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b) 5(3xn + 1 - yn - 1) + 3(xn + 1 + 5yn - 1) - 5(3xn + 1 + 2yn - 1) - (3n + 1 - 10)
= 15xn + 1 - 5yn - 1 + 3xn + 1 + 15yn - 1 - 15xn + 1 - 10yn - 1 - 3n + 1 - 10
= (15xn + 1 + 3xn + 1 - 15xn + 1 - 3n + 1) + (15yn - 1 - 5yn - 1 - 10yn - 1) - 10
= xn + 1(15 + 3 - 15 - 3) + yn - 1(15 - 5 - 10) - 10
= 0 - 0 - 10 = -10 (đpcm)
a) h(x) = (x + 1)(x2 - x + 1) - (x - 1)(x2 + x + 1)
= x3 - x2 + x + x2 - x + 1 - x3 - x2 - x + x2 + x + 1
= (x3 - x3) - (x2 - x2 + x2 - x2) + (x - x - x + x) + (1 + 1)
= 1 + 1
= 2 (đpcm)
a) h(x) = ( x + 1 )( x2 - x + 1 ) - ( x - 1 )( x2 + x + 1 )
= ( x3 + 13 ) - ( x3 - 13 )
= x3 + 1 - x3 + 1
= 2
Vậy h(x) không phụ thuộc vào biến ( đpcm )
b) 5( 3xn+1 - yn-1 ) + 3( xn+1 + 5yn-1 ) - 5( 3xn+1 + 2yn-1 ) - ( 3xn+1 - 10 )
= 15xn+1 - 5yn-1 + 3xn+1 + 15yn-1 - 15xn+1 - 10yn-1 - 3xn+1 + 10
= ( 15xn+1 + 3xn+1 - 15xn+1 - 3xn+1 ) + ( -5yn-1 + 15yn-1 - 10yn-1 ) + 10
= 0 + 0 + 10 = 10
Vậy giá trị của biểu thức không phụ thuộc vào biến ( đpcm )
\(3x\left(x+5\right)-\left(18+3x\right)\left(x-1\right)-1\)
\(=3x^2+15x-18x+18-3x^2+3x-1\)
\(=18-1\)
\(=17\)
\(\Rightarrow\)\(3x\left(x+5\right)-\left(18+3x\right)\left(x-1\right)-1\)không phụ thuộc vào biến
đpcm
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
Bài 1.
x = 14
=> 13 = x - 1 ; 15 = x + 1 ; 16 = x + 2 ; 29 = 2x + 1
Thế vào N(x) ta được :
x5 - ( x + 1 )x4 + ( x + 2 )x3 - ( 2x + 1 )x2 + ( x - 1 )x
= x5 - x5 - x4 + x4 + 2x3 - 2x3 - x2 + x2 - x
= -x = -14
Bài 2.
a) ( 1 - x - 2x3 + 3x2 )( 1 - x + 2x3 - 3x2 )
= [ ( 1 - x ) - ( 2x3 - 3x2 ) ][ ( 1 - x ) + ( 2x3 - 3x2 ) ]
= ( 1 - x )2 - ( 2x3 - 3x2 )2
= 1 - 2x + x2 - [ ( 2x3 )2 - 2.2x3.3x2 + ( 3x2 )2 ]
= x2 - 2x + 1 - ( 4x6 - 12x5 + 9x4 )
= x2 - 2x + 1 - 4x6 + 12x5 - 9x4
= -4x6 + 12x5 - 9x4 + x2 - 2x + 1
b) ( x - y + z )2 + ( z - y )2 + 2( x - y + z )( y - z )
= ( x - y + z )2 + ( z - y )2 - 2( x - y + z )( z - y )
= [ ( x - y + z ) - ( z - y ) ]2
= ( x - y + z - z + y )2
= x2
Ta có:
\(A=\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=\left[3x\left(2x+11\right)-5\left(2x+11\right)\right]-\left[2x\left(3x+7\right)+3\left(3x+7\right)\right]\)
\(=\left[\left(6x^2+33x\right)-\left(10x+55\right)\right]-\left[\left(6x^2+14x\right)+\left(9x+21\right)\right]\)
\(=\left[6x^2+23x-55\right]-\left[6x^2+23x+21\right]\)
\(=-55-21=-76\)
Vậy biểu thức A không phụ thuộc vào biến x, y.
