5-3x\right)}{12}\)
\(\Leftrightarrow\) 6x + 6 - x + 3 = 36 - 20 - 12x
\(\Leftrightarrow\) 6x - x + 12x = -6 - 3 + 36 - 20
\(\Leftrightarrow\) 17x = 7
\(\Leftrightarrow\) x = \(\frac{7}{17}\)
Vậy S = {\(\frac{7}{17}\)}
3) x - \(\frac{x+1}{3}\) = \(\frac{2x-1}{5}\) Mc : 15
\(\Leftrightarrow\) \(\frac{15.x}{15}-\frac{5.\left(x+1\right)}{15}=\frac{3.\left(2x-1\right)}{15}\)
\(\Leftrightarrow\) 15x - 5x - 5 = 6x - 3
\(\Leftrightarrow\) 15x - 5x - 6x = 5 - 3
\(\Leftrightarrow\) 4x = 2
\(\Leftrightarrow\) x = \(\frac{2}{4}=\frac{1}{2}\)
Vậy S = {\(\frac{1}{2}\)}
4) \(\frac{2x+7}{3}-\frac{x-2}{4}=-2\) Mc : 12
\(\Leftrightarrow\) \(\frac{4.\left(2x+7\right)}{12}-\frac{3.\left(x-2\right)}{12}=\frac{12.\left(-2\right)}{12}\)
\(\Leftrightarrow\) 8x + 28 -3x + 6 = -24
\(\Leftrightarrow\) 8x - 3x = -28 - 6 -24
\(\Leftrightarrow\) 5x = -58
\(\Leftrightarrow\) x = -11,6
Vậy S = {-11,6}
5) \(\frac{2x-3}{4}-\frac{4x-5}{3}=\frac{5-x}{6}\) Mc : 12
\(\Leftrightarrow\) \(\frac{3.\left(2x-3\right)}{12}-\frac{4.\left(4x-5\right)}{12}=\frac{2.\left(5-x\right)}{12}\)
\(\Leftrightarrow\) 6x - 9 - 16x + 20 = 10 - 2x
\(\Leftrightarrow\) 6x - 16x + 2x = 9 - 20 + 10
\(\Leftrightarrow\) -8x = -1
\(\Leftrightarrow\) x = \(\frac{1}{8}\)
Vậy S = {\(\frac{1}{8}\)}
6) \(\frac{12x+1}{4}=\frac{9x+1}{3}-\frac{3-5x}{12}\) Mc : 12
\(\Leftrightarrow\frac{3.\left(12x+1\right)}{12}=\frac{4.\left(9x+1\right)}{12}-\frac{3-5x}{12}\)
\(\Leftrightarrow\) 36x + 3 = 36x + 4 - 3 + 5x
\(\Leftrightarrow\) 36x - 36x - 5x = -3 + 4 - 3
\(\Leftrightarrow\) -5x = -2
\(\Leftrightarrow x=\frac{2}{5}\)
7) \(\frac{x+6}{4}\) - \(\frac{x-2}{6}-\frac{x+1}{3}=0\) Mc : 12
\(\Leftrightarrow\) \(\frac{3.\left(x+6\right)}{12}-\frac{2.\left(x-2\right)}{12}-\frac{4.\left(x+1\right)}{12}=0\)
\(\Leftrightarrow\) 3x + 18 - 2x + 4 - 4x - 4 = 0
\(\Leftrightarrow\) 3x - 2x - 4x = -18 - 4 + 4
\(\Leftrightarrow\) -3x = -18
\(\Leftrightarrow\) x = 6
Vậy S = {6}
8) x\(^2\) - x - 6 = 0
\(\Leftrightarrow\) x\(^2\) + 2x - 3x - 6 = 0
\(\Leftrightarrow\) x.(x + 2) - 3.(x + 2) = 0
\(\Leftrightarrow\) (x - 3).(x + 2) = 0
\(\Leftrightarrow\) x - 3 = 0 hoặc x + 2 = 0
\(\Leftrightarrow\) x = 3 hoặc x = -2
Vậy S = {3; -2}
