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Bài 1.
a) x( 8x - 2 ) - 8x2 + 12 = 0
<=> 8x2 - 2x - 8x2 + 12 = 0
<=> 12 - 2x = 0
<=> 2x = 12
<=> x = 6
b) x( 4x - 5 ) - ( 2x + 1 )2 = 0
<=> 4x2 - 5x - ( 4x2 + 4x + 1 ) = 0
<=> 4x2 - 5x - 4x2 - 4x - 1 = 0
<=> -9x - 1 = 0
<=> -9x = 1
<=> x = -1/9
c) ( 5 - 2x )( 2x + 7 ) = ( 2x - 5 )( 2x + 5 )
<=> -4x2 - 4x + 35 = 4x2 - 25
<=> -4x2 - 4x + 35 - 4x2 + 25 = 0
<=> -8x2 - 4x + 60 = 0
<=> -8x2 + 20x - 24x + 60 = 0
<=> -4x( 2x - 5 ) - 12( 2x - 5 ) = 0
<=> ( 2x - 5 )( -4x - 12 ) = 0
<=> \(\orbr{\begin{cases}2x-5=0\\-4x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)
d) 64x2 - 49 = 0
<=> ( 8x )2 - 72 = 0
<=> ( 8x - 7 )( 8x + 7 ) = 0
<=> \(\orbr{\begin{cases}8x-7=0\\8x+7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{7}{8}\end{cases}}\)
e) ( x2 + 6x + 9 )( x2 + 8x + 7 ) = 0
<=> ( x + 3 )2( x2 + x + 7x + 7 ) = 0
<=> ( x + 3 )2 [ x( x + 1 ) + 7( x + 1 ) ] = 0
<=> ( x + 3 )2( x + 1 )( x + 7 ) = 0
<=> x = -3 hoặc x = -1 hoặc x = -7
g) ( x2 + 1 )( x2 - 8x + 7 ) = 0
Vì x2 + 1 ≥ 1 > 0 với mọi x
=> x2 - 8x + 7 = 0
=> x2 - x - 7x + 7 = 0
=> x( x - 1 ) - 7( x - 1 ) = 0
=> ( x - 1 )( x - 7 ) = 0
=> \(\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=7\end{cases}}\)
Bài 2.
a) ( x - 1 )2 - ( x - 2 )( x + 2 )
= x2 - 2x + 1 - ( x2 - 4 )
= x2 - 2x + 1 - x2 + 4
= -2x + 5
b) ( 3x + 5 )2 + ( 26x + 10 )( 2 - 3x ) + ( 2 - 3x )2
= 9x2 + 30x + 25 - 78x2 + 22x + 20 + 9x2 - 12x + 4
= ( 9x2 - 78x2 + 9x2 ) + ( 30x + 22x - 12x ) + ( 25 + 20 + 4 )
= -60x2 + 40x2 + 49
d) ( x + y )2 - ( x + y - 2 )2
= [ x + y - ( x + y - 2 ) ][ x + y + ( x + y - 2 ) ]
= ( x + y - x - y + 2 )( x + y + x + y - 2 )
= 2( 2x + 2y - 2 )
= 4x + 4y - 4
Bài 3.