1. (x + 2)(x2 - 2x + 4) - (x3 + 2x2) = 5
=> x(x2 - 2x + 4) + 2(x2 - 2x + 4) - x3 - 2x2 - 5 = 0
=> x3 - 2x2 + 4x + 2x2 - 4x + 8 - x3 - 2x2 - 5 = 0
=> (x3 - x3) + (-2x2 + 2x2 - 2x2) + (4x - 4x) + (8 - 5) = 0
=> -2x2 + 3 = 0
=> -2x2 = -3
=> x2 = 3/2
=> x = \(\pm\sqrt{\frac{3}{2}}\)
2. \(\left(x+5\right)^2-6=0\)
=> x2 + 10x + 25 - 6 = 0
=> x2 + 10x + 19 = 0
=> x vô nghiệm(do mình không để căn nên ghi vô nghiệm thôi nhá)
3. \(\left(x+3\right)\left(x^2-3x+9\right)-x^3=2x\)
=> x(x2 - 3x + 9) + 3(x2 - 3x + 9) - x3 - 2x = 0
=> x3 - 3x2 + 9x + 3x2 - 9x + 27 - x3 - 2x = 0
=> (x3 - x3) + (-3x2 + 3x2) + (9x - 9x - 2x) + 27 = 0
=> -2x + 27 = 0
=> -2x = -27
=> x = 27/2
4. \(\left(x-2\right)^3-x^3+6x^2=7\)
=> x3 - 6x2 + 12x - 8 - x3 + 6x2 = 7
=> (x3 - x3) + (-6x2 + 6x2) + 12x - 8 = 7
=> 12x - 8 = 7
=> 12x = 15
=> x = 5/4
5. \(3\left(x-2\right)^2+9\left(x-1\right)-3\left(x^2+x-3\right)=12\)
=> 3x2 - 12x + 12 + 9x - 9 - 3x2 - 3x + 9 = 12
=> (3x2 - 3x2) + (-12x + 9x - 3x) + (12 - 9 + 9) = 12
=> -6x + 12 = 12
=> -6x = 0
=> x = 0
6. \(\left(4x+3\right)^2-\left(4x-3\right)^2-5x-2=0\)
=> 48x - 5x - 2 = 0
=> 43x - 2 = 0
=> 43x = 2
=> x = 2/43
Còn bài cuối tự làm :>
Anh Sang làm cầu kì quá ;-;
1. ( x + 2 )( x2 - 2x + 4 ) - ( x3 + 2x2 ) = 5
<=> x3 + 8 - x3 - 2x2 = 5
<=> 8 - 2x2 = 5
<=> 2x2 = 3
<=> x2 = 3/2
<=> \(x^2=\left(\pm\sqrt{\frac{3}{2}}\right)^2\)
<=> \(x=\pm\sqrt{\frac{3}{2}}\)
2. ( x + 5 )2 - 6 = 0
<=> ( x + 5 )2 - ( √6 )2 = 0
<=> ( x + 5 - √6 )( x + 5 + √6 ) = 0
<=> \(\orbr{\begin{cases}x+5-\sqrt{6}=0\\x+5+\sqrt{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{6}-5\\x=-\sqrt{6}-5\end{cases}}\)
3. ( x + 3 )( x2 - 3x + 9 ) - x3 = 2x
<=> x3 + 27 - x3 = 2x
<=> 27 = 2x
<=> x = 27/2
4. ( x - 2 )3 - x3 + 6x2 = 7
<=> x3 - 6x2 + 12x - 8 - x3 + 6x2 = 7
<=> 12x - 8 = 7
<=> 12x = 15
<=> x = 15/12 = 5/4
5. 3( x - 2 )2 + 9( x - 1 ) - 3( x2 + x - 3 ) = 12
<=> 3( x2 - 4x + 4 ) + 9x - 9 - 3x2 - 3x + 9 = 12
<=> 3x2 - 12x + 12 + 6x - 3x2 = 12
<=> -6x + 12 = 12
<=> -6x = 0
<=> x = 0
6. ( 4x + 3 )2 - ( 4x - 3 )2 - 5x - 2 = 0
<=> 16x2 + 24x + 9 - ( 16x2 - 24x + 9 ) - 5x - 2 = 0
<=> 16x2 + 24x + 9 - 16x2 + 24x - 9 - 5x - 2 = 0
<=> 43x - 2 = 0
<=> 43x = 2
