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\(1)\)
\(a)\)\(A=5-8x-x^2\)
\(A=-\left(x^2+8x+16\right)+21\)
\(A=-\left(x+4\right)^2+21\le21\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(-\left(x+4\right)^2=0\)
\(\Leftrightarrow\)\(x=-4\)
Vậy GTLN của \(A\) là \(21\) khi \(x=-4\)
\(b)\)\(B=5-x^2+2x-4y^2-4y\)
\(-B=\left(x^2-2x+1\right)+\left(4y^2+4y+1\right)-7\)
\(-B=\left(x-1\right)^2+\left(2y+1\right)^2-7\ge-7\)
\(B=-\left(x-1\right)^2-\left(2y+1\right)^2+7\le7\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}-\left(x-1\right)^2=0\\-\left(2y+1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{-1}{2}\end{cases}}}\)
Vậy GTLN của \(B\) là \(7\) khi \(x=1\) và \(y=\frac{-1}{2}\)
Chúc bạn học tốt ~
\(2)\)\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(............\)
\(2A=\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(2A=3^{128}-1\)
\(A=\frac{2^{128}-1}{3}\)
Chúc bạn học tốt ~
\(D=50^2-49.51\)
\(\Leftrightarrow D=50^2-\left(50-1\right)\left(50+1\right)\)
\(\Leftrightarrow D=50^2-50^2+1=1\)
\(C=39^2+78.61+61^2\)
\(\Leftrightarrow C=39^2+2.39.61+61^2\)
\(\Leftrightarrow C=\left(39+61\right)^2=100^2=10000\)
Bài làm
a) 4x - 8y
<=> 4( x - 2y )
b) 12x( x - 2y ) - 8y( x - 2y )
<=> ( 12x - 8y )( x - 2y )
<=> 4( 3x - 2y )( x - 2y )
c) 2x + 2y - x2 - xy
= 2( x + y ) - x( x + y )
= ( x + y )( 2 - x )
d) x2 - 4y2
<=> ( x - 2y )( x + 2y )
e) x3 + x2y - 4x - 4y
<=> x2( x + y ) - 4( x + y )
<=> ( x - 2 )( x + 2 )( x + y )
g) 3x2 - 6xy + 3y2 - 12x3
<=>3( x2 - 3xy + y2 - 4x3 )
# Học tốt #
a)4(x-2y)
b)(x-2y)(12x-8y)
=4(x-2y)(3x-2y)
c)2(x+y)-x(x+y)
=(2-x)(x+y)
d)(x-2y)(x+2y)
e)x2(x+y)-4(x+y)
=(x+y)(x2-4)
=(x+y)(x-2)(x+2)
g)3(x2-2xy+y2-4z3)
=3[(x-y)2-4z3]
????????????phải là 4z2chứ nhỉ.....
d)
$8y-2y^3+8xy^2-8x^2y=2y(4-y^2+4xy-4x^2)$
$=2y[4-(4x^2-4xy+y^2)]=2y[2^2-(2x-y)^2]$
$=2y(2-2x+y)(2+2x-y)$
e)
$4x^2(x-3)+9(3-x)=4x^2(x-3)-9(x-3)=(x-3)(4x^2-9)=(x-3)[(2x)^2-3^2]$
$=(x-3)(2x-3)(2x+3)$
f)
$x^4y^4+64=(x^2y^2)^2+8^2=(x^2y^2)^2+2.x^2y^2.8+8^2-16x^2y^2$
$=(x^2y^2+8)^2-(4xy)^2=(x^2y^2+8-4xy)(x^2y^2+8+4xy)$
a)
$4b-6a^2b+8a-3ab^2$
$=(4b-3ab^2)+(8a-6a^2b)$
$=b(4-3ab)+2a(4-3ab)=(b+2a)(4-3ab)$
b)
$4-4y^2+12xy-9x^2=4-(9x^2-12xy+4y^2)$
$=2^2-(3x-2y)^2=(2-3x+2y)(2+3x-2y)$
c)
$x^2-3x-10=x^2-5x+2x-10=x(x-5)+2(x-5)=(x+2)(x-5)$
Bài 1 : (x + 5)3 - x3 - 125
= (x + 5 - x)[(x + 5)2 + x(x + 5) + x2] - 125
= 5(x2 + 10x + 25 + x2 + 5x + x2)
= 5(3x2 + 15x + 25) - 125
= 5(3x2 + 15x + 25 - 25)
= 5(3x2 + 15x)
1.\(\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)+8y^3\)
\(=\left(3x-2y\right)\left[\left(3x\right)^2+3x.2y+\left(2y\right)^2\right]+8y^3\)
\(=\left(3x\right)^3-\left(2y\right)^3+8y^3\)
\(=27x^3-8y^3+8y^3\)
\(=27x^3\)
2.Ta có:
\(\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+3a^2b+3ab^2+b^3-\left[\left(3ab.a\right)+\left(3ab.b\right)\right]\)
\(=a^3+3a^2b+3ab^2+b^3-\left[3a^2b+3ab^2\right]\)
\(=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)
\(=a^3+\left(3a^2b-3a^2b\right)+\left(3ab^2-3ab^2\right)+b^3\)
\(=a^3+b^3\)
Vậy suy ra \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)