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\(\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}=\frac{x}{2\left(x-1\right)}+\frac{x^2+1}{2\left(1-x^2\right)}\)
\(=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x^2-1\right)}\)
\(=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x\left(x+1\right)-\left(x^2+1\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x-1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{1}{2\left(x+1\right)}\)
a,
\(\Leftrightarrow A=\left(\frac{x+1}{\left(x+1\right)\left(x-1\right)}+\frac{x}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(\Leftrightarrow A=\frac{2x+1}{\left(x+1\right)\left(x-1\right)}\cdot\frac{\left(x+1\right)^2}{2x+1}\)
\(\Leftrightarrow A=\frac{x+1}{x-1}\)
b, dùng máy tính kq là-3
\(3x^3-\frac{3}{2}x^2-x^3-\frac{1}{2}x+\frac{1}{2}x+2=2x^3-\frac{3}{2}x^2+2\)
\(2x^2-10x-3x-2x^2=26\)
-13x=26
x=-2
1. \(B=\left(x-2\right)\left(x+2\right)\left(x+3\right)-\left(x+1\right)^3\)
\(=\left(x^2-4\right)\left(x+3\right)-\left(x^3+3x^2+3x+1\right)\)
\(=x^3+3x^2-4x-12-x^3-3x^2-3x-1\)
\(=-7x-13\)
2. \(64-x^2-y^2+2xy=64-\left(x^2+y^2-2xy\right)\)
\(=64-\left(x-y\right)^2=\left(8+x-y\right)\left(8-x+y\right)\)
3. \(2x^3-x^2+2x-1=0\)
\(\Leftrightarrow x^2.\left(2x-1\right)+\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2+1\right)=0\)
Vì \(x^2\ge0\)\(\Rightarrow x^2+1>0\)
\(\Rightarrow2x-1=0\)\(\Rightarrow2x=1\)\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
Bài 1.
B = ( x - 2 )( x + 2 )( x + 3 ) - ( x + 1 )3
= ( x2 - 4 )( x + 3 ) - ( x3 + 3x2 + 3x + 1 )
= x3 + 3x2 - 4x - 12 - x3 - 3x2 - 3x - 1
= -7x - 13
Bài 2.
64 - x2 - y2 + 2xy
= 64 - ( x2 - 2xy + y2 )
= 82 - ( x - y )2
= ( 8 - x + y )( 8 + x - y )
Bài 3.
2x3 - x2 + 2x - 1 = 0
<=> ( 2x3 - x2 ) + ( 2x - 1 ) = 0
<=> x2( 2x - 1 ) + 1( 2x - 1 ) = 0
<=> ( 2x - 1 )( x2 + 1 ) = 0
<=> \(\orbr{\begin{cases}2x-1=0\\x^2+1=0\end{cases}}\Leftrightarrow x=\frac{1}{2}\)( vì x2 + 1 ≥ 1 > 0 ∀ x )
\(\frac{x^4+x^3-x-1}{x^4+x^3+2x^2+x+1}\)
\(=\frac{\left(x^4+x^3\right)-\left(x+1\right)}{\left(x^4+x^3\right)+\left(x+1\right)+2x^2}\)
\(=\frac{x^3\left(x+1\right)-\left(x+1\right)}{x^3\left(x+1\right)+\left(x+1\right)+2x^2}\)
\(=\frac{\left(x^3-1\right)\left(x+1\right)}{\left(x^3+1\right)\left(x+1\right)+2x^2}\)
\(=\frac{\left(x^3-1\right)}{\left(x^3+1\right)+2x^2}\)
\(=\frac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)+2x^2}\)
a) \(\left(3x-5\right)\left(2x+3\right)-\left(2x-3\right)\left(3x+7\right)-2x\left(x-4\right)\)
\(=\left(6x^2-x-15\right)-\left(6x^2+5x-21\right)-\left(2x^2-8x\right)\)
\(=6x^2-x-15-6x^2-5x+21-2x^2+8x\)
