Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=6x^4-9x^3+3x^2-4x^3+6x^2-2x+10x^2-15x+5\)
\(=6x^4-13x^3+19x^2-17x+5\)
b: \(=6x^4-\dfrac{9}{4}x^3-\dfrac{9}{2}x^2-\dfrac{8}{3}x^3+x^2+2x-\dfrac{20}{3}x^2+\dfrac{5}{2}x+5\)
\(=6x^4-\dfrac{59}{12}x^3-\dfrac{67}{6}x^2+\dfrac{9}{2}x+5\)
c: \(=3x^4-\dfrac{9}{8}x^3-\dfrac{3}{4}x^2+8x^3-3x^2-6x-\dfrac{4}{3}x^2+\dfrac{1}{2}x+1\)
\(=3x^4-\dfrac{55}{8}x^3-\dfrac{25}{12}x^2-\dfrac{11}{2}x+1\)
\(x^2-\left(y-3\right)^2-4x+4\)
\(=x^2-\left(y^2-6y+9\right)-4x+4\)
\(=x^2-y^2+6y-9-4x+4\)
\(=\left(x^2-4x+4\right)-\left(y^2-6y+9\right)\)
\(=\left(x-2\right)^2-\left(y-3\right)^2\)
\(=\left[\left(x-2\right)-\left(y-3\right)\right]\left[\left(x-2\right)+\left(y-3\right)\right]\)
\(=\left(x-y+5\right)\left(x+y-5\right)\)
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
a, A = (x-1)(x+6) (x+2)(x+3)
= (x^2 + 5x -6 ) (x^2 + 5x + 6)
Đặt t = x^2 +5x
A= (t-6)(t+6)
= t^2 - 36
GTNN của A là -36 khi và ck t= 0
<=> x^2 +5x = 0
<=> x=0 hoặc x=-5
Vậy...
Bài 1:
\(B=\dfrac{4\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\dfrac{\left(x^2-25\right)}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)
\(=\dfrac{4\left(x+3\right)^2}{\left(3x+5-2x\right)\left(3x+5+2x\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{\left(4x+15-x\right)\left(4x+15+x\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{5\left(x-5\right)\left(x+1\right)}-\dfrac{3\left(x+3\right)\left(x+1\right)}{15\left(x+5\right)\left(x+3\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{x+5}{5\left(x+1\right)}-\dfrac{x+1}{5\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+5\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+1\right)^2}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{4\left(x^2+6x+9\right)-\left(x^2+10x+25\right)-\left(x^2+2x+1\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{4x^2+24x+36-x^2-10x-25-x^2-2x-1}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2x^2+12x+10}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x^2+6x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x^2+5x+x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x+5\right)\left(x+1\right)}{5\left(x+5\right)\left(x+1\right)}=\dfrac{2}{5}\)
đc bn , nhg mà đề bài câu a b2 sao tự nhiên lại có " n "
bn xem lại đề đi
+) \(E=x^2-6x+9+x^2-22x+121=2x^2-28x+130\)
\(\Rightarrow2E=4x^2-56x+242=\left(4x^2-56x+196\right)+46=\left(2x-14\right)^2+46\)
Vì \(\left(2x-14\right)^2\ge0\Rightarrow2E=\left(2x-14\right)^2+46\ge46\Rightarrow E\ge23\)
Dấu "=" xảy ra khi x=7
Vậy Emin=23 khi x=7
+) \(F=\frac{-2}{x^2-2x+5}=\frac{-2}{x^2-2x+1+4}=\frac{-2}{\left(x-1\right)^2+4}\)
Vì \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+4\ge4\Rightarrow F=\frac{-2}{\left(x-1\right)^2+4}\le-\frac{2}{4}=-\frac{1}{2}\)
Dấu "=" xảy ra khi x=1
Vậy Fmin=-1/2 khi x=1
+) \(G=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left(x^2-6x+x-6\right)\left(x^2-3x-2x+6\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\)
Đặt x2-5x=t, ta được:
\(G=\left(t-6\right)\left(t+6\right)=t^2-36=\left(x^2-5x\right)^2-36\)
Vì \(\left(x^2-5x\right)^2\ge0\Rightarrow G=\left(x^2-5x\right)^2-36\ge36\)
Dấu "=" xảy ra khi x=0 hoặc x=5
Vậy Gmin=36 khi x=0 hoặc x=5
câu 1:
thực hiện phép chia được thương là 2x2+2x+1
ta có: 2x2+2x+1=2x2+2x+\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)
=2(x2+x+\(\dfrac{1}{4}\))+\(\dfrac{1}{2}\)=2(x+\(\dfrac{1}{2}\))2+\(\dfrac{1}{2}\)
vì 2(x+\(\dfrac{1}{2}\))2\(\ge\)0 nên 2(x+\(\dfrac{1}{2}\))2+\(\dfrac{1}{2}\)\(\ge\)\(\dfrac{1}{2}\)
dấu = xảy ra khi x=\(\dfrac{-1}{2}\)
câu 2:
\(\dfrac{\left(x+2\right)P}{x-2}=\dfrac{\left(x-1\right)Q}{x^2-4}\Leftrightarrow\left(x^2-4\right)\left(x+2\right)P=\left(x-2\right)\left(x-1\right)Q\Leftrightarrow\left(x+2\right)^2P=\left(x-1\right)Q\)
\(\Leftrightarrow\)P=x-1
Q=(x+2)2