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a)A(x) = 3x^3 - 4x^4 - 2x^3 + 4x^4 - 5x + 3
=x^3-5x+3
bậc:3
hệ số tự do:3
hệ số cao nhất :3
B(x) = 5x^3 - 4x^2 - 5x^3 - 4x^2 - 5x - 3
=-8x^2-5x+3
bậc:2
hệ số tự do:3
hệ số cao nhất:3
b)A(x)+B(x)=x^3-8^2+10x+6
câu b mik ko đặt tính theo hàng dọc đc thông cảm nha
a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
a) \(M(x) = A(x) + B(x) \\= 4{x^4} + 6{x^2} - 7{x^3} - 5x - 6 - 5{x^2} + 7{x^3} + 5x + 4 - 4{x^4} \\=(4x^4-4x^4)+(-7x^3+7x^3)+(6x^2-5x^2)+(-5x+5x)+(-6+4)\\= {x^2} - 2.\)
b) \(A(x) = B(x) + C(x) \Rightarrow C(x) = A(x) - B(x)\)
\(\begin{array}{l}C(x) = A(x) - B(x)\\ = 4{x^4} + 6{x^2} - 7{x^3} - 5x - 6 - ( - 5{x^2} + 7{x^3} + 5x + 4 - 4{x^4})\\ = 4{x^4} + 6{x^2} - 7{x^3} - 5x - 6 + 5{x^2} - 7{x^3} - 5x - 4 + 4{x^4}\\ =(4x^4+4x^4)+(-7x^3-7x^3)+(6x^2+5x^2)+(-5x-5x)+(-6-4)\\= 8{x^4} - 14{x^3} + 11{x^2} - 10x - 10\end{array}\)
a) A(x) = 5x4 - 5 + 6x3 + x4 - 5x - 12
= (5x4 + x4) + (- 5 - 12) + 6x3 - 5x
= 6x4 - 17 + 6x3 - 5x
= 6x4 + 6x3 - 5x - 17
B(x) = 8x4 + 2x3 - 2x4 + 4x3 - 5x - 15 - 2x2
= (8x4 - 2x4) + (2x3 + 4x3) - 5x - 15 - 2x2
= 4x4 + 6x3 - 5x - 15 - 2x2
= 4x4 + 6x3 - 2x2 - 5x - 15
b) C(x) = A(x) - B(x)
= 6x4 + 6x3 - 5x - 17 - (4x4 + 6x3 - 2x2 - 5x - 15)
= 6x4 + 6x3 - 5x - 17 - 4x4 - 6x3 + 2x2 + 5x + 15
= ( 6x4 - 4x4) + ( 6x3 - 6x3) + (- 5x + 5x) + (-17 + 15) + 2x2
= 2x4 - 2 + 2x2
= 2x4 + 2x2 - 2
a) ( 5x + 3) - ( x -1 ) = 0
\(\Leftrightarrow\)5x + 3 - x +1 =0
\(\Leftrightarrow\)4x +4 = 0
\(\Leftrightarrow\)4x = -4 \(\Leftrightarrow\)x = \(\frac{-4}{4}\) =-1
b) (3x -2 ) - ( 5x + 4) = ( x - 3) - ( x +5 )
\(\Leftrightarrow\)3x -2 - 5x -4 = x-3 - x -5
\(\Leftrightarrow\)3x - 5x - x + x = -3 -5 +2 +4
\(\Leftrightarrow\)-2x = -2 \(\Leftrightarrow\)x =\(\frac{-2}{-2}\)= 1
\(\dfrac{3}{5}x-\dfrac{25}{4}=\dfrac{1}{2}:x+\dfrac{5}{4}\)
=>\(\dfrac{3}{5}x-\dfrac{25}{4}-\dfrac{1}{2x}+\dfrac{5}{4}\)
=>\(\dfrac{3}{5}x-\dfrac{1}{2x}=\dfrac{30}{4}\)
=>\(\dfrac{3x}{5}-\dfrac{1}{2x}=\dfrac{30}{4}\)
=>\(\dfrac{6x^2-5}{10x}=\dfrac{30}{4}\)
=>\(6x^2-5=10x\cdot\dfrac{30}{4}=5x\cdot15=75x\)
=>\(6x^2-75x-5=0\)
=>\(x=\dfrac{75\pm\sqrt{5745}}{12}\)
a)
A(x)= 5x^4 - 3 + 2x^2 - 6x + 7x^2 - x^4
A(x)= 4x^4 + 9x^2 - 6x - 3.
Bậc: 4.
B= -9x^2 + x - 3 - 4x^4 + 5x^3
B(x)= -4x^4 + 5x^3 - 9x^2 + x - 3
b)
N(x) = A(x) + B(x)= ( 4x^4 + 9x^2 - 6x - 3 ) + (-4x^4 + 5x^3 - 9x^2 + x - 3)
N(x)= 5x^3 - 5x - 6
M(x) = A(x) - B(x)= ( 4x^4 + 9x^2 - 6x - 3 ) -
(-4x^4 + 5x^3 - 9x^2 + x - 3)
M(x)= 8x^4 - 5x^3 + 18x^2 - 7x.
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(\Leftrightarrow5^{x+3+1}-3.5^{x+3}=2.5^{11}\)
\(\Leftrightarrow5^{x+3}.5-3.5^{x+3}=2.5^{11}\)
\(\Leftrightarrow5^{x+3}.\left(5-3\right)=5^{11}.2\)
\(\Leftrightarrow5^{x+3}.2=5^{11}.2\)
\(\Leftrightarrow5^{x+3}=5^{11}\)
\(\Leftrightarrow x+3=11\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\)