A = 3x2 + 18x + 33
= 3( x2 + 6x + 9 ) + 6
= 3( x + 3 )2 + 6 ≥ 6 ∀ x
Đẳng thức xảy ra <=> x + 3 = 0 => x = -3
=> MinA = 6 <=> x = -3
B = x2 - 6x + 10 + y2
= ( x2 - 6x + 9 ) + y2 + 1
= ( x - 3 )2 + y2 + 1 ≥ 1 ∀ x,y
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-3=0\\y^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=0\end{cases}}\)
=> MinB = 1 <=> x = 3 ; y = 0
C = ( 2x - 1 )2 + ( x + 2 )2
= 4x2 - 4x + 1 + x2 + 4x + 4
= 5x2 + 5 ≥ 5 ∀ x
Đẳng thức xảy ra <=> 5x2 = 0 => x = 0
=> MinC = 5 <=> x = 0
D = -2/7x2 - 8x + 7 ( sửa thành tìm Max )
Để D đạt GTLN => 7x2 - 8x + 7 đạt GTNN
7x2 - 8x + 7
= 7( x2 - 8/7x + 16/49 ) + 33/7
= 7( x - 4/7 )2 + 33/7 ≥ 33/7 ∀ x
Đẳng thức xảy ra <=> x - 4/7 = 0 => x = 4/7
=> MaxC = \(\frac{-2}{\frac{33}{7}}=-\frac{14}{33}\)<=> x = 4/7
Bài 1 :
a) \(x^4-4x^2-4x-1\)
\(=x^4-\left(4x^2+4x+1\right)\)
\(=x^4-\left(2x+1\right)^2\)
\(=\left(x^2-2x-1\right)\left(x^2+2x+1\right)\)
b) \(x^2+2x-15\)
\(=x^2+2x+1-16\)
\(=\left(x+1\right)^2-4^2\)
\(=\left(x+1+4\right)\left(x+1-4\right)=\left(x+5\right)\left(x-3\right)\)
c) \(x^3y-2x^2y^2+5xy\)
\(=xy\left(x^2-2xy+5\right)\)
B2:
a) \(2\left(x-1\right)^2-\left(2x+3\right)\left(2x-3\right)\)
\(=2\left(x^2-2x+1\right)-\left(4x^2-9\right)\)
\(=2x^2-4x+2-4x^2+9\)
\(=-2x^2-4x+11\)
b) \(\left(x+3\right)^2-2\left(x+3\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left(x+3-x+3\right)^2=6^2=36\)
c) \(4\left(x-1\right)\left(x+3\right)+5\left(2x+1\right)^2-2\left(5-3x\right)^2\)
\(=4\left(x^2+2x-3\right)+5\left(4x^2+4x+1\right)-2\left(9x^2-30x+25\right)\)
\(=4x^2+8x-12+20x^2+20x+5-18x^2+60x-50\)
\(=6x^2+88x-57\)
Mấy câu trên dễ
\(M=4a^2-6a+12\)
\(M=\left(2a\right)^2-2\cdot2a\cdot\frac{3}{2}+\left(\frac{3}{2}\right)^2+\frac{39}{4}\)
\(M=\left(2a-\frac{3}{2}\right)^2+\frac{39}{4}\ge\frac{39}{4}\forall x\left(đpcm\right)\)
1. a) 2x2y - 3xy2 - 6x + 9y = 2x( xy - 3 ) - 3y ( xy - 3) = ( 2x - 3y)(xy - 3)
b) x2 - 2x + 8 = x2 - 2x + 12 - 1 + 9 = ( x - 1 )2 + 32 ( xem lại đề bài )
2. a) ( 2x - 1) 2 - (2x-1)(2x+3) = 5
(2x-1)(2x-1-2x-3) = 5
-4(2x-1) = 5
2x - 1 = -1,25
2x = -0,25
x= -0,125
b) x(x-9 ) = 0
x= 0 hoặc x = 9
c, ko hiểu
3, M = (2a)2 - 2.2a.1,5 + ( 1,5)2 + 9,75
M= ( 2a - 1,5)2 + 9,75
Vì ( 2a - 1,5 )2 \(\ge\)0 \(\forall x\)
\(\Rightarrow\)( 2a - 1,5)2 + 9,75 \(\ge9,75\forall x\)
Vậy biểu thức trên luôn dương
4.a) \(2x^2-10x-3x-2x^2-26=0\)
