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\(A=-3\left(x+1\right)^2+7\le7\)
\(A_{max}=7\) khi \(x=-1\)
\(B=-\left(x-\frac{3}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\)
\(B_{max}=\frac{5}{4}\) khi \(x=\frac{3}{2}\)
\(C=-x^2-2x+2=-\left(x+1\right)^2+3\le3\)
\(C_{max}=3\) khi \(x=-1\)
\(D=-\left[\left(x+2y\right)^2+\left(x-1\right)^2-4\right]=-\left(x+2y\right)^2-\left(x-1\right)^2+4\le4\)
\(D_{max}=4\) khi \(\left\{{}\begin{matrix}x=1\\y=-\frac{1}{2}\end{matrix}\right.\)
\(A=x^2+3x+7\)
\(=x^2+2.1,5x+2,25+4,75\)
\(=\left(x+1,5\right)^2+4,75\ge4,75\)
Vậy \(A_{min}=4,75\Leftrightarrow x=-1,5\)
\(B=2x^2-8x\)
\(=2\left(x^2-4x\right)\)
\(=2\left(x^2-4x+4-4\right)\)
\(=2\left[\left(x-2\right)^2-4\right]\)
\(=2\left(x-2\right)^2-8\ge-8\)
Vậy \(B_{min}=-8\Leftrightarrow x=2\)
\(\text{a) }3x^2y^2:x^2=3y^2\)
\(\text{b) }\left(x^5+4x^3-6x^2\right):4x^2\\ =\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)
\(\text{c) }\left(x^3-8\right):\left(x^2+2x+4\right)\\ =\left(x-2\right)\left(x^2+2x+4\right):\left(x^2+2x+4\right)\\ =x-2\)
\(\text{d) }\left(3x^2-6x\right):\left(2-x\right)\\ =3x\left(x-2\right):\left(2-x\right)\\ =-3x\left(2-x\right):\left(2-x\right)\\ =-3x\)
\(\text{e) }\left(x^3+2x^2-2x-1\right):\left(x^2+3x+1\right)\\ =\left(x^3+3x^2-x^2+x-3x-1\right):\left(x^2+3x+1\right)\\ =\left[\left(x^3+3x^2+x\right)-\left(x^2+3x+1\right)\right]:\left(x^2+3x+1\right)\\ =\left[x\left(x^2+3x+1\right)-\left(x^2+3x-1\right)\right]:\left(x^2+3x+1\right)\\ =\left(x-1\right)\left(x^2+3x+1\right):\left(x^2+3x+1\right)\\ =x-1\)
a) 3x2y2 : x2 = 3y2
b)( x5 + 4x3 - 6x2 ) : 4x2
=\(\dfrac{1}{4}\)x3+ x - \(\dfrac{3}{2}\)
A/\(\left(2x^3+y^2-7xy\right)4xy^2.\)
\(=8x^4y^2+4xy^4-28x^2y^3\)
B/\(\left(2x^3-x-1\right)\left(5x-2\right)\)
\(=10x^4-5x^2-5x-4x^3+2x+2\)
\(=10x^4-5x^3-3x-4x^3+2\)
C/\(\left(2x^2-3\right)\left(4x^4+6x^2+9\right)\)
\(=\left(2x^2-3\right)\left(2x+3\right)^2\)
D/\(\left(3x^2-2y\right)^3-\left(2x^2-y\right)^3\)
( Bài này áp dụng hằng đẳng thức là làm được ạ )
Bài 1 :
a ) \(2x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(2x+2\right)=2\left(x+1\right)^2\)
b ) \(y^2\left(x^2+y\right)-zx^2-zy=y^2\left(x^2+y\right)-z\left(x^2+y\right)=\left(x^2+y\right)\left(y^2-z\right)\)
c ) \(4x\left(x-2y\right)+8y\left(2y-x\right)=4x\left(x-2y\right)-8y\left(x-2y\right)=4\left(x-2y\right)^2\)
d ) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)=\left(x+1\right)\left(3x-2x^2+7\right)\)
e ) \(x^2-6xy+9y^2=\left(x-3x\right)^2\)
Bài 1 :
f ) \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)
g ) \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
h ) \(125x^3+y^6=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
a: \(A=2x^2-2xy-y^2+2xy=2x^2-y^2\)
\(=2\cdot\dfrac{4}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
b: \(B=5x^2-20xy-4y^2+20xy=5x^2-4y^2\)
\(=5\cdot\dfrac{1}{25}-4\cdot\dfrac{1}{4}\)
=1/5-1=-4/5
