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Bài a,b,c,e,g,i thì đặt điều kiện rồi bình phương 2 vế rồi giải, bài j chuyển vế rồi bình phương
Chỉ trình bày lời giải, tự tìm điều kiện nha :v
d) \(\sqrt{x+2\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-1}+1=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Rightarrow x-1=1\Leftrightarrow x=2\)
f) \(\sqrt{x+4\sqrt{x-4}}=2\)
\(\Leftrightarrow\sqrt{x-4+2.2\sqrt{x-4}+4}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-4}+2=2\)
\(\Leftrightarrow\sqrt{x-4}=0\)
\(\Rightarrow x-4=0\Leftrightarrow x=4\)
câu a nè = (4x-1)(2x-3)
câu f = (x+y+z) ( x^ 2 + y^2 + z^2 +xy + yz + zx)
a) \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
b) \(-x^2+2xy-y^2=-\left(x-y\right)^2\)
c) \(-4x^4-4x^2=-4x^2\left(x^2-1\right)=-4x^2\left(x-1\right)\left(x+1\right)\)
d) \(\dfrac{1}{9}x^2-\dfrac{2}{3}x+1=\left(\dfrac{1}{3}x-1\right)^2\)
e) \(\left(4x^2+1\right)^2-16x^2=\left(4x^2+1+4x^2\right)\left(4x^2+1-4x^2\right)=8x^2+1\)
f) \(16x^2-\left(x^2+4\right)^2=\left(4x^2+x^2+4\right)\left(4x^2-x^2-4\right)=\left(5x^2+4\right)\left(3x^2-4\right)\)
g) \(x^2+6x^2+12x+8=\left(x+2\right)^3\)
h) \(27x^3-54x^2+36x-8=\left(3x-2\right)^3\)
i) \(x^3-\dfrac{3}{2}x^2+\dfrac{3}{4}x-\dfrac{1}{8}=\left(x-\dfrac{1}{2}\right)^3\)
k) \(0,125x^3-0,75x^2+1,5x-1=\left(0,5-1\right)^3\)
Bài 1:
a) -16 +(x-3)2
<=> (x-3)2-16
<=> (x-3)2 -42
<=> (x-3-4)(x-3+4)
<=> (x-7)(x+1)
b) 64+16y+y2
<=> y2 + 2.8.y + 82
<=> (y+8)2
c) \(\dfrac{1}{8}-8x^3\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^3-\left(2x\right)^3\)
\(\Leftrightarrow\left(\dfrac{1}{2}-2x\right)\left(\dfrac{1}{4}+x+4x^2\right)\)
d)\(x^2-x+\dfrac{1}{4}\)
\(\Leftrightarrow x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2\)
e) x4 + 4x2 + 4
<=> (x2)2 + 2.2.x2 +22
<=> (x2 + 2)2
g)\(8x^3+60x^2y+150xy^2+125y^3\)
\(\Leftrightarrow\left(2x+5y\right)^3\)
a) \(A=x^2-2x-6\)
\(A=\left(x^2-2x+1\right)-7\)
\(A=\left(x-1\right)^2-7\)
Mà \(\left(x-1\right)^2\) luôn \(\ge\)\(0\) => GTNN của biểu thức là -7 với \(\left(x-1\right)^2=0\) tức x=1
a: \(=x^2-2x+1-7=\left(x-1\right)^2-7>=-7\)
Dấu '=' xảy ra khi x=1
b: \(=4x^2-4x+1+6=\left(2x-1\right)^2+6>=6\)
Dấu '=' xảy ra khi x=1/2
c: \(=9x^2-6x+1-1=\left(3x-1\right)^2-1>=-1\)
Dấu '=' xảy ra khi x=1/3
d: \(=x^2+12x+36-36=\left(x+6\right)^2-36>=-36\)
Dấu '=' xảy ra khi x=-6
e: \(=x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}>=-\dfrac{9}{4}\)
Dấu '=' xảy ra khi x=3/2
Dài dữ trời :V Về sau gửi từng bài một thôi, nhìn hoa mắt quá @@
B1: Phân tích thành nhân tử:
a) \(6x^2+9x=3x\left(2x+3\right)\)
b) \(4x^2+8x=4x\left(x+2\right)\)
c) \(5x^2+10x=5x\left(x+2\right)\)
d) \(2x^2-8x=2x\left(x-4\right)\)
e) \(5x-15y=5\left(x-3y\right)\)
f) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
g) \(x^2-2x+1-4y^2=\left(x-1\right)^2-4y^2\)
\(=\left(x-1-2y\right)\left(x-1+2y\right)\)
h) \(x^2-100=\left(x-10\right)\left(x+10\right)\)
i) \(9x^2-18x+9=\left(3x-3\right)^2\)
k) \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
l) \(x^2+6xy^2+9y^4=\left(x+3y\right)^2\)
m) \(4xy-4x^2-y^2=-\left(4x^2-4xy+y^2\right)\)
\(=-\left(2x-y\right)^2\)
n) \(\left(x-15\right)^2-16=\left(x-15-16\right)\left(x-15+16\right)\)
\(=\left(x-31\right)\left(x+1\right)\)
o) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3+x\right)\)
\(=\left(2+x\right)\left(8+x\right)\)
p) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)
\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)
\(=\left(5x-5\right)\left(9x-3\right)\)
Bài 1 :
