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mik cần gấp
Bài 2:
a: \(A=-3\left(x^2-\dfrac{4}{3}x+\dfrac{1}{3}\right)\)
\(=-3\left(x^2-2\cdot x\cdot\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{1}{9}\right)\)
\(=-3\left(x-\dfrac{2}{3}\right)^2+\dfrac{1}{3}\le\dfrac{1}{3}\)
Dấu '=' xảy ra khi x=2/3
b: \(B=-x^2+5x+3\)
\(=-\left(x^2-5x-3\right)\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{37}{4}\right)\)
\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\)
Dấu '=' xảy ra khi x=5/2
a, \(A=5x-x^2=-x^2+5x=-x^2+2x\cdot2,5-\dfrac{25}{4}+\dfrac{25}{4}\)
\(=-\left(x-2,5\right)^2+\dfrac{25}{4}\)
Có: \(-\left(x-2,5\right)^2\le0\forall x\)
=> \(-\left(x-2,5\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)
''='' xảy ra khi \(x-2,5=0\Rightarrow x=2,5\)
Vậy \(A_{MAX}=\dfrac{25}{4}\Leftrightarrow x=2,5\)
b, \(B=x-x^2=x^2-x=x^2-2\cdot x\cdot\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)
Lập luận như câu a
c, \(C=4x-x^2+3=-x^2+2\cdot x\cdot2-4+7\)
\(=-\left(x-2\right)^2+7\)
Vì \(-\left(x-2\right)^2\le0\forall x\)
=> \(-\left(x-2\right)^2+7\le7\)
Dấu ''='' xảy ra khi và chỉ khi x = 2
Vậy \(C_{MAX}=7\Leftrightarrow x=2\)
d, \(D=-x^2+6x-11=-x^2+2\cdot x\cdot3-9-2\)
\(=-\left(x-3\right)^2-2\)
Vì \(-\left(x-3\right)^2\le0\forall x\)
=> \(-\left(x-3\right)^2-2\le-2\)
Dấu ''='' xảy ra khi và chỉ khi x - 3 = 0 => x = 3
Vậy \(D_{MAX}=-2\Leftrightarrow x=3\)
e, \(E=5-8x-x^2=-x^2-8x+5=-x^2-2\cdot x\cdot4-16+21\)
\(=-\left(x+4\right)^2+21\)
Lập luận như trên
f, \(F=4x-x^2+1=-x^2+4x+1=-x^2+2\cdot x\cdot2-4+5\)
\(=-\left(x-2\right)^2+5\)
Tượng tự mấy ý trc
mik cần gấp
Bài 1:
\(a,\dfrac{1}{2}x^2y^2\left(2x+y\right)\left(x^2-xy+1\right)=\left(x^3y^2+\dfrac{1}{2}x^2y^3\right)\left(x^2-xy+1\right)=x^5y^2-x^4y^3+x^3y^2+\dfrac{1}{2}x^3y^3-\dfrac{1}{2}x^3y^4+\dfrac{1}{2}x^2y^3\)
\(b,\left(\dfrac{1}{2}x-1\right)\left(2x-3\right)=x^2-\dfrac{3}{2}x-2x+3=x^2-\dfrac{7}{2}x+3\)\(c,\left(x-7\right)\left(x-5\right)=x^2-5x-7x+35=x^2-12x+35\)\(f,\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)\left(4x-1\right)=\left(x^2-\dfrac{1}{4}\right)\left(4x-1\right)=4x^3-x^2-x+\dfrac{1}{4}\)Bài 2 ,
\(\left(x-1\right)\left(x^2+x+1\right)=x^3+x^2+x-x^2-x-1=x^3-1\Rightarrowđpcm\)\(b,\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)=x^4+x^3y+x^2y^2+y^3x+x^3y-x^2y^2-xy^3-y^4=x^4-y^4\)
(1) đa thức A\(⋮̸\) B vì \(7x⋮̸\)3x2
(2) đa thức A\(⋮̸\) B vì 2ab3c2 \(⋮̸\) -5a2bc2
Bài 1:
a, \(2x\left(y-z\right)+5y\left(z-y\right)=2x\left(y-z\right)-5y\left(y-z\right)\)
\(=\left(y-z\right)\left(2x-5y\right)\)
b, \(x^3-3x^2+3x-1=x^3-x^2-2x^2+2x+x-1\)
\(=x^2.\left(x-1\right)-2x.\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+1\right)=\left(x-1\right)\left(x^2-x-x+1\right)\)
\(=\left(x-1\right)\left(x-1\right)^2=\left(x-1\right)^3\)
c, \(7x^2-7xy-4x+4y=7x.\left(x-y\right)-4.\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-4\right)\)
d, \(x^2-6x+8=x^2-2x-4x+8\)
\(=x.\left(x-2\right)-4\left(x-2\right)=\left(x-2\right)\left(x-4\right)\)
Chúc bạn học tốt!!!
