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Bài 2 :
a. A = 2 ( x3 + y3 ) - 3 ( x2 + y2 ) với x + y = 1
=> A = 2 ( x + y ) ( x2 - xy + y2 ) - 3 [ ( x + y )2 - 2xy ]
=> A = 2 [ ( x + y )2 - 3xy ] - 3 ( 1 - 2xy )
=> A = 2 ( 1 - 3xy ) - 3 + 6xy
=> A = 2 - 6xy - 3 + 6xy
=> A = - 1
B = x3 + y3 + 3xy với x + y = 1
=> B = ( x3 + 3x2y + 3xy2 + y3 ) - ( 3x2y + 3xy2 - 3xy )
=> B = ( x + y )3 - 3xy ( x + y - 1 )
=> B = 13 - 3xy . 0
=> B = 1
Bài 1.
a) ( x - 1 )3 + ( 2 - x )( 4 + 2x + x2 ) + 3x( x + 2 ) = 16
<=> x3 - 3x2 + 3x - 1 + 8 - x3 + 3x2 + 6x = 16
<=> 9x + 7 = 16
<=> 9x = 9
<=> x = 1
b) ( x + 2 )( x2 - 2x + 4 ) - x( x2 - 2 ) = 15
<=> x3 + 8 - x3 + 2x = 15
<=> 2x + 8 = 15
<=> 2x = 7
<=> x = 7/2
c) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 9( x + 1 )2 = 15
<=> ( x - 3 )[ ( x - 3 )2 - ( x2 + 3x + 9 ) + 9( x2 + 2x + 1 ) = 15
<=> ( x - 3 )( x2 - 6x + 9 - x2 - 3x - 9 ) + 9x2 + 18x + 9 = 15
<=> ( x - 3 ).(-9x) + 9x2 + 18x + 9 = 15
<=> -9x2 + 27x + 9x2 + 18x + 9 = 15
<=> 45x + 9 = 15
<=> 45x = 6
<=> x = 6/45 = 2/15
d) x( x - 5 )( x + 5 ) - ( x + 2 )( x2 - 2x + 4 ) = 3
<=> x( x2 - 25 ) - ( x3 + 8 ) = 3
<=> x3 - 25x - x3 - 8 = 3
<=> -25x - 8 = 3
<=. -25x = 11
<=> x = -11/25
Bài 2.
a) A = 2( x3 + y3 ) - 3( x2 + y2 )
= 2( x + y )( x2 - xy + y2 ) - 3x2 - 3y2
= 2( x2 - xy + y2 ) - 3x2 - 3y2
= 2x2 - 2xy + 2y2 - 3x2 - 3y2
= -x2 - 2xy - y2
= -( x2 + 2xy + y2 )
= -( x + y )2
= -(1)2 = -1
b) B = x3 + y3 + 3xy
= x3 + 3x2y + 3xy2 + y3 - 3x2y - 3xy2 + 3xy
= ( x3 + 3x2y + 3xy2 + y3 ) - ( 3x2y + 3xy2 - 3xy )
= ( x + y )3 - 3xy( x + y - 1 )
= 13 - 3xy( 1 - 1 )
= 1 - 3xy.0
= 1
\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
\(x^3-2x^2+4x+2x^2-4x+8-x^3+2x=15\)
\(2x+8=15\)
\(2x=7\)
\(x=\frac{7}{2}\)
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)
\(\Leftrightarrow9x+7=17\)
\(\Leftrightarrow9x=10\)
\(\Leftrightarrow x=\frac{10}{9}\)
Bài 1:
a) \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)
\(\Rightarrow x^3-3x^2+3x-1+2^3-x^3+3x^2+6x=17\)
\(\Rightarrow9x+7=17\)
\(\Rightarrow9x=17-7=10\)
\(\Rightarrow x=\dfrac{10}{9}\)
b) \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
\(\Rightarrow x^3+2^3-x^3+2x=15\)
\(\Rightarrow8+2x=15\)
\(\Rightarrow2x=15-8=7\)
\(\Rightarrow x=\dfrac{7}{2}\)
c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
\(\Rightarrow x^3-3x^2.3+3x.3^2-3^3-x^3+3^3+9\left(x^2+2x+1\right)=15\)
\(\Rightarrow-9x^2+27x+9x^2+18x+9=15\)
\(\Rightarrow45x+9=15\)
\(\Rightarrow45x=6\)
\(\Rightarrow x=\dfrac{6}{45}=\dfrac{2}{15}\)
d) \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
\(\Rightarrow x\left(x^2-5^2\right)-x^3-2^3=3\)
\(\Rightarrow x^3-25x-x^3-8=3\)
\(\Rightarrow-25x-8=3\)
\(\Rightarrow-25x=3+8=11\)
\(\Rightarrow x=-\dfrac{11}{25}\)
Bài 2:
a) Ta có:
\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\)
\(B=2^{16}-1\)
Vì 216 - 1 < 216
=> B < A
b) Ta có:
\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{128}-1\right)\)
Vì 1/2( 3128 - 1) < 3128 - 1
=> A < B
Ít thôi -..-