c \(C=x^3+6x^2+12x+8=\left(x+2\right)^3=\left(-9\right)^3=-729\)
d: \(D=20x^3-10x^2+5x-20x^2+10x+4\)
\(=20x^3-30x^2+15x+4\)
\(=20\cdot5^3-30\cdot5^2+15\cdot2+4=1784\)
a) \(5x-10x^2\) = \(5x\left(1-2x\right)\)
b) Mạn phép sửa đề:
\(\dfrac{1}{2}x\left(x^2-4\right)+4\left(x+2\right)\) = \(\left(x+2\right)\left[\dfrac{1}{2}x\left(x-2\right)+4\right]\)
= \(\left(x+2\right)\left(\dfrac{1}{2}x^2-x+4\right)\)
c) \(x^4-y^6=\left(x^2-y^3\right)\left(x^2+y^3\right)\)
e) \(x^3-4x^2+4x-1=x^3-x^2-3x^2+3x+x-1\)
= \(x^2\left(x-1\right)-3x\left(x-1\right)+\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2-3x+1\right)\)
g) \(x^4+6x^3-12x^2-8x\)
= \(x\left(x^3-2x^2+8x^2-16x+4x-8\right)\)
= \(x\left[x^2\left(x-2\right)+8x\left(x-2\right)+4\left(x-2\right)\right]\)
= \(x\left(x-2\right)\left(x^2+8x+4\right)\)
h) \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\) (*)
Đặt \(x^2+4x+8=a\) => (*) trở thành:
\(a^2+3ax+2x^2\) = \(a^2+ãx+2ax+x^2\)
= \(a\left(a+x\right)+2x\left(a+x\right)\)
= \(\left(a+x\right)\left(a+2x\right)\) (1)
Thay \(a=x^2+4x+8\) vào (1) ta được:
\(\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)
=\(\left(x^2+5x+8\right)\left(x^2+2x+4x+8\right)\)
= \(\left(x^2+5x+8\right)\left[x\left(x+2\right)+4\left(x+2\right)\right]\)
= \(\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)
P/s: Còn câu f đang suy nghĩ!
\(A=-x^2+6x+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left[\left(x-3\right)^2-10\right]\)
\(=-\left(x-3\right)^2+10\le10\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy GTLN của A là : \(10\Leftrightarrow x=3\)
\(B=-x^2-4x-2\)
\(=-\left(x^2+4x+2\right)\)
\(=-\left(x^2+4x+4-2\right)\)
\(=-\left[\left(x+2\right)^2-2\right]\)
\(=-\left(x+2\right)^2+2\le2\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy GTLN của B là : \(2\Leftrightarrow x=-2\)
C ) Sai đề
\(D=\left(2-x\right)\left(3x+4\right)\)
\(=6x-3x^2+8-4x\)
\(=-3x^2+2x+8\)
\(=-3\left(x^2-\dfrac{2}{3}x-\dfrac{8}{3}\right)\)
\(=-3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}-\dfrac{25}{9}\right)\)
\(=-3\left[\left(x-\dfrac{1}{3}\right)^2-\dfrac{25}{9}\right]\)
\(=-3\left(x-\dfrac{1}{3}\right)^2+\dfrac{25}{3}\le\dfrac{25}{3}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy GTLN của D là : \(\dfrac{25}{3}\Leftrightarrow x=\dfrac{1}{3}\)
\(E=-8x^2+4xy-y^2+3\)
\(=-8x^2+4xy-\dfrac{y^2}{2}-\dfrac{y^2}{2}+3\)
\(=-2\left[4x^2-2xy+\dfrac{y^2}{4}\right]-\dfrac{y^2}{2}+3\)
\(=-2\left(2x-\dfrac{y}{2}\right)^2-\dfrac{y^2}{2}+3\le3\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-\dfrac{y}{2}\right)^2=0\\\dfrac{y^2}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-\dfrac{y}{2}=0\\y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{y}{2}\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=0\\y=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy GTLN của E là : \(3\Leftrightarrow x=y=0\)
bn ơi có thể giúp mk lm câu c đc k đề mk vt nhầm đề đúng là \(-2x^2-3x+5\)