<=> x = 2/43
7, ( 4x + 7 )( 2 - 3x ) - ( 6x + 2 )( 5 - 2x ) = 0
<=> -12x2 - 13x + 14 - ( -12x2 + 26x + 10 ) = 0
<=> -12x2 - 13x + 14 + 12x2 - 26x - 10 = 0
<=> -39x + 4 = 0
<=> -39x = -4
<=> x = 4/39
P = 3x2 - 2x + 3y2 - 2y + 6xy - 100
= 3( x2 + 2xy + y2 ) - 2( x + y ) - 100
= 3( x + y )2 - 2( x + y ) - 100
Với x + y = 5
=> P = 3.52 - 2.5 - 100 = 75 - 10 - 100 = -35
Q = x3 + y3 - 2x2 - 2y2 + 3xy( x + y ) - 4xy + 3( x + y ) + 10
= x3 + y3 - 2x2 - 2y2 + 3x2y + 3xy2 - 4xy + 3( x + y ) + 10
= ( x3 + 3x2y + 3xy2 + y3 ) - ( 2x2 + 4xy + 2y2 ) + 3( x + y )
= ( x + y )3 - 2( x2 + 2xy + y2 ) + 3( x + y ) + 10
= ( x + y )3 - 2( x + y )2 + 3( x + y ) + 10
Với x + y = 5
=> Q = 53 - 2.52 + 3.5 + 10 = 100
a. \(P=3x^2-2x+3y^2-2y+6xy-100\)
\(\Leftrightarrow P=\left(3x^2+6xy+3y^2\right)-\left(2x+2y\right)-100\)
\(\Leftrightarrow P=3\left(x+y\right)^2-2\left(x+y\right)-100\)
\(\Leftrightarrow P=3.5^2-2.5-100\)
\(\Leftrightarrow P=-35\)
b. \(Q=x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)
\(\Leftrightarrow Q=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(2x^2+4xy+2y^2\right)+3\left(x+y\right)+10\)
\(\Leftrightarrow Q=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10\)
\(\Leftrightarrow Q=5^3-2.5^2+3.5+10\)
\(\Leftrightarrow Q=100\)
1. ( 2x + y )( 4x2 - 2xy + y2 ) - 8x3 - y3 - 16
= [ ( 2x )3 + y3 ] - 8x3 - y3 - 16
= 8x3 + y3 - 8x3 - y3 - 16
= -16 ( đpcm )
2. ( 3x + 2y )2 + ( 3x + 2y )2 - 18x2 - 8y2 + 3
= 2( 3x + 2y )2 - 18x2 - 8y2 + 3
= 2( 9x2 + 12xy + 4y2 ) - 18x2 - 8y2 + 3
= 18x2 + 24xy + 8y2 - 18x2 - 8y2 + 3
= 24xy + 3 ( có phụ thuộc vào biến )
3. ( -x - 3 )3 + ( x + 9 )( x2 + 27 ) + 19
= -x3 - 9x2 - 27x - 27 + x3 + 9x2 + 27x + 243 + 19
= -27 + 243 + 19 = 235 ( đpcm )
4. ( x - 2 )3 - x( x + 1 )( x - 1 ) + 13( x - 4 )
= x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 13x - 52
= x3 - 6x2 + 12x - 8 - x3 + x + 13x - 52
= -6x2 + 26x - 60 ( có phụ thuộc vào biến )
1. (2x+y).(4x2-2xy+y2)-8x3-y3-16
=(2x)3+y3-8x3-y3-16
=-16
Vậy đa thức trên kh phụ thuộc vào biến x
2. (3x+2y)2+(3x+2y)2-18x2-8y2+3
=(9x2+12xy+4y2)+(9x2+12xy+4y2)-18x2-8y2+3
=9x2+12xy+4y2+9x2+12xy+4y2-18x2-8y2+3
=24xy+3
Vậy đa thức trên phụ thuộc biến x