x+2-(x-5)=(7+2x)(3x-1)-(6x-5)=1
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Trả lời:
Bài 1:
a, \(-2x\left(x-3\right)=-2x^2+6x\)
b, \(-4xy\left(x-3xy^2\right)=-4x^2y+12x^2y^3\)
d, \(\left(x-3\right)\left(x-4\right)=x^2-4x-3x+12=x^2-7x+12\)
e, \(\left(x-y\right)\left(x+y\right)=x^2-y^2\)
Bài 2:
a, \(2\left(x-1\right)-3\left(2x-2\right)=2x-2-6x+6=-4x+4\)
b, \(x\left(x-y\right)-y\left(y-x\right)=x^2-xy-y^2+xy=x^2-y^2\)
c, \(3x\left(x+1\right)-2x\left(x-3\right)=3x^2+3x-2x^2+6x=x^2+9x\)
d, \(-4x\left(x-1\right)+\left(2x+1\right)\left(2x+5\right)=-4x^2+4x+4x^2+10x+2x+5=16x+5\)
e, \(\left(x-2\right)\left(x-3\right)-\left(x-1\right)\left(x-4\right)=x^2-5x+6-x^2+5x-4=2\)
f, \(\left(x-4\right)^2=x^2-8x+16\)
Bài 3:
a, \(4\left(x+3\right)-5\left(x-1\right)=-7\)
\(\Leftrightarrow4x+12-5x+5=-7\)
\(\Leftrightarrow17-x=-7\)
\(\Leftrightarrow-x=-24\)
\(\Leftrightarrow x=24\)
Vậy x = 24 là nghiệm của pt.
b, \(x\left(x-5\right)-\left(x+2\right)x=9\)
\(\Leftrightarrow x^2-5x-x^2-2x=9\)
\(\Leftrightarrow-7x=9\)
\(\Leftrightarrow x=-\frac{9}{7}\)
Vậy x = - 9/7 là nghiệm của pt.
a) 4x2 + 4xy + y2
= (2x + y)2
b) (2x + 1)2 - (x - 1)2
= (2x + 1 + x - 1)(2x + 1 - x + 1)
= 3x(x + 2)
c) 9 - 6x + x2 - y2
= (x2 - 6x + 9) - y2
= (x - 3)2 - y2
= (x - y - 3)(x + y - 3)
d) (-x - 2) + 3(x2 - 4)
= -(x + 2) + 3(x - 2)(x + 2)
= (x + 2)(3x - 7)
e) 5x2- 10xy2 + 5y4
= 5(x2 - 2xy2 + y4)
= 5(x - y2)2
f) \(\frac{x^4}{2}-2x^2=\frac{x^4-4x^2}{2}=\frac{x^2\left(x^2-4\right)}{2}=\frac{x^2\left(x-2\right)\left(x+2\right)}{2}\)
g) 49(x - 4)2 - 9(x + 2)2
= (7x - 28)2 - (3x + 6)2
= (10x - 22)(4x - 34)
h) (x2 + y2 - 5)2 - 2(xy + 2)2
= \(\left(x^2+y^2-5\right)^2-\left(\sqrt{2}xy+2\sqrt{2}\right)^2\)
\(=\left(x^2+y^2+2\sqrt{xy}+2\sqrt{2}-5\right)\left(x^2+y^2-\sqrt{2}xy-2\sqrt{2}-5\right)\)
a, \(\left(y-2\right)\left(y+2\right)\left(y^2+4\right)-\left(y+3\right)\left(y-3\right)\left(y^2+9\right)\)
\(=\left(y^2-4\right)\left(y^2+4\right)-\left(y^2-9\right)\left(y^2+9\right)\)
\(=y^4-16-y^4+81=65\)
b, \(2\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)-2\left(x^6-y^6\right)\)
\(=2\left(x^3-y^3\right)\left(x^3+y^3\right)-2\left(x^6-y^6\right)\)
\(=2\left(x^6-y^6\right)-2\left(x^6-y^6\right)=0\)
a) A = ( x - 2y )3 + ( x + 2y )3 - 2x ( x2 + y )=
= x3 - 6x2y + 12xy2 - 8y3 + x3 + 6x2y + 12xy2 + 8y3 - 2x3 - 2xy
= 24xy2 - 2xy
b) B = ( x - 1 )( x + y ) ( x - y ) - x2( x - 1 )=
= ( x -1 )( x2 - y2 ) - x2 ( x - 1 )
= ( x - 1 )( x2 - y2 - x2 )
= -y2 ( x - 1 )
c ) C = ( x + 2)2 - 2( x + 2 )( x - 8 ) + ( x - 8 ) 2 =
= ( x + 2 - x + 8 ) 2
= 102
= 100
HOk tốt!!!!!!!!!!
Bài 2 : a) \(2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}}\)
b) \(2\left(2x-1\right)+6x\left(2x-1\right)=0\Leftrightarrow\left(2x-1\right)\left(2+6x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\2+6x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=1\\6x=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{2}{6}=-\frac{1}{3}\end{cases}}}\)
c) \(\left(x-3\right)^2-\left(2x+6\right)^2=0\Leftrightarrow\left(x-3-2x-6\right)\left(x-3+2x+6\right)=0\)
\(\Leftrightarrow\left(-x-9\right)\left(3x+3\right)=0\Leftrightarrow\orbr{\begin{cases}-x-9=0\\3x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-9\\x=-1\end{cases}}}\)
Tự KL cho các phần
Cho biểu thức P = (4x−x21−4x2 1−x):(4x2−x41−4x2 +1)
a) Rút gọn P
= (x^21+4x^2-3x)/(x^41-1)
b) Tìm x để P =< 0
b) Tìm x để P ≤0
ĐKXĐ : \(\hept{\begin{cases}x^2-2\ge0\\7-x^2\ge0\end{cases}}\Leftrightarrow\sqrt{2}\le x\le\sqrt{7}\)
Áp dụng bất đẳng thức Bunhiacopxki
Ta có N = \(\sqrt{x^2+1}+\sqrt{2\left(x^2-2\right)}+\sqrt{3\left(7-x^2\right)}\)
\(=1.\sqrt{x^2+1}+1.\sqrt{2\left(x^2-2\right)}+1.\sqrt{3\left(7-x^2\right)}\)
\(\le\sqrt{\left(1^2+1^2+1^2\right)\left[x^2+1+2\left(x^2-2\right)+3\left(7-x^2\right)\right]}\)
\(=\sqrt{3.18}=\sqrt{54}\)
Dấu "=" xảy ra <=> \(\frac{1}{x^2+1}=\frac{1}{2\left(x^2-2\right)}=\frac{1}{3\left(7-x^2\right)}\)
<=> x2 + 1 = 2x2 - 4
<=> x = \(\sqrt{5}\)(tm)
Vậy Max N = \(\sqrt{54}\Leftrightarrow x=\sqrt{5}\)
chịu bó tay
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