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\(4x^2+12x+10=\left(4x^2+12x+9\right)+1=\left(2x+3\right)^2+1\ge1\)
\(25x^2+5x+1=\left(25x^2+5x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(5x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Câu a : \(x^2-3x+3=\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Xem lại đề câu a .
Nếu không áp dụng BĐT thì chuyển vế cũng được nhưng hơi dài :
Mình thử làm thôi nhé :
\(\frac{1}{1+a^2}+\frac{1}{1+b^2}-\frac{2}{1+ab}\)
\(=\frac{2+a^2+b^2}{\left(1+a^2\right)\left(1+b^2\right)}-\frac{2}{\left(1+ab\right)}\)
\(=\frac{2+a^2+b^2-2\left(1+a^2\right)\left(1+b^2\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\)
\(=\frac{2+a^2+b^2-2-2b^2-2a^2-2\left(ab\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\)
\(=\frac{-\left(a^2+b^2+2a^2b^2\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\)
....
Giải bất mà không được dùng bất ? Vô lý thế ??
Bài Đạt chưa làm hết,mình làm nốt nha !
a , \(-q^3+12q^2x-48qx^2+64x^3\)
\(=-\left(q^3-12q^2x+48qx^2-64x^3\right)\)
\(=\)\(-\left(q-4x\right)^3\)
b , x2 + 2xy - y2 - 9
= - ( x2 - 2xy + y2 ) - 9
= - ( x - y )2 - 9
= ( - x + y - 3 ) ( x - y + 3 )
3 , 1 - m2 + 2mn - n2
= 1 - ( m2 - 2mn + n2 )
= 1 - ( m - n )2
= ( 1 - m + n ) ( 1 + m - n )
4 , x3 - 8 + 6a2 - 12a
= x3 + 6a2 - 12a + 8
= x3 + 6a2 - 12a + 4 + 4
= x3 + ( 6a2 - 12a + 4 ) + 4
= x3 + ( 3a - 2 )2 + 4
= ( x + 3a - 2 + 2 ) ( x2 + 3a + 2 + 2 )
( Mai làm tiếp mấy ý sau '-' muộn rồi ~ )
5 , x2 - 2xy + y2 - xz - yz
= ( x2 - 2xy + y2 ) - ( xz + yz )
= ( x - y )2 - z ( x + y )
= ( x - y ) 2 - z ( x - y )
= ( x - y ) ( x - y - z )
6 , x2 - 4xy + 4y 2 - z2 + 4z - 4t2
=( x2 - 4xy + 4y 2 ) - (z2 - 4z +4 ) . t2
= ( x - y )2 - ( z - 2 )2 . t2
= ( x - y - z - 2 ) ( x - y + z - 2 ) t2
7 , 25 - 4x2 - 4xy - y2
= 25 + ( - 4x2 - 4xy + y2 )
= 25 + ( 2x - y )2
= ( 5 + 2x - y ) ( 5 + 2x + y )
8 ,
x3 + y3 + z3 - 3xyz
= (x+y)3 - 3xy (x - y ) + z3 - 3xyz
= [ ( x + y)3 + z3 ] - 3xy ( x + y + z )
= ( x + y + z )3 - 3z ( x + y )( x + y + z ) - 3xy ( x - y - z )
= ( x + y + z )[( x + y + z )2 - 3z ( x + y ) - 3xy ]
= ( x + y + z )( x2 + y2 + z2 + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= ( x + y + z)(x2 + y2 + z2 - xy - xz - yz)
c) \(\left(ax+by\right)^2\le\left(a^2+b^2\right)\left(x^2+y^2\right)\)
\(\Leftrightarrow\)\(\left(ax\right)^2+2axby+\left(by\right)^2\le\left(ax\right)^2+\left(ay\right)^2+\left(bx\right)^2+\left(by\right)^2\)
\(\Leftrightarrow\)\(2axby\le\left(ay\right)^2+\left(bx\right)^2\)
\(\Leftrightarrow\)\(\left(ay\right)^2-2axby+\left(bx\right)^2\ge0\)
