Rút gọn biểu thức sau:
A= (x - 3)2 - 2.( 2 - x).(x + 2) - (x+1)2.
help me please!!!!!
Rút gọn biểu thức sau
a)(x^2-1)^3-(x^4+x^2+1)(x^2-1)
b)(x^4-3x^2+9)(x^2+3-(3+x^2)^3
c)(x-3)^3-(x-3)(x^2+3x+9)+6(x+1)^2
Help me chiều nay mk đi học rồi
a) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=\left(x^2-1\right)\left[\left(x^2-1\right)^2-\left(x^4+x^2+1\right)\right]\)
\(=\left(x^2-1\right)\left(x^4-2x^2+1-x^4-x^2-1\right)=\left(x^2-1\right)\left(-3x^2\right)\)
\(=-3x^4+3x^2=3\left(x^2-x^4\right)=3\left(x-x^2\right)\left(x+x^2\right)=\left(3x-3x^2\right)\left(x+x^2\right).\)
b)\(\left(x^4-3x^2+9\right)\left(x^2+3-\left(3+x^2\right)\right)^3=\left(x^4-3x^2+9\right).0^3=0\)
c)\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=\left(x-3\right)^3-\left(x^3-3^3\right)+6\left(x^2+2x+1\right)\)
\(=\left(x-3\right)^3-\left[\left(x-3\right)^3+3.x.3.\left(x-3\right)\right]+6x^2+12x+6\)
\(=6x^2+12x+6-9x\left(x-3\right)=6x^2+12x+6-9x^2+27x\)
\(=39x-3x^2+6=3\left(13x-x^2+2\right).\)
Rút gọn biểu thức A = 2 - x +1/3 |3x+2|
HELP ME !
Tìm điều kiện xác định củ A và rút gọn biểu thức
A = \(\left(\frac{1}{\sqrt{x}-2}-\frac{\sqrt{x}}{x-4}\right).\left(2x^2-9x+4\right)\)
--------Help me please !------Thank you
Rút gọn biểu thức:
a, ( x+1 )^2 - ( x-1 )^2 - 3 . ( x + 1 ) . ( x - 1)
b, 5 . ( x + 2 ) . ( x - 2 ) - 1/2 . ( 6 - 8x ) + 17
HELP ME
a) \(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+3\)
\(=4x+3\)
b) \(5\left(x+2\right)\left(x-2\right)-\frac{1}{2}\left(6-8x\right)+17\)
\(=5\left(x^2-4\right)-3+4x+17\)
\(=5x^2-20-3+4x+17\)
\(=5x^2-6+4x\)
4x^2-28x+49
Rút gọn r tính giá trị của biểu thức khi x = 4 help me please !!
\(4x^2-28x+49=\left(2x\right)^2-2\cdot2x\cdot7+7^2=\left(2x-7\right)^2\)
Khi x=4 thì \(4x^2-28x+49=\left(2x-7\right)^2=\left(2\cdot4-7\right)^2=1\)
Rút gọn biểu thức sau: 2Q=2|x+1|-3|x-1|
Help me. Mai mình phải nộp rồi.
Rút gọn:
a) -3x(x+2)^2 + (x+3)(x-1)(x+1) - (2x-3)^2
b) (x-3)(x+3)(x+2)- (x-1)(x2-3) - 5x(x+4)2-(x-5)2
please help me!!
Cho biểu thức \(A=\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\) với x khác -1 và 1 . Rút gọn .
Help me !
\(A=\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\)
\(=\frac{\left(1-x\right)\left(1+x+x^2\right)-x+x^2}{1-x}.\frac{\left(1-x\right)-x^2\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}\)
\(=\frac{\left(1-x\right)\left(1+x+x^2\right)-x\left(1-x\right)}{1-x}.\frac{\left(1-x\right)\left(1-x^2\right)}{\left(1-x\right)\left(1+x\right)}\)
\(=\frac{\left(1-x\right)\left(1+x^2\right)}{1-x}.\frac{\left(1-x\right)\left(1-x\right)\left(1+x\right)}{\left(1-x\right)\left(1+x\right)}\)
\(=\left(1+x^2\right)\left(1-x\right)\)
\(=-x^3+x^2-x+1\)
Ta có : \(A=\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\)
\(=\left(\frac{\left(1-x\right)\left(1+x+x^2\right)}{\left(1-x\right)}-x\right):\frac{\left(1-x\right)\left(1+x\right)}{\left(1-x\right)-\left(x^2-x^3\right)}\)
\(=\left(\left(1+x+x^2\right)-x\right):\frac{\left(1-x\right)\left(1+x\right)}{\left(1-x\right)-x^2\left(x-1\right)}\)
\(=\left(1+x^2\right):\frac{\left(1-x\right)\left(1+x\right)}{\left(1-x\right)\left(1-x^2\right)}\)
\(=\left(1+x^2\right):\frac{\left(1-x\right)\left(1+x\right)}{\left(1-x\right)\left(1-x\right)\left(x+1\right)}\)
\(=\left(1+x^2\right):\frac{1}{1-x}\)
\(=\left(1+x^2\right)\left(1-x\right)\)
RÚT GỌN BIỂU THỨC
D=2.(1+x)-5.|2-x|
help me