\(-13x-26=0\Rightarrow-13\left(x+2\right)=0\)
\(\Rightarrow x=-2\)
b) \(2\left(x+5\right)-x^2-5x=0\)
\(2x+10-x^2-5x=0\Leftrightarrow-x^2-3x+10=0\)
\(-\left(x^2+3x-10\right)=0\)
\(-\left(x^2-2x+5x-10\right)=-\left(x\left(x-2\right)+5\left(x-2\right)\right)=0\)
\(-\left(x-2\right)\left(x+5\right)=0\)
\(\left\{{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
c) \(\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\left(x-8\right)\left(3x+2\right)=0\)
\(\left\{{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
d) \(x^3+x^2-4x-4=0\)
\(x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\left(x+1\right)\left(x^2-4\right)=\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\x-2=0\\x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)
g) \(\left(x-1\right)\left(2x+3-x\right)=0\)
\(\left(x-1\right)\left(x+3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
h) \(x^2-4x+8-2x+1=x^2-6x+9=0\)
\(\left(x-3\right)^2=0\Rightarrow x=3\)
Bài 1:
a) 2x^2 -3x + 1 = 2x^2 -2x -x +1 = 2x.(x-1) - (x-1) = (x-1).(2x-1)
b) 2x^3y - 2xy^3 - 4xy^2 - 2xy = 2xy.(x^2 - y^2 - 2y -1) = 2xy.[ x^2 - (y^2 + 2y+1)] = 2xy.[x^2 - (y+1)^2]
= 2xy.(x-y-1).(x+y+1)
c) (x^2 + x+3).(x^2 + x +5) - 8 = (x^2+x+4-1).(x^2+x+4+1) - 8 = (x^2+x+4)^2 - 1 - 8 = (x^2+x+4)^2 - 3^2
= (x^2+x+4-3).(x^2+x+4+3) = (x^2+x+1).(x^2+x+7)
Bài 2:
a) (x+2).(x^2-2x+4) - (x^3+2x) = 0
x^3 + 8 - x^3 - 2x = 0
8 - 2x = 0
x = 4
b) x^2 - 2x - 8 = 0
x^2 +2x - 4x - 8 = 0
x.(x+2) - 4.(x+2) = 0
(x+2).(x-4) = 0
...
bn tự làm tiếp nha
Bài2: phân tích đa thức thành nhân tử
\(a,x^2-y^2-2x+2y\)
\(=\left(x-y\right)\left(y+x-2\right)\)
\(b,x^3-5x^2+x-5\)
\(=x^2\left(x-5\right)+\left(x-5\right)\)
\(=\left(x+x-5\right)\left(x-x-5\right)\)
\(c,x^2-2xy+y^2-9\)
\(=\left(x^2-y^2\right)-3^2\)
\(=\left(x-y+3\right)\left(x-y-3\right)\)
chúc bạn học tốt !
a) A = (3x - 5)(2x + 11) - (2x + 3)(3x + 7)
A = 6x^2 + 33x - 10x - 55 - 6x^2 - 23x - 21
A = -76
b) B = 4x(3x - 2) - 3x(4x + 1)
B = 12x^2 - 8x - 12x^2 - 3x
B = -11x
c) C = (x + 3)(x - 2) - (x - 1)^2
C = x^2 + x - 6 - x^2 + 2x - 1
C = 3x - 7
1)a)3(2x-1)(3x-1)-(2x-3)(9x-1)=0
<=>18x2-15x+1-18x2+29x-3=0
<=>14x-2=0
<=>14x=2
<=>x=1/7
b)4(x+1)2+(2x-1)2-8(x-1)(x+1)=11
<=>4x2+8x+4+4x2-4x+1-8x2+8=11
<=>4x+13=11
<=>4x=11-13
<=>4x=-2
<=>x=-1/2
c)Sai đề phải là dấu - chứ không phải +
(x-3)(x2+3x+9)-x(x-2)(x+2)=1
<=>x3-27-x3+4x=1
