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a ) \(\left(5x+2y\right)^2=25x^2+20xy+4y^2\)
b ) \(\left(-3x+2\right)^2=9x^2-12x+4\)
c ) \(\left(\dfrac{2}{3}x+\dfrac{1}{3}y\right)^2=\dfrac{4}{9}x^2+\dfrac{4}{9}xy+\dfrac{1}{9}y^2\)
d ) \(\left(2x-\dfrac{5}{2}y\right)^2=4x^2-10xy+\dfrac{25}{4}y^2\)
e ) \(\left(x+\dfrac{4}{3}y^2\right)^2=x^2+\dfrac{8}{3}xy^2+\dfrac{16}{9}y^4\)
f ) \(\left(2x^2+\dfrac{5}{3}y\right)^2=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)
\(x^2+4x+1=0\)
\(\Rightarrow x^2+4x+4-3=0\)
\(\Rightarrow\left(x+2\right)^2-3=0\)
\(\Rightarrow\left(x+2+\sqrt{3}\right)\left(x+2-\sqrt{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-\sqrt{3}-2\\x=\sqrt{3}-2\end{matrix}\right.\)
Thay vào pt rồi tính
a.
\(\left(2x-1\right)^3+6\left(3x-1\right)^3=2\left(x+1\right)^3+6\left(x+2\right)^3\)
\(\Leftrightarrow\left(2x\right)^3-3.\left(2x\right)^2.1+3.2x.1+1^3+6.\left[\left(3x\right)^3-3.\left(3x\right)^2.1+3.3x.1+1^3\right]=2\left(x^3+3x^2+3x+1\right)+6\left(x^2+3.x^2.2+3.x.2^2+2^3\right)\)
a) \(\dfrac{1}{8}x^3y^3-27=\left(\dfrac{1}{2}xy\right)^3-3^3=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{1}{4}x^2y^2+\dfrac{1}{6}xy+9\right)\)
b)\(\dfrac{8}{125}x^3+27y^3=\left(\dfrac{2}{5}x\right)^3+\left(3y\right)^3=\left(\dfrac{2}{5}x+3y\right)\left(\dfrac{4}{25}x^2-\dfrac{6}{5}xy+9y^2\right)\)
c) \(0.008x^6-27y^3=\left(0.2x^2\right)^3-\left(3y\right)^3=\left(0.2x^2-3y\right)\left(0.04x^4+\dfrac{3}{5}x^2y+9y^2\right)\)
d)\(\left(2x+y\right)^3-\left(x-y\right)^3=\left(2x+y-x+y\right)[\left(2x+y\right)^2+\left(2x+y\right)\left(x-y\right)+\left(x-y\right)^2]\\ =\left(x+2y\right)\left(4x^2+4xy+y^2+2x^2-2xy+xy-y^2+x^2-2xy+y^2\right)\\ =\left(x+2y\right)\left(6x^2+xy+y^2\right)\)
Bài 1:
a) \(\dfrac{1}{8}x^3y^3-27\)
\(=\left(\dfrac{1}{2}xy\right)^3-3^3\)
\(=\left(\dfrac{1}{2}xy-3\right)\left[\left(\dfrac{1}{2}xy\right)^2+\dfrac{1}{2}xy.3+3^2\right]\)
\(=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{1}{4}xy+\dfrac{3}{2}xy+9\right)\)
\(=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{7}{4}xy+9\right)\)
b) \(\dfrac{8}{125}x^3+\dfrac{1}{8}y^3\)
\(=\left(\dfrac{2}{5}x\right)^3+\left(\dfrac{1}{2}y\right)^3\)
\(=\left(\dfrac{2}{5}x+\dfrac{1}{2}y\right)\left[\left(\dfrac{2}{5}x\right)^2-\dfrac{2}{5}x.\dfrac{1}{2}y+\left(\dfrac{1}{2}y\right)^2\right]\)
\(=\left(\dfrac{2}{5}x+\dfrac{1}{2}y\right)\left(\dfrac{4}{25}x-\dfrac{1}{5}xy+\dfrac{1}{4}y\right)\)
c) \(0.008x^6-27y^3\)
\(=\left(\dfrac{1}{5}x^2\right)^3-\left(3y\right)^3\)
\(=\left(\dfrac{1}{5}x^2-3y\right)\left[\left(\dfrac{1}{5}x^2\right)^2+\dfrac{1}{5}x^2.3y+\left(3y\right)^2\right]\)
\(=\left(\dfrac{1}{5}x^2-3y\right)\left(\dfrac{1}{25}x^4+\dfrac{3}{5}x^2y+9y^2\right)\)
d) \(\left(2x+y\right)^3-\left(x-y\right)^3\)
\(=\left[\left(2x+y\right)-\left(x-y\right)\right]\left[\left(2x+y\right)^2+\left(2x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=\left(2x+y-x+y\right)\left(4x^2+4xy+y^2+2x^3-2xy+xy-y^2+x^2-2xy+y^2\right)\)
