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a, ĐKXĐ: \(x\ne0;x\ne\pm1\)
\(P=\left(\frac{2x}{x^2-1}+\frac{x-1}{2x+2}\right):\frac{x+1}{2x}=\left(\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{2\left(x+1\right)}\right):\frac{x+1}{2x}\)
\(=\left(\frac{2x.2}{2\left(x-1\right)\left(x+1\right)}+\frac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\right):\frac{x+1}{2x}\)
\(=\frac{4x+x^2-2x+1}{2\left(x-1\right)\left(x+1\right)}:\frac{x+1}{2x}=\frac{x^2+2x+1}{2\left(x-1\right)\left(x+1\right)}=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}\cdot\frac{2x}{x+1}=\frac{x}{x-1}\)
b,Để \(P=2\Leftrightarrow\frac{x}{x-1}=2\Leftrightarrow2\left(x-1\right)=x\Leftrightarrow2x-2-x=0\Leftrightarrow x-2=0\Leftrightarrow x=2\left(tmđk\right)\)
Vậy để P=2 <=> x=2
a) (2x+1)^2+2(4x^2-2)+(2x-1)^2=4x2+4x+1+8x2-4+4x2-4x+1=16x2-2
a) \(A=3x\left(x^2-2x+3\right)-x^2.\left(3x-2\right)+5\left(x^2-x\right)\)
\(=3x^3-6x^2+9x-3x^3+2x^2+5x^2-5x\)
\(=x^2+4x\)
Thay \(x=5\)vào biểu thức ta có: \(A=5^2+4.5=25+20=45\)
b) \(B=x\left(x^2+xy+y^2\right)-y\left(x^2+xy+y^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)
Thay \(x=10\); \(y=-1\)vào biểu thức ta có:
\(B=10^3-\left(-1\right)^3=1000+1=1001\)
a) = x3 + 9x2 + 27x + 27 - 9x3 -6x2 - x + 8x3 +1 -3x2 =54
26x +28 = 54
26x = 54-28 = 26
x = 1
b) = x3 - 9x2 + 27x -27 - x3 +27 +6x2 + 12x + 6 +3x2 = -33
39x +6 = -33
39x = -33-6 = -39
x = -1
a) 2(2x+1)(3x-1)+(2x+1)2+(3x-1)2 = (2x+1+3x-1)2=(5x)2
b) (x-3)(x+3) - (x-3)2= x2-9-x2+6x-9=6x
c) (x2 -1)(x+2) - (x-2)(x2+2x+4)= x3+2x2-x-2-x3+8=2x2-x+6
Đúng hông tar, hình như lag lag chỗ nào đó thì phải á '-'?
Bài làm :
a)=[(2x+1) + (3x-1)]2 = (2x+1+3x-1)2 = (5x)2 = 25x2
b)=(x-3) . [(x+3) - (x-3)] = (x-3)(x+3-x+3) =(x-3) . 6 = 6x - 18
c)= (x2 -1)(x+2) - (x-2)(x+2)2 =(x+2)[(x2 - 1) - (x2 -4 )] = (x+2). 3 = 3x + 6
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
1) -3x( x + 2 )2 + ( x + 3 )( x - 1 )( x + 1 ) - ( 2x - 3 )2
= -3x( x2 + 4x + 4 ) + ( x + 3 )( x2 - 1 ) - ( 4x2 - 12x + 9 )
= -3x3 - 12x2 - 12x + x3 + 3x2 - x -3 - 4x2 + 12x - 9
= ( -3x3 + x3 ) + ( -12x2 + 3x2 - 4x2 ) + ( -12x - x + 12x ) + ( -3 - 9 )
= -2x3 - 13x2 - x - 12
2) ( x - 3 )( x + 3 )( x + 2 ) - ( x - 1 )( x2 - 3 ) - 5x( x + 4 )2 - ( x - 5 )2
= ( x2 - 9 )( x + 2 ) - ( x3 - x2 - 3x + 3 ) - 5x( x2 + 8x + 16 ) - ( x2 - 10x + 25 )
= x3 + 2x2 - 9x - 18 - x3 + x2 + 3x - 3 - 5x3 - 40x2 - 80x - x2 + 10x - 25
= ( x3 - x3 - 5x3 ) + ( 2x2 + x2 - 40x2 - x2 ) + ( -9x + 3x - 80x + 10x ) + ( -18 - 3 - 25 )
= -5x3 - 38x2 - 76x - 46
3) 2x( x - 4 )2 - ( x + 5 )( x - 2 )( x + 2 ) + 2( x + 5 )2 + ( x - 5 )2
= 2x( x2 - 8x + 16 ) - ( x + 5 )( x2 - 4 ) + 2( x2 + 10x + 25 ) + x2 - 10x + 25
= 2x3 - 16x2 + 32x - ( x3 + 5x2 - 4x - 20 ) + 2x2 + 20x + 50 + x2 - 10x + 25
= 2x3 - 16x2 + 32x - x3 - 5x2 + 4x + 20 + 2x2 + 20x + 50 + x2 - 10x + 25
= ( 2x3 - x3 ) + ( -16x2 - 5x2 + 2x2 + x2 ) + ( 32x + 4x + 20x - 10x ) + ( 20 + 50 + 25 )
= x3 - 18x2 + 46x + 95
a) P=\(\frac{x^2+x}{x^2-2x+1}:\left(\frac{x+1}{x}-\frac{1}{1-x}+\frac{2-x^2}{x^2-x}\right)\left(x\ne\pm1;x\ne0\right)\)
P=\(\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x\left(x-1\right)}\right)\)
P=\(\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}+\frac{x}{x\left(x-1\right)}+\frac{2-x^2}{x\left(x-1\right)}\right)\)
P=\(\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\frac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
P=\(\frac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\frac{x\left(x-1\right)}{x+1}=\frac{x\left(x+1\right)x\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}=\frac{x^2}{x-1}\)
vậy P=\(\frac{x^2}{x-1}\left(x\ne\pm1;x\ne0\right)\)
b) ta có \(P=\frac{x^2}{x-1}\left(x\ne\pm1;x\ne0\right)\)
để P<1 => \(\frac{x^2}{x-1}< 1\)
\(\Leftrightarrow\frac{x^2}{x-1}-1< 0\Leftrightarrow\frac{x^2-x+1}{x-1}< 0\Leftrightarrow\frac{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}{x-1}< 0\)
thấy \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
vậy để P-1<0 thì x-1<0
=> x<1. kết hợp với điều kiện ta được \(\hept{\begin{cases}x< 1\\x\ne0\\x\ne-1\end{cases}}\)thì P<1