a) ( 3x + 2 )( 2x + 9 ) - ( x + 3 )( 6x + 1 ) = ( x + 1 )2 - ( x + 2 )( x - 2 )
<=> 6x2 + 31x + 18 - ( 6x2 + 19x + 3 ) = x2 + 2x + 1 - ( x2 - 4 )
<=> 6x2 + 31x + 18 - 6x2 - 19x - 3 = x2 + 2x + 1 - x2 + 4
<=> 12x + 15 = 2x + 5
<=> 12x - 2x = 5 - 15
<=> 10x = -10
<=> x = -1
b) ( 2x + 3 )( x - 4 ) + ( x - 5 )( x - 2 ) = ( 3x - 5 )( x - 4 )
<=> 2x2 - 5x - 12 + x2 - 7x + 10 = 3x2 - 17x + 20
<=> 3x2 - 12x - 2 = 3x2 - 17x + 20
<=> 3x2 - 12x - 3x2 + 17x = 20 + 2
<=> 5x = 22
<=> x = 22/5
c) ( x + 2 )3 - ( x - 2 )3 - 12x( x - 1 ) = -8
<=> x3 + 6x2 + 12x + 8 - ( x3 - 6x2 + 12x - 8 ) - 12x2 + 12x = -8
<=> x3 + 6x2 + 12x + 8 - x3 + 6x2 - 12x + 8 - 12x2 + 12x = -8
<=> 12x + 16 = -8
<=> 12x = -24
<=> x = -2
d) ( 3x - 1 )2 - 5( x + 1 ) + 6x - 3.2x + 1 - ( x - 1 )2 = 16
<=> 9x2 - 6x + 1 - 5x - 5 + 6x - 6x + 1 - ( x2 - 2x + 1 ) = 16
<=> 9x2 - 11x - 3 - x2 + 2x - 1 = 16
<=> 8x2 - 9x - 4 = 16
<=> 8x2 - 9x - 4 - 16 = 0
<=> 8x2 - 9x - 20 = 0
( Đến đây bạn có hai sự lựa chọn : 1 là vô nghiệm
2 là nghiệm vô tỉ =) )
a) (3x + 2)(2x + 9) - (x + 3)(6x + 1) = (x + 1)2 - (x + 2)(x - 2)
=> 3x(2x + 9) + 2(2x + 9) - x(6x + 1) - 3(6x + 1) = x2 + 2x + 1 - x(x - 2) - 2(x - 2)
=> 6x2 + 27x + 4x + 18 - 6x2 - x - 18x - 3 = x2 + 2x + 1 - x2 + 2x - 2x + 4
=> (6x2 - 6x2) + (27x + 4x - x - 18x) + (18 - 3) = (x2 - x2) + (2x + 2x - 2x) + (1 + 4)
=> 12x + 15 = 2x + 5
=> 12x + 15 - 2x - 5 = 0
=> 10x + 10 = 0
=> 10x = -10 => x = -1
b) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
=> 2x(x - 4) + 3(x - 4) + x(x - 2) - 5(x - 2) = 3x(x - 4) - 5(x - 4)
=> 2x2 - 8x + 3x - 12 + x2 - 2x - 5x + 10 = 3x2 - 12x - 5x + 20
=> (2x2 + x2) + (-8x + 3x - 2x - 5x) + (-12 + 10) = 3x2 - 17x + 20
=> 3x2 - 12x - 2 = 3x2 - 17x + 20
=> 3x2 - 12x - 2 - 3x2 + 17x - 20 = 0
=> (3x2 - 3x2) + (-12x + 17x) + (-2 - 20) = 0
=> 5x - 22 = 0
=> 5x = 22 => x = 22/5
c) (x + 2)3 - (x - 2)3 - 12x(x - 1) = -8
=> x3 + 6x2 + 12x + 8 - (x3 - 6x2 + 12x - 8) - 12x2 + 12x = -8
=> x3 + 6x2 + 12x + 8 -x3 + 6x2 - 12x + 8 - 12x2 + 12x = -8
=> (x3 - x3) + (6x2 + 6x2 - 12x2) + (12x - 12x + 12x) + (8 + 8) = -8
=> 12x + 16 = -8
=> 12x = -24
=> x = -2
Còn bài cuối làm nốt
\(12\left(x-2\right)\left(x+2\right)-3\left(2x+3\right)^2\) \(=52\)
\(12\left(x^2-4\right)-3\left(4x^2+12x+9\right)\) \(=52\)
\(12x^2-48-12x^2-36x-27\) \(=52\)
\(-36x-75=52\)
\(-36x=127\)
\(x=\frac{-127}{36}\)
\(\left(2x+1\right)^2-4\left(x-1\right)\left(x+1\right)\) \(+2x=5\)
\(4x^2+4x+1-4\left(x^2-1\right)\) \(+2x=5\)
\(4x^2+4x-1-4x^2+4+2x=5\)
\(6x+3=5\)
\(6x=2\)
\(x=3\)
\(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)\) \(+6\left(x-1\right)^2=15\)
\(x^3-6x^2+12x-8-\left(x-3\right)\left(x+3\right)^2\) \(+6\left(x^2-2x+1\right)=15\)
\(x^3-6x^2+12x-8-\left(x^2-9\right)\left(x+3\right)\) \(+6x^2-12x+6=15\)
\(x^3-2\) \(-\left(x^3+3x^2-9x-27\right)\)\(=15\)
\(x^3-2-x^3-3x^2+9x+27=15\)
\(-3x^2+9x+25=15\)
\(-3x^2+9x+10=0\)
\(-3\left(x^2-3x-\frac{10}{3}\right)\) \(=0\)
\(x=\frac{9+\sqrt{201}}{6}\)
các câu còn lại tương tự
I don't now
sorry
...................