\(=\left(x-2y\right)\left(4x^2+2x^3+xy\right)\)
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)
Bài 1:
\(A=\left(x-y\right)\left(x^2+xy+y^2\right)+2y^3\)
\(A=x^3-y^3+2y^3\)
\(A=x^3+y^3\)
Thay \(x=\dfrac{2}{3},y=\dfrac{1}{3}\) vào A, ta có:
\(A=\left(\dfrac{2}{3}\right)^3+\left(\dfrac{1}{3}\right)^3=\dfrac{8}{27}+\dfrac{1}{27}=\dfrac{9}{27}=\dfrac{1}{3}\)
1) \(\left(x-3\right)\left(x-5\right)+44\)
\(=x^2-3x-5x+15+44\)
\(=x^2-8x+59\)
\(=x^2-2.x.4+4^2+43\)
\(=\left(x-4\right)^2+43\ge43>0\)
\(\rightarrowĐPCM.\)
2) \(x^2+y^2-8x+4y+31\)
\(=\left(x^2-8x\right)+\left(y^2+4y\right)+31\)
\(=\left(x^2-2.x.4+4^2\right)-16+\left(y^2+2.y.2+2^2\right)-4+31\)
\(=\left(x-4\right)^2+\left(y+2\right)^2+11\ge11>0\)
\(\rightarrowĐPCM.\)
3)\(16x^2+6x+25\)
\(=16\left(x^2+\dfrac{3}{8}x+\dfrac{25}{16}\right)\)
\(=16\left(x^2+2.x.\dfrac{3}{16}+\dfrac{9}{256}-\dfrac{9}{256}+\dfrac{25}{16}\right)\)
\(=16\left[\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{256}\right]\)
\(=16\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{16}>0\)
-> ĐPCM.
4) Tương tự câu 3)
5) \(x^2+\dfrac{2}{3}x+\dfrac{1}{2}\)
\(=x^2+2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{2}\)
\(=\left(x+\dfrac{1}{3}\right)^2+\dfrac{7}{18}>0\)
-> ĐPCM.
6) Tương tự câu 5)
7) 8) 9) Tương tự câu 3).
Vì dài quá nên mình chỉ có thể trả lời được mấy câu thôi
Bài 1:
27x3 - 8 : (6x + 9x2 +4)
= (3x - 2) (9x2 + 6x + 4) : (9x2 + 6x + 4)
= 3x - 2
Bài 3:
a, 81x4 + 4 = (9x2)2 + 36x2 + 4 - 36x2
= (9x2 + 2)2 - (6x)2
= (9x2 + 6x + 2)(9x2 - 6x + 2)
b, x2 + 8x + 15 = x2 + 3x + 5x + 15
= x(x + 3) + 5(x + 3)
= (x + 3)(x + 5)
c, x2 - x - 12 = x2 + 3x - 4x - 12
= x(x + 3) - 4(x + 3)
= (x + 3) (x - 4)
Câu 1:
(27x3 - 8) : (6x + 9x2 + 4)
= (3x - 2)(9x2 + 6x + 4) : (6x + 9x2 + 4)
= 3x - 2
Câu 2:
a) (3x - 5)(2x+ 11) - (2x + 3)(3x + 7)
= 6x2 + 33x - 10x - 55 - 6x2 - 14x - 9x - 21
= -76
⇒ đccm
b) (2x + 3)(4x2 - 6x + 9) - 2(4x3 - 1)
= 8x3 + 27 - 8x3 + 2
= 29
⇒ đccm
Câu 3:
a) 81x4 + 4
= (9x2)2 + 22
= (9x2 + 2)2 - (6x)2
= (9x2 - 6x + 2)(9x2 + 6x + 2)
b) x2 + 8x + 15
= x2 + 3x + 5x + 15
= x(x + 3) + 5(x + 3)
= (x + 3)(x + 5)
c) x2 - x - 12
= x2 - 4x + 3x - 12
= x(x - 4) + 3(x - 4)
= (x - 4)(x + 3)
\(P=\left[\left(\dfrac{-1}{3}\right)^2x^3+\left(2x^2\right)^2+\dfrac{1}{2}\right]-\left[x\left(\dfrac{1}{3}x\right)^2+\dfrac{3}{2^3}+x^4\right]+\left(y-2013\right)^2=\left(\dfrac{1}{9}x^3+4x^4+\dfrac{1}{2}\right)-\left(\dfrac{1}{9}x^3+x^4+\dfrac{3}{8}\right)+\left(y-2013\right)^2=3x^4+\dfrac{1}{8}+\left(y-2013\right)^2\ge\dfrac{1}{8}\).
Dấu "=" xảy ra khi x = 0; y = 2013.
đây là những món quà mà bn sẽ nhận đc: 1: áo quần 2: tiền 3: đc nhiều người yêu quý 4: may mắn cả 5: luôn vui vẻ trong cuộc sống 6: đc crush thích thầm 7: học giỏi 8: trở nên xinh đẹp phật sẽ ban cho bn những điều này nếu cậu gửi tin nhắn này cho 25 người,