\(\Leftrightarrow\)\(\left(ay-bx\right)^2\ge0\) luôn đúng
Dấu "=" xảy ra \(\Leftrightarrow\)\(\frac{a}{x}=\frac{b}{y}\)
Bài 1
a) \(\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x\left(x^2-xy+y^2\right)+y\left(x^2-xy+y^2\right)\)
\(=x^3-x^2y+xy^2+x^2y-xy^2+y^3\)
\(=x^3+y^3\left(Đpcm\right)\)
b) \(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x\left(x^2+xy+y^2\right)-y\left(x^2+xy+y^2\right)\)
\(=x^3+x^2y+xy^2-x^2y-xy^2-y^3\)
\(=x^3-y^3\left(Đpcm\right)\)
Bài 2
a) \(16x^2-24xy+9y^2\)
\(=\left(4x\right)^2-2.4x.3y+\left(3y\right)^2\)
\(=\left(4x-3y\right)^2\)
b) \(\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
Bài 3
a) \(\left(x+2\right)\left(x^2-2x+4\right)+x\left(x-5\right)\left(x+5\right)=-17\)
\(\Rightarrow x^3+2^3+x\left(x^2-5^2\right)=-17\)
\(\Rightarrow x^3+8+x^3-25x=-17\)
\(\Rightarrow2x^3-25x=-17-8=-25\)
Hình như câu này đề sai rồi đấy bạn 
b) \(25x^2-2=0\)
\(\Rightarrow25x^2=2\)
\(\Rightarrow x^2=\dfrac{2}{25}\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{2}{25}}\\x=-\sqrt{\dfrac{2}{25}}\end{matrix}\right.\)
1.
\(a.\left(x+y\right).\left(x^2-xy+y^2\right)=x^3-x^2y+xy^2+x^2y-xy^2+y^3=x^3+y^3\)\(b.\left(x-y\right)\left(x^2+xy+y^2\right)=x^3+x^2y+xy^2-x^2y-xy^2-y^3=x^3-y^3\)2.
\(a.16x^2-24xy+9y^2=\left(4x\right)^2-2.4x.3y+\left(3y\right)^2=\left(4x-3y\right)^2\)\(b.\left(x-2\right)^2-y^2=\left(x-2-y\right)\left(x-2+y\right)\)
3.
\(b.25x^2-2=0\)
\(\Leftrightarrow25x^2=2\Leftrightarrow x^2=\dfrac{2}{25}\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{2}{25}}\\x=-\sqrt{\dfrac{2}{25}}\end{matrix}\right.\)
1.a (3x-2y)2= (3x)2 - 2. 3x . 2y - (2y)2 = 9x2 - 12xy - 4y2
2.b (2x - 1/2)2 = (2x)2 - 2.2x.1/2 - (1/2)2= 4x2 - 2 - 1/4
3.c (x/2 - y) (x/2+y)= (x/2)2 - (y)2 = x/4 - y2
Bài 1 :
\(\left(3x-2y\right)^2=9x^2-12xy+4y^2\)
\(\left(2x-\frac{1}{2}\right)^2=4x^2-4x+\frac{1}{4}\)
\(\left(\frac{x}{2}-y\right)\left(\frac{x}{2}+y\right)=\frac{x^2}{4}-y^2\)
\(\left(x+\frac{1}{3}\right)^3=x^3+x^2+\frac{1}{3}x+\frac{1}{27}\)
\(\left(x-2\right)\left(x^2+2x+2^2\right)=x^3-8\)
Lời giải:
c) Sửa đề: \(x^2-3x+3\geq 0,75\)
Ta có:
\(x^2-3x+3=x^2-2.\frac{3}{2}x+3=x^2-2.\frac{3}{2}x+(\frac{3}{2})^2+0,75\)
\(=(x-\frac{3}{2})^2+0,75\)
Vì \((x-\frac{3}{2})^2\geq 0, \forall x\Rightarrow x^2-3x+3=(x-\frac{3}{2})^2+0,75\geq 0,75\)
Ta có đpcm
d) Không có dấu "=" bạn nhé.
\(m^2+n^2+5+2mn-4m-4n\)
\(=(m^2+2mn+n^2)-4(m+n)+5\)
\(=(m+n)^2-2.2(m+n)+5\)
\(=(m+n)^2-2.2(m+n)+2^2+1\)
\(=(m+n-2)^2+1\)
Vì \((m+n-2)^2\geq 0, \forall m,n\)
\(\Rightarrow m^2+n^2+5+2mn-4m-4n=(m+n-2)^2+1\geq 0+1>0\)