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\(2.A=x\left(x^2-y\right)-x^2\left(x+y\right)+y\left(x^2-x\right)=x^3-xy-x^3-x^2y+x^2y-xy=-2xy\\ Thayx=\frac{1}{2};y=-100vàoAđược:A=-2.\frac{1}{2}.\left(-100\right)=100\)
\(3.x\left(5-2x\right)+2x\left(x-1\right)=15\Leftrightarrow5x-2x^2+2x^2-2x=15\Leftrightarrow3x=15\Leftrightarrow x=5\)
bài 1:
a)\(A=x^3+y^3+xy=1^3+\left(-1\right)^3+1.\left(-1\right)=1-1-1=-1\)
b)\(B=\sqrt{x^2+y^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=\left|10\right|=10\)
c)\(C=10x+10y+15=10\left(x+y\right)+15=10.1+15=25\)
d)\(D=x^2y+y^2x+5=xy\left(x+y\right)+5=xy.0+5=5\)
e)\(E=4x+7x^2y^2+3y^4+5y^2=?????\)
Bài 2:
bạn chỉ cần tìm nhân tử chung r gộp lại dưới dạng tích
VD: 10x+5xy=5x(2+y)
a) x2 - 5x - y2 -5y
= ( x2 - y2 ) + ( -5x - 5y)
= ( x - y ) ( x + y) - 5( x + y )
= ( x + y ) ( x - y -5)
b) x3 + 2x2 - 4x - 8
= x2 ( x + 2 ) - 4 ( x + 2 )
= ( x +2 ) ( x2 -4 )
= ( x+2)2 ( x-2)
Bai 2 :
a, \(A=\left(x+3\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)
\(=x^2+6x+9+x^2-4x+4-2\left(x^2-2x+3x-6\right)\)
\(=2x^2+2x+13-2x^2-2x+12=25\)
b, \(B=\left(x-2\right)^2-x\left(x-1\right)\left(x-3\right)+3x^2-9x+8\)
\(=x^2-4x+4-x\left(x^2-3x-x+3\right)+3x^2-9x+8\)
\(=4x^2-13x+12-x^3+4x^2-3x=-16x+12-x^3\)
Bài 1 :
\(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
Vậy \(MIN_A=-36\) . Dấu \("="\) xảy ra khi \(x^2+5x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Bài 2 :
a ) \(x+y=5\Rightarrow\left(x+y\right)^2=25\)
\(\Leftrightarrow x^2+2xy+y^2=25\)
\(\Leftrightarrow x^2+y^2=25-2.6=13\)
\(B=x^2-4x+1\)
\(B=x^2-4x+4-3\)
\(B=\left(x-2\right)^2-3\ge-3\)
"="<=>x=2
\(C=\dfrac{-4}{x^2-4x+10}\)
Ta có:\(x^2-4x+10=x^2-4x+4+6=\left(x-2\right)^2+6\ge6\)
\(\Rightarrow\dfrac{-4}{x^2-4x+10}\ge-\dfrac{4}{6}=-\dfrac{2}{3}\)
"="<=>x=2
D\(\ge-\dfrac{8}{3}\)<=>x=0,5(tương tự)
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
a) 5x - 15y = 5(x - 3y)
b) \(\dfrac{3}{5}\)x2 + 5x4 - x2 - y
= \(\dfrac{3}{5}\)x2 + 5x2.x2 - x2 - y
= x2(\(\dfrac{3}{5}\) + 5x2 -1) - y
c) 14x2y2 - 21xy2 + 28x2y
= 7xy.xy - 7xy.3y + 7xy.4x
= 7xy(xy - 3y + 4x)
= 7xy[(xy - 3y) + 4x]
= 7xy[y(x - 3) +4x]
d) \(\dfrac{2}{7}x\)(3y - 1) - \(\dfrac{2}{7}y\)(3y - 1)
= (3y - 1).(\(\dfrac{2}{7}x\) - \(\dfrac{2}{7}y\) )
= (3y - 1).[\(\dfrac{2}{7}\)(x - y)]
e) x3 - 3x2 + 3x - 1
= x2.x - 3x.x + 3.x - 1
= x(x2-3x+3) - 1
g) 27x3 + \(\dfrac{1}{8}\)
= (3x)3 + \(\left(\dfrac{1}{2}\right)^3\)
= (3x + \(\dfrac{1}{2}\)).(9x2 - \(\dfrac{3}{2}\)x + \(\dfrac{1}{4}\))
h) (x+y)3 - (x-y)3
= 2(3x2y) + 2y3
f) (x+y)2 - 4x2
= -3x2 + y(2x + y)
BÀI 1:
a) \(ĐKXĐ:\) \(\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}}\) \(\Leftrightarrow\)\(\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)
b) \(A=\left(\frac{2}{x-2}-\frac{2}{x+2}\right).\frac{x^2+4x+4}{8}\)
\(=\left(\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right).\frac{\left(x+2\right)^2}{8}\)
\(=\frac{2x+4-2x+4}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)^2}{8}\)
\(=\frac{x+2}{x-2}\)
c) \(A=0\) \(\Rightarrow\)\(\frac{x+2}{x-2}=0\)
\(\Leftrightarrow\) \(x+2=0\)
\(\Leftrightarrow\)\(x=-2\) (loại vì ko thỏa mãn ĐKXĐ)
Vậy ko tìm đc x để A = 0
p/s: bn đăng từng bài ra đc ko, mk lm cho
Bài 1.
a) 5(4x - y)
= 20x - 5y
b) (x + 2)(x - 2) - (x - 3)(x + 1)
= x2 - 4 - [(x - 1) - 2][(x - 1) + 2)]
= x2 - 4 - [(x - 1)2 - 4]
= x2 - 4 - (x - 1)2 + 4
= x2 - x2 + 2x - 1
= 2x - 1
Bài 2.
a) x - y + 5x - 5y
= (x + 5x) - (y + 5y)
= 6x - 6y
= 6(x - y)
b) 3x2 - 6xy + 3y2 - 12z2
= 3(x2 - 2xy + y2 - 4z2)
= 3[(x2 - 2xy + y2) - 4z2]
= 3[(x - y)2 - 4z2]
= 3(x - y + z)(x - y - z)
Bài 3.
(x3- y3) : (x2 + xy + y2)
= (x - y)(x2 + xy + y2) : (x2 + xy + y2)
= x - y
Thay x = \(\dfrac{2}{3}\); y = \(\dfrac{1}{3}\) vào biểu thức đại số ta có:
\(\dfrac{2}{3}\)- \(\dfrac{1}{3}\)= \(\dfrac{1}{3}\)
Vậy (x3- y3) : (x2 + xy + y2) = \(\dfrac{1}{3}\) tại x = \(\dfrac{2}{3}\) và y = \(\dfrac{1}{3}\)