\(=-2x^2+2x+6\)
\(=-2\left(x^2-x-3\right)\)
b) \(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
\(=\left(x^2+2\right)^2-\left(x^2-4\right)\left(x^2+4\right)\)
\(=\left(x^2+2\right)^2-\left(x^4-16\right)\)
\(=\left(x^4+4x^2+4\right)-\left(x^4-16\right)\)
\(=x^4+4x^2+4-x^4+16\)
\(=4x^2+20\)
\(=4\left(x^2+5\right)\)
c) \(\left(2x-y\right)^2-2\left(x+3y\right)^2-\left(1+3x\right)\left(3x-1\right)\)
\(=\left(4x^2-4xy+y^2\right)-2\left(x^2+6xy+9y^2\right)-\left(9x^2-1\right)\)
\(=4x^2-4xy+y^2-2x^2-16xy-18y^2-9x^2+1\)
\(=-7x^2-20xy-17y^2+1\)
d) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)
\(=\left(x^6-3x^4+3x^2-1\right)-\left(x^6-1\right)\)
\(=x^6-3x^4+3x^2-1-x^6+1\)
\(=-3x^4+3x^2\)
\(=-3x^2\left(x^2-1\right)\)
\(=-3x^2\left(x-1\right)\left(x+1\right)\)
e) \(\left(2x-1\right)^2-2\left(4x^2-1\right)+\left(2x+1\right)^2\)
\(=\left(2x-1\right)^2-2\left(2x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\)
\(=\left[\left(2x-1\right)-\left(2x+1\right)\right]^2\)
\(=\left(2x-1-2x-1\right)^2\)
\(=\left(-2\right)^2=4\)
g) \(\left(x-y+z\right)^2+\left(y-z\right)^2-2\left(x-y+z\right)\left(z-y\right)\)
\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left(x-y+z+y+z\right)^2\)
\(=\left(x+2z\right)^2\)
h) \(\left(2x+3\right)^2+\left(2x+5\right)^2-\left(4x+6\right)\left(2x+5\right)\)
\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\)
\(=\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\)
\(=\left(2x+3-2x-5\right)^2\)
\(=\left(-2\right)^2=4\)
i) \(5x^2-\dfrac{10x^3+15x^2-5x}{-5x}-3\left(x+1\right)\)
\(=5x^2-\dfrac{-5x\left(-2x^2-3x+1\right)}{-5x}-3\left(x+1\right)\)
\(=5x^2-\left(-2x^2-3x+1\right)-3\left(x+1\right)\)
\(=5x^2+2x^2+3x-1-3x-3\)
\(=7x^2-4\)
a: Ta có: \(\left(x-2\right)\left(x+3\right)-\left(x+\frac12\right)^2\)
\(=x^2+3x-2x-6-\left(x^2+x+\frac14\right)\)
\(=x^2+x-6-x^2-x-\frac14=-6-\frac14=-\frac{25}{4}\)
b: Ta có: \(\left(2x-1\right)\left(2x+2\right)-\left(2x+\frac12\right)^2\)
\(=4x^2+4x-2x-2-4x^2-2x-\frac14\)
\(=-2-\frac14=-\frac94\)
\(\left(\right. x - 2 \left.\right) \left(\right. x + 3 \left.\right) - \left(\right. x + \frac{1}{2} \left.\right)^{2}\)
\(=\left(\right.x^2+x-6\left.\right)-\left(\right.x^2+x+\frac{1}{4}\left.\right)\) \(= x^{2} + x - 6 - x^{2} - x - \frac{1}{4}\)\(=-6-\frac14\)
\(=-\frac{25}{4}\)
\(\left(\right. 2 x - 1 \left.\right) \left(\right. 2 x + 2 \left.\right) - \left(\right. 2 x + \frac{1}{2} \left.\right)^{2}\)
\(=\left(\right.4x^2+2x-2\left.\right)-\left(\right.4x^2+2x+\frac{1}{4}\left.\right)\) \(= 4 x^{2} + 2 x - 2 - 4 x^{2} - 2 x - \frac{1}{4}\)\(=\) \(-2-\frac14\)
\(-\frac94\)