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a)
\(\begin{array}{l}P(x) = 5{x^3} + 2{x^4} - {x^2} + 3{x^2} - {x^3} - 2{x^4} - 4{x^3}\\ = \left( {2{x^4} - 2{x^4}} \right) + \left( {5{x^3} - {x^3} - 4{x^3}} \right) + \left( { - {x^2} + 3{x^2}} \right)\\ = 0 + 0 + 2{x^2}\\ = 2{x^2}\\Q(x) = 3x - 4{x^3} + 8{x^2} - 5x + 4{x^3} + 5\\ = \left( { - 4{x^3} + 4{x^3}} \right) + 8{x^2} + \left( {3x - 5x} \right) + 5\\ = 0 + 8{x^2} + ( - 2x) + 5\\ = 8{x^2} - 2x + 5\end{array}\)
b) P(1) = 2.12 = 2
P(0) = 2. 02 = 0
Q(-1) = 8.(-1)2 – 2.(-1) +5 = 8 +2 +5 =15
Q(0) = 8.02 – 2.0 + 5 = 5
a)
\(\begin{array}{l}A(x) = {x^3} + \dfrac{3}{2}x - 7{x^4} + \dfrac{1}{2}x - 4{x^2} + 9\\ = - 7{x^4} + {x^3} - 4{x^2} + \left( {\dfrac{3}{2}x + \dfrac{1}{2}x} \right) + 9\\ = - 7{x^4} + {x^3} - 4{x^2} + 2x + 9\\B(x) = {x^5} - 3{x^2} + 8{x^4} - 5{x^2} - {x^5} + x - 7\\ = \left( {{x^5} - {x^5}} \right) + 8{x^4} + \left( { - 3{x^2} - 5{x^2}} \right) + x - 7\\ = 0 + 8{x^4} + ( - 8{x^2}) + x - 7\\ = 8{x^4} - 8{x^2} + x - 7\end{array}\)
b) * Đa thức A(x):
+ Bậc của đa thức là: 4
+ Hệ số cao nhất là: -7
+ Hệ số tự do là: 9
* Đa thức B(x):
+ Bậc của đa thức là: 4
+ Hệ số cao nhất là: 8
+ Hệ số tự do là: -7
\(\begin{array}{l}a)3{x^7}:\dfrac{1}{2}{x^4} = (3:\dfrac{1}{2}).({x^7}:{x^4}) = 6{x^3}\\b)( - 2x):x = [( - 2):1].(x:x) = - 2\\c)0,25{x^5}:( - 5{x^2}) = [0,25:( - 5)].({x^5}:{x^2}) = - 0,05.{x^3}\end{array}\)
\(\begin{array}{l}a)x + 7,25 = 15,75\\x = 15,75 - 7,25\\x = 8,5\end{array}\)
Vậy x = 8,5
\(\begin{array}{l}b)\left( { - \frac{1}{3}} \right) - x = \frac{{17}}{6}\\\left( { - \frac{1}{3}} \right) - \frac{{17}}{6} = x\\\frac{{ - 2}}{6} - \frac{{17}}{6} = x\\\frac{{ - 19}}{6} = x\\x = \frac{{ - 19}}{6}\end{array}\)
Vậy \(x = \frac{{ - 19}}{6}\)
Chú ý: A = B và B = A là tương đương nhau
\(\begin{array}{l}a)2x + \frac{1}{2} = \frac{7}{9}\\2x = \frac{7}{9} - \frac{1}{2}\\2x = \frac{{14}}{{18}} - \frac{9}{{18}}\\2x = \frac{5}{{18}}\\x = \frac{5}{{18}}:2\\x = \frac{5}{{18}}.\frac{1}{2}\\x = \frac{5}{{36}}\end{array}\)
Vậy \(x = \frac{5}{{36}}\)
\(\begin{array}{l}b)\frac{3}{4} - 6x = \frac{7}{{13}}\\ 6x = \frac{3}{{4}} - \frac{7}{13}\\ 6x = \frac{{39}}{{52}} - \frac{{28}}{{52}}\\ 6x = \frac{{11}}{{52}}\\x = \frac{{11}}{{52}}:6\\x = \frac{{11}}{{52}}.\frac{{1}}{6}\\x = \frac{{11}}{{312}}\end{array}\)
Vậy \(x = \frac{{11}}{{312}}\)
P(x)-Q(x)
=6x^3+8x^2+5x-2+9x^3-6x^2-2x-3
=15x^3+2x^2+2x-5
\(P\left(x\right)-Q\left(x\right)=\left(6x^3+8x^2+5x-2\right)-\left(-9x^3+6x^2+3+2x\right)\)
\(=6x^3+8x^2+5x-2+9x^3-6x^2-3-2x\)
\(=6x^3+9x^3+8x^2-6x^2+5x-2x-2-3\)
\(=15x^3+2x^2+3x-5\)
\(\begin{array}{l}a)x + 0,25 = \frac{1}{2}\\x = \frac{1}{2} - 0,25\\x = \frac{1}{2} - \frac{1}{4}\\x = \frac{2}{4} - \frac{1}{4}\\x = \frac{1}{4}\end{array}\)
Vậy \(x = \frac{1}{4}\)
\(\begin{array}{l}b)x - \left( { - \frac{5}{7}} \right) = \frac{9}{{14}}\\x = \frac{9}{{14}} + \left( { - \frac{5}{7}} \right)\\x = \frac{9}{{14}} + \left( { - \frac{{10}}{{14}}} \right)\\x = \frac{{ - 1}}{{14}}\end{array}\)
Vậy \(x = \frac{{ - 1}}{{14}}\)
\(\begin{array}{l}a)A = 3x - 4{x^4} + {x^3}\\ = - 4{x^4} + {x^3} + 3x\\b)B = - 2{x^3} - 5{x^2} + 2{x^3} + 4x + {x^2} - 5\\ = ( - 2{x^3} + 2{x^3}) + \left( { - 5{x^2} + {x^2}} \right) + 4x - 5\\ = 0 + ( - 4{x^2}) + 4x - 5\\ = - 4{x^2} + 4x - 5\\c)C = {x^5} - \dfrac{1}{2}{x^3} + \dfrac{3}{4}x - {x^5} + 6{x^2} - 2\\ = \left( {