Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^2+20x+100=\left(x+10\right)^2\)
\(16x^2+24xy+\left(3y\right)^2=\left(4x+3y\right)^2\)
\(y^2-14y+49=\left(y-7\right)^2\)
\(a,\) \(x^2+20x+100=\left(x+10\right)^2\)
\(b,\) \(16x^2+24x+9=\left(4x+3\right)^2\)
\(c,\) \(y^2-14x+49=\left(y-7\right)^2\)
`a)x^2+20x+100=(x+10)^2`
`b)16x^2+24xy+9y^2=(4x+3y)^2`
`c)y^2-14y+49=(y-7)^2`
`d)9x^2-42xy+49y^2=(3x-7y)^2`
a, \(x^2+2x.10+100=\left(x+10\right)^2\)
\(b,16x^2+2.4x.3y+9y^2=\left(4x+3y\right)^2\)
\(c,y^2-14y+49=\left(y-7\right)^2\)
\(d,9x^2-2.3x.7x+49y^2=\left(3x-7y\right)^2\)
1.
\(\left(x+y\right)^2=\left(\dfrac{1}{2}.2x+\dfrac{1}{3}.3y\right)^2\le\left(\dfrac{1}{4}+\dfrac{1}{9}\right)\left(4x^2+9y^2\right)=\dfrac{169}{36}\)
\(\Rightarrow-\dfrac{13}{6}\le x+y\le\dfrac{13}{6}\)
Dấu "=" lần lượt xảy ra tại \(\left(-\dfrac{3}{2};-\dfrac{2}{3}\right)\) và \(\left(\dfrac{3}{2};\dfrac{2}{3}\right)\)
2.
\(\left(y-2x\right)^2=\left(\dfrac{1}{4}.4y+\left(-\dfrac{1}{3}\right).6x\right)^2\le\left(\dfrac{1}{16}+\dfrac{1}{9}\right)\left(16y^2+36x^2\right)=\dfrac{25}{16}\)
\(\Rightarrow\left|y-2x\right|\le\dfrac{5}{4}\)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\mp\dfrac{2}{5};\pm\dfrac{9}{20}\right)\)
3.
\(B^2=\left(6.\sqrt{x-1}+8\sqrt{3-x}\right)^2\le\left(6^2+8^2\right)\left(x-1+3-x\right)=200\)
\(\Rightarrow B\le2\sqrt{10}\)
Dấu "=" xảy ra khi \(\dfrac{\sqrt{x-1}}{6}=\dfrac{\sqrt{3-x}}{8}\Leftrightarrow x=\dfrac{43}{25}\)
\(B=6\sqrt{x-1}+6\sqrt{3-x}+2\sqrt{3-x}\ge6\sqrt{x-1}+6\sqrt{3-x}\)
\(B\ge6\left(\sqrt{x-1}+\sqrt{3-x}\right)\ge6\sqrt{x-1+3-x}=6\sqrt{2}\)
\(B_{min}=6\sqrt{2}\) khi \(\sqrt{3-x}=0\Rightarrow x=3\)
4.
\(49=\left(3a+4b\right)^2=\left(\sqrt{3}.\sqrt{3}a+2.2b\right)^2\le\left(3+4\right)\left(3a^2+4b^2\right)\)
\(\Rightarrow3a^2+4b^2\ge\dfrac{49}{7}=7\)
Dấu "=" xảy ra khi \(a=b=1\)
thì 
c: \(C=\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)
=4xy
d: \(D=\left(\dfrac{x}{2}-y\right)\left(\dfrac{x}{2}+y\right)=\dfrac{1}{4}x^2-y^2\)
a) x2+20x+*
=> x2 +2 x 5x2+52
= (x+5)2
b) 16x2+24xy+*
=> (4x)2+2 x 4x x 3+32
= (4x + 3)2
c) y2 -*+49
=> y2 - 2y72+72
= (y-7)2
d) * - 42xy + 49y2
= (3x)2 + 2 x 7y3x + (7y)2
= (3x+7y)2
a) x2 + 20x + *
= x2 + 2.x.10 + 102
= x2 + 20x + 100
b) 16x2 + 24xy +*
= 16x2 +2.x.12y + (12y)2
= 16x2 +24xy + 144y2
c) y2 - * + 49
= y2 - * +72
= y2 - 2.y.7 + 49
= y2 - 14y + 49
d) * - 42xy + 49y2
= * - 42xy + (7y)2
= * - 2.3x.7y + (7y)2
= (3x)2 - 42xy + (7y)2
= 9x2 - 42xy + 49y2
\(\left(3x\right)^2-42xy+49y^2=\left(3x-7y\right)^2\)
\(\left(3x+y\right)\left(3x-y\right)=9x^2-y^2\)
\(\left(5y-3x\right)\left(5y+3x\right)=25y^2-9x^2\)