a ) \(6x^2+9x=3x\left(x+3\right)\)
b ) \(4x^2+8x=4x\left(x+2\right)\)
c ) \(5x^2+10x=5x\left(x+2\right)\)
d ) \(2x^2-8x=2x\left(x-4\right)\)
e ) \(5x-15y=5\left(x-3y\right)\)
f ) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
g ) \(x^2-2x+1-4y^2=\left(x-1\right)^2-\left(2y\right)^2=\left(x-1-2y\right)\left(x-1+2y\right)\)
h ) \(x^2-100=x^2-10^2=\left(x-10\right)\left(x+10\right)\)
i ) \(9x^2-18x+9=\left(3x-3\right)^2\)
k ) \(x^3-8=\left(x-2\right)\left(x^2+2x+2^2\right)\)
l ) \(x^2+6xy^2+9y^4=\left(x+3y^2\right)^2\)
m ) \(4xy-4x^2-y^2=-\left(2x-y\right)^2\)
n ) \(\left(x-15\right)^2=x^2-30x+15^2\)
o ) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)
p ) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)\)
Bài 2 :
a ) \(3x^3-6x^2+3x^2y-6xy=3x\left(x^2-2x+xy-2y\right)\)
b ) \(x^2-2x+xy-2y=x\left(x-2\right)+y\left(x-2\right)=\left(x-2\right)\left(x+y\right)\)
c ) \(2x+x^2-2y-2xy=......................\)
d ) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
e ) \(x^2+y^2-2xy-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
f )\(2xy-x^2-y^2+9=-\left(x-y\right)^2+9=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)
Bài 3:
a) ta có: \(A=x^2+4x+9\)
\(=x^2+4x+4+5=\left(x+2\right)^2+5\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+5\ge5\forall x\)
Dấu '=' xảy ra khi
\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy: GTNN của đa thức \(A=x^2+4x+9\) là 5 khi x=-2
b) Ta có: \(B=2x^2-20x+53\)
\(=2\left(x^2-10x+\frac{53}{2}\right)\)
\(=2\left(x^2-10x+25+\frac{3}{2}\right)\)
\(=2\left[\left(x-5\right)^2+\frac{3}{2}\right]\)
\(=2\left(x-5\right)^2+2\cdot\frac{3}{2}\)
\(=2\left(x-5\right)^2+3\)
Ta có: \(\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi
\(2\left(x-5\right)^2=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy: GTNN của đa thức \(B=2x^2-20x+53\) là 3 khi x=5
c) Ta có : \(M=1+6x-x^2\)
\(=-x^2+6x+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left[\left(x-3\right)^2-10\right]\)
\(=-\left(x-3\right)^2+10\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-3\right)^2+10\le10\forall x\)
Dấu '=' xảy ra khi
\(-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: GTLN của đa thức \(M=1+6x-x^2\) là 10 khi x=3
Bài 2:
a) \(\left(x+y\right)^2+\left(x^2-y^2\right)\)
\(=\left(x+y\right)^2+\left(x-y\right).\left(x+y\right)\)
\(=\left(x+y\right).\left(x+y+x-y\right)\)
\(=\left(x+y\right).2x\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left[x-y-\left(z-t\right)\right].\left(x-y+z-t\right)\)
\(=\left(x-y-z+t\right).\left(x-y+z-t\right)\)
Chúc bạn học tốt!
a ) \(x^3+1=x^3+1^3=\left(x+1\right)^3-3x\left(x+1\right)\)
b ) \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)
c ) \(x^3-8=x^3-2^3=\left(x-2\right)^3+3xy\left(x-2\right)\)
d ) \(x-4=\left(\sqrt{x}\right)^2-2^2=\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)\)
e ) \(x-1=\left(\sqrt{x}\right)^2-1^2=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
f )\(x\sqrt{x}+1=\left(\sqrt{x}\right)^3+1^3=\left(\sqrt{x}+1\right)-3\sqrt{x}\left(\sqrt{x}+1\right)\)
g ) \(x\sqrt{x}-1=\left(\sqrt{x}\right)^3-1^3=\left(\sqrt{x}-1\right)+3\sqrt{x}\left(\sqrt{x}-1\right)\)
h ) \(8-12x+6x^2-x^3=2^3-3.2^2x+3.2x^2-x^3=\left(2-x\right)^3\)
a) x3 + 1 = ( x + 1 ).( x2 - x + 1 )
b) x3 + y3 = ( x + y ).( x2 - xy + y2 )
c) x3 - 8 = x3 - 23 = ( x - 2 ).( x2 + 2x + 4 )
h) 8 - 12x + 6x2 - x3 = - x3 + 8 + 6x2 - 12x
= -( x3- 8 ) + 6x.( x - 2 )
= -( x - 2 ).( x2 + 2x +4 ) + 6x.( x - 2 )
= -( x - 2 ).( x2 + 2x + 4 + 6x )
= -( x - 2 ).( x2 + 10x + 4 )
= -( x - 2 )3