1)
a) \(2x\left(y-z\right)+5y\left(z-y\right)\)
\(=2x\left(y-z\right)-5y\left(y-z\right)\)
\(=\left(y-z\right)\left(2x-5y\right)\)
b) \(x^3-3x^2+3x-1\)
\(=x^3-3.x^2.1+3.x.1^2-1^3\)
\(=\left(x-1\right)^3\)
c) \(7x^2-7xy-4x+4y\)
\(=7x\left(x-y\right)-4\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-4\right)\)
d) \(x^2-6x+8\)
\(=x^2-4x-2x+8\)
\(=x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
2)
a) \(\left(5x^2+3x-1\right)\left(x+3\right)\)
\(=5x^3+3x^2-x+15x^2+9x-3\)
\(=5x^3+3x^2+15x^2-x+9x-3\)
\(=5x^3+18x^2+8x-3\)
b) \(\left(x^3+2x^2+3x-1\right):\left(x^2-2\right)\)
\(=x+2+\dfrac{5x+3}{x^2-2}\)
Bài 2:
a: \(x^2-16-\left(x+4\right)=0\)
=>(x+4)(x-4)-(x+4)=0
=>(x+4)(x-5)=0
=>x=5 hoặc x=-4
b: \(\left(3x-1\right)^2-\left(9x^2-1\right)=0\)
\(\Leftrightarrow9x^2-6x+1-9x^2+1=0\)
=>-6x+2=0
=>-6x=-2
hay x=1/3
c: \(4x^2+9=-12x^2\)
\(\Leftrightarrow4x^2+12x^2=-9\)
\(\Leftrightarrow16x^2=-9\)(vô lý)
Do đó: \(x\in\varnothing\)
d: \(4x^2-5x+1=0\)
\(\Leftrightarrow4x^2-4x-x+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)=0\)
=>x=1 hoặc x=1/4
e: \(4x^2-4x+3=0\)
\(\Leftrightarrow4x^2-4x+1+2=0\)
\(\Leftrightarrow\left(2x-1\right)^2=-2\)(vô lý)
Do đó: \(x\in\varnothing\)
1,(2x + 3 ) \(^{^{ }2}\)=\(\left(2x\right)^2+2.2x.3+3^2\)
=\(4x^2+12x+9\)
2, ( 3x + 2y )\(^2=\left(3x\right)^2+2.3x.2y+\left(2y\right)^2\)
=\(9x^2+12xy+4y^2\)
3,(3a -1 )\(^2=\left(3a\right)^2-2.3a.1+1^2\)
\(=9a^2-6a+1\)
4, (a - 2 )\(^2=a^2-2.a.2+2^2\)
=\(a^2-4a+4\)
5, ( 1 - 5a )\(^2=1^2-2.1.5a+\left(5a\right)^2\)
=\(1-10a+25a\)
6, ( x - 4 )\(^3=x^3-3x^24+3x4^2-4^3\)
=\(x^3-12x^2+48x-64\)
\(\left(3a-1\right)^2=9a^2-6a+1\)
\(\left(a-2\right)^2=a^2-4a+4\)
\(\left(1-5a\right)^2=1-10a+25a^2\)
\(\left(3a-2b\right)^2=9a^2-12ab+4a^2\)
\(\left(4-3a\right)^2=16-24a+9a^2\)
\(\left(5a-4b\right)^2=25a^2-40ab+16b^2\)
\(\left(5a-3b\right)\left(5a+3b\right)=25a^2-9b^2\)
\(\left(3x+1\right)\left(3x-1\right)=9x^2-1\)
\(\left(5x^2-2\right)\left(5x^2+2\right)=25x^4-4\)
\(\left(2a+\dfrac{1}{2}\right)\left(2a-\dfrac{1}{2}\right)=4a^2-\dfrac{1}{4}\)
\(\left(3x^2-y\right)\left(3x^2+y\right)=9x^4-y^2\)
\(\left(\dfrac{1}{2}x-1\right)\left(\dfrac{1}{2}x+1\right)=\dfrac{1}{4}x^2-1\)
\(\left(\dfrac{3}{4}x+2\right)\left(\dfrac{3}{4}x-2\right)=\dfrac{9}{16}x^2-4\)
\(\left(5x-\dfrac{3}{2}\right)\left(5x+\dfrac{3}{2}\right)=25x^2-\dfrac{9}{4}\)
\(\left(2a^2-7\right)\left(2a^2+7\right)=4a^2-49\)