<=>4x=1+27
<=>4x=28
<=>x=7
2)a)(2x-3y)(2x+3y)-4(x-y)2-8xy
=4x2-9y2-4x2+8xy-4y2-8xy
=-13y2
b)(x-2)3-x(x+1)(x-1)+6x(x-3)
=x3-6x2+12x+8-x3+x+6x2-18x
=8-5x
c)(x-2)(x2-2x+4)(x+2)(x2+2x+4)
=(x-2)(x2+2x+4)(x+2)(x2-2x+4)
=(x3-8)(x3+8)
=x6-64
Bài 1:
a) \(x^2+9y^2-y^4-6xy\)
\(=\left(x^2-6xy+9y^2\right)-y^4\)
\(=\left[x^2-2.x.3y+\left(3y\right)^2\right]-\left(y^2\right)^2\)
\(=\left(x-3y\right)^2-\left(y^2\right)^2\)
\(=\left(x-3y-y^2\right)\left(x-3y+y^2\right)\)
b) \(2x^2-x-28\)
\(=2x^2-8x+7x-28\)
\(=2x\left(x-4\right)+7\left(x-4\right)\)
\(=\left(x-4\right)\left(2x+7\right)\)
Bài 2:
a) \(2x\left(x^2-2x+3\right)-2x^3\)
\(=2x\left(x^2-2x+3-x^2\right)\)
\(=2x\left(3-2x\right)\)
b) \(2x\left(x-3\right)-\left(x+5\right)\left(2x-1\right)\)
\(=\left(2x^2-6x\right)-\left(2x^2+9x-5\right)\)
\(=2x^2-6x-2x^2-9x+5\)
\(=-15x+5\)
\(=-5\left(3x-1\right)\)
c) \(\left(5-x\right)^2+\left(x+5\right)^2-\left(2x+10\right)\left(x-5\right)\)
\(=\left(x-5\right)^2-2\left(x+5\right)\left(x-5\right)+\left(x+5\right)^2\)
\(=\left[\left(x-5\right)-\left(x+5\right)\right]^2\)
\(=\left(x-5-x-5\right)^2\)
\(=\left(-10\right)^2=100\)
Bài 3:
a) \(x-2=\left(x-2\right)^2\)
\(\Rightarrow\left(x-2\right)-\left(x-2\right)^2=0\)
\(\Rightarrow\left(x-2\right)\left(1-x+2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(3-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\3-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
b) \(\left(-3x+9\right)x^2-7x+21=0\)
\(\Rightarrow-3\left(x-3\right)x^2-7\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(-3x^2-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\-3x^2-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-\dfrac{7}{3}\end{matrix}\right.\)
Mà x2 > 0 hoặc x2 = 0 với mọi x
=> x2 = -7/3 không thỏa mãn
=> x= 3
Phân tích đa thức
a, x^2+9y^2-y^4-6xy
=(x^2-6xy+9y^2)-y^4
=(x-3y)^2-y^4
=(x-3y-y^2)(x-3y+y^2)
b, 2x^2-x-28
=(2x^2-8x)+(7x-28)
=2x(x-4)+7(x-4)
=(x-4)(2x+7)
Rút gọn
a,2x(x^2-2x+3)-2x^3
=2x(x^2-2x+3-x^2)
=2x(-2x+3)
b,2x(x-3)-(x+5)(2x-1)
=2x^2-6x-2x^2-9x+5
=-15x+5
=-5(3x-1)
c,(5-x)^2+(x+5)^2-(2x+10)(x-5)
Ta có:(5-x)^2=(x-5)^2
=(x-5)^2-2(x+5)(x-5)+(x+5)^2
=(x-5-x-5)^2
=100
Tìm x
a,x-2=(x-2)^2=0
=>x-2=0=>x=2
b,(-3x+9)x^2-7x+21=0
=>-3(x-3)x^2-7(x-3)=0
=>(x-3)(-3x^2-7)=0
=>\(\left[{}\begin{matrix}x-3=0=>x=3\\-3x^2-7=0=>x=\sqrt{\dfrac{-7}{3}}\end{matrix}\right.\)