nha
b) \(\left(3x-2\right)\left(x+1\right)^2\left(3x+8\right)=-16\)
\(\Leftrightarrow\)\(\left(3x-2\right)\left(3x+3\right)^2\left(3x+8\right)+144=0\)
Đặt: \(3x+3=a\)pt trở thành:
\(\left(a-5\right)a^2\left(a+5\right)+144=0\)
\(\Leftrightarrow\)\(a^4-25a^2+144=0\)
\(\Leftrightarrow\)\(\left(a-4\right)\left(a-3\right)\left(a+3\right)\left(a+4\right)=0\)
đến đây bạn tìm a rồi tính x
c) \(\left(4x-5\right)\left(2x-3\right)\left(x-1\right)=9\)
\(\Leftrightarrow\)\(\left(4x-5\right)\left(4x-6\right)\left(4x-4\right)-72=0\)
Đặt \(4x-5=a\)pt trở thành:
\(a\left(a-1\right)\left(a+1\right)-72=0\)
\(\Leftrightarrow\)\(a^3-a-72=0\)
p/s: ktra lại đề
d) \(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)
\(\Leftrightarrow\)\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2-4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)=0\)
\(\Leftrightarrow\)\(\left[\left(2x^2+x-2013\right)-2\left(x^2-5x-2012\right)\right]^2=0\)
\(\Leftrightarrow\)\(\left(11x+2011\right)^2=0\)
đến đây làm nốt
\(12\left(x-2\right)\left(x+2\right)-3\left(2x+3\right)^2\)=52\(\Leftrightarrow12\left(x^2-2^2\right)-3\left(4x^2+12x+9\right)=52\)
\(\Leftrightarrow12x^2-48-12x^2-36x-27-52=0\)
\(\Leftrightarrow-36x-127=0\)
\(\Leftrightarrow x=-3.52\)
Bạn học hằng đẳng thức chưa bạn , bạn chỉ cần nắp chúng vào là làm đc thôi
c ) Ta có : 2(x2 - 9) - (x - 3)(x + 4) = 0
=> 2(x - 3)(x + 3) - ( x - 3)(x + 4) = 0
=> (x - 3) [2(x + 3) - (x + 4)] = 0
=> (x - 3)(2x + 3 - x + 4) = 0
=> (x - 3)(x + 7) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-7\end{cases}}\)
ta có 2(x2-9)+(x-3)(x+4) =0
=>\(2\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(x+4\right)=0\)
=>\(\left(x-3\right)\left[2\left(x+3\right)-\left(x-4\right)\right]=0\)
=>\(\left(x-3\right)\left(2x+3-x+4\right)=0\)
=>\(\left(x-3\right)\left(x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\Rightarrow x=3\\x+7=0\Rightarrow x=-7\end{cases}}\)
a) (x - 1)3 + (2 - x)(4 + 2x + x2) + 3x(x + 2) = 16
x3 - 3x2 + 3x - 1 + 8 - x3 + 3x2 + 6x - 16 = 0
9x - 9 = 0
9x = 9
x = 1
Vậy x ∈ {1}
b) ( x + 2)(x2 - 2x + 4) - x(x2 - 2) = 16
x3 + 8 - x3 + 2x - 16 = 0
2x - 8 = 0
2x = 8
x = 4
Vậy x ∈ {4}
c) x(x - 5)(x + 5) - (x + 2)(x2 - 2x + 4) = 17
x3 - 25x - x3 - 8 - 17 = 0
-25x - 25 = 0
-25x = 25
x = -1
Vậy x ∈ {1}
d) (x - 3)3 - (x - 3)(x2 + 3x + 9) + 9(x + 1)2 = 15
x3 - 9x2 + 27x - 27 - x3 + 27 + 9x2 + 18x + 9 - 15 = 0
45x - 6 = 0
45x = 6
x = \(\frac{2}{15}\)
Vậy x ∈ {\(\frac{2}{15}\)}