Đặt \(x^2+x=t\)
\(\Rightarrow t^2+4t-12=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=2\\x^2+x=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+6=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)
(x+1)(x+2)(x+3)(x+4)=40
\(\left(x^2+x\right)4\left(x^2+x\right)=12\)
Tính:
\(a,\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(b,\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
Bài 1:cho phương trình
a,\(\left(x-1\right)^3-x\left(x-1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)
b,\(\dfrac{\left(x+10\right)\left(x+4\right)}{12}-\dfrac{\left(x+4\right)\left(2-x\right)}{4}=\dfrac{\left(x+10\right)\left(x-2\right)}{3}\)
c,\(\dfrac{2\left(x-3\right)}{7}+\dfrac{x-5}{3}=\dfrac{13x+4}{21}\)
d,\(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{5}\)
e,\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
Chứng minh biểu thức sau không phụ thuộc vào giá trị của biến :
\(A=x.\left(5x-3\right)-x^2.\left(x-1\right)+x.\left(x^2-6x\right)-10+3x+x.\left(x^2+x+1\right)-x^2.\left(x+1\right)-x+5\)
\(B=3.\left(2x-1\right)-5.\left(x-3\right)+6.\left(3x-4\right)-19x+x.\left(3x+12\right)-\left(7x-20\right)+x^2.\left(2x-3\right)-x.\left(2x^2+5\right)\)
Giải phương trình:
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+12\right)=165x^2\)
Giaỉ các phương trình sau:
a) \(\left(x^2+11x+12\right)\left(x^2+9x+20\right)\left(x^2+13x+42\right)=36\left(x^2+11x+30\right)\left(x^2+11x+31\right)\)
b) \(20\left(\frac{x-2}{x+1}\right)^2-5\left(\frac{x+2}{x-1}\right)^2+48\cdot\frac{x^2-4}{x^2-1}=0\)
Bài 2: tìm x, biết:
a,\(\left(2x-1\right)^2_{ }-19=45\)
b,\(\left(x-3\right)^2-x\left(x-7\right)=12\)
c,\(x^2-4+3x\left(x-2\right)=0\)
d,\(x^2-9=3\left(x-3\right)\)
e,\(3\left(3x^2+1\right)=6-2\left(3x+2\right)\)
h,\(2x\left(x+3\right)-4\left(x+3\right)=0\)
SAVE ME!!!THANK YOU
Thực hiện phép tính:
\(a,\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)
\(b,\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
a, (x-1)3 - x(x-1)2 = 5(2-x) - 11(x+2)
b, (x-2)3 + (3x-1)(3x+1) = (x+1)3
c, \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\)
d, \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)
e, \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
a) rút gọn biểu thức: \(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2x^2y+2xy^2\)
b) tìm x biết: \(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=12\)
