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Câu 1:
\(Tacó\)
\(\frac{2}{2x-1}+\frac{4x^2+1}{4x^2-1}-\frac{1}{2x+1}=\frac{2}{2x-1}+\frac{4x^2+1}{\left(2x+1\right)\left(2x-1\right)}-\frac{1}{2x+1}\)
\(=\frac{4x+2}{\left(2x+1\right)\left(2x-1\right)}+\frac{4x^2+1}{\left(2x+1\right)\left(2x-1\right)}-\frac{2x-1}{\left(2x+1\right)\left(2x-1\right)}\)
\(=\frac{4x+2+4x^2+1-2x+1}{\left(2x+1\right)\left(2x-1\right)}=\frac{2x\left(2x+1\right)+4}{\left(2x+1\right)\left(2x-1\right)}=\frac{2x+4}{2x-1}\)
\(b,x=\frac{1}{2}\Rightarrow2x-1=0\left(loại\right)\)
..... 2 câu sau easy
Theo mk nghĩ thì đề bài fải như thế này:
\(\left(4x^5+2x^4+4x^3-x^2-1\right):\left(2x^3+x-1\right)\)
Kết quả của phép chia trên là: \(2x^2+x+1\)
Ta có: \(2x^2+x+1=2\left(x^2+\frac{1}{2}x+\frac{1}{2}\right)\)
\(=2\left(x^2+\frac{1}{2}x+\frac{1}{16}+\frac{7}{16}\right)\)
\(=2\left(x+\frac{1}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\forall x\)
=> Min = 7/8 tại \(2\left(x+\frac{1}{4}\right)^2=0\Rightarrow x=-\frac{1}{4}\)
=.= hok tốt!!
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)
\(Q=\left(\frac{2x-x^2}{2x^2+8}-\frac{2x^2}{x^3-2x^2+4x-8}\right).\left(\frac{2}{x^2}+\frac{1-x}{x}\right)\)
\(\Leftrightarrow Q=\left(\frac{x\left(2-x\right)}{2\left(x^2+4\right)}-\frac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right).\frac{2+x\left(1-x\right)}{x^2}\)
\(\Leftrightarrow Q=\frac{-x\left(x-2\right)^2-4x^2}{2\left(x-2\right)\left(x^2+4\right)}.\frac{2+x-x^2}{x^2}\)
\(\Leftrightarrow Q=\frac{x\left(x^2-4x+4\right)-4x^2}{2\left(x-2\right)\left(x^2+4\right)}.\frac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(\Leftrightarrow Q=\frac{x\left(x^2+4\right)}{2\left(x^2+4\right)}.\frac{x+1}{x^2}\)
\(\Leftrightarrow Q=\frac{x+1}{2x}\)
b) Để \(Q\inℤ\)
\(\Leftrightarrow x+1⋮2x\)
\(\Leftrightarrow2\left(x+1\right)⋮2x\)
\(\Leftrightarrow2x+2⋮2x\)
\(\Leftrightarrow2⋮2x\)
\(\Leftrightarrow2x\inƯ\left(2\right)\)
\(\Leftrightarrow2x\in\left\{\pm1;\pm2\right\}\)
\(\Leftrightarrow x\in\left\{\pm\frac{1}{2};\pm1\right\}\)
Mà \(x\inℤ\)
Vậy để \(Q\inℤ\Leftrightarrow x\in\left\{1;-1\right\}\)
\(x^2-\left(y-3\right)^2-4x+4\)
\(=x^2-\left(y^2-6y+9\right)-4x+4\)
\(=x^2-y^2+6y-9-4x+4\)
\(=\left(x^2-4x+4\right)-\left(y^2-6y+9\right)\)
\(=\left(x-2\right)^2-\left(y-3\right)^2\)
\(=\left[\left(x-2\right)-\left(y-3\right)\right]\left[\left(x-2\right)+\left(y-3\right)\right]\)
\(=\left(x-y+5\right)\left(x+y-5\right)\)
1.
x2 - ( y - 3 )2 - 4x + 4
= ( x2 - 4x + 4 ) - ( y - 3 )2
= ( x - 2 )2 - ( y - 3 )2
= [ ( x - 2 ) - ( y - 3 ) ][ ( x - 2 ) + ( y - 3 ) ]
= ( x - 2 - y + 3 )( x - 2 + y - 3 )
= ( x - y + 1 )( x + y - 5 )
2.
a) Ta có : 2x4 + 8x3 + 9x2 - 4x - 5
= 2x4 + 10x2 - x2 + 8x3 - 4x - 5
= ( 2x4 - x2 ) + ( 8x3 - 4x ) + ( 10x2 - 5 )
= x2( 2x2 - 1 ) + 4x( 2x2 - 1 ) + 5( 2x2 - 1 )
= ( 2x2 - 1 )( x2 + 4x + 5 )
=>(2x4 + 8x3 + 9x2 - 4x - 5) : ( 2x2 - 1 ) = x2 + 4x + 5
b) Ta có : x2 + 4x + 5 = ( x2 + 4x + 4 ) + 1 = ( x + 2 )2 + 1 ≥ 1 > 0 ∀ x
=> đpcm
Bài 1:
a. A = x^2 - 5x - 1
\(=x^2-5x+\frac{25}{4}-\frac{29}{4}\)
\(=x^2-5x+\left(\frac{5}{2}\right)^2-\frac{29}{4}\)
\(=\left(x-\frac{5}{2}\right)^2-\frac{29}{4}\ge0-\frac{29}{4}=-\frac{29}{4}\)
Dấu = khi x=5/2
Vậy MinC=-29/4 khi x=5/2
2. Tìm x:
a. ( 2x - 3 )^2 - ( 4x + 1 )( 4x - 1 ) = ( 2x - 1 ).( 3 - 7x )
=>4x2-12x+9+1-16x2=-14x2+13x-3
=>-12x2-12x+10=-14x2+13x-3
=>2x2-25x+13=0
\(\Rightarrow2\left(x-\frac{25}{4}\right)^2-\frac{521}{8}=0\)
\(\Rightarrow\left(x-\frac{25}{4}\right)^2=\frac{521}{16}\)
\(\Rightarrow x-\frac{25}{4}=\pm\sqrt{\frac{521}{16}}\)
\(\Rightarrow x=\frac{25}{4}\pm\frac{\sqrt{521}}{4}\)
c. 4.( x - 3 ) - ( x + 2 ) = 0
=>4x-12-x-2=0
=>3x-14=0
=>3x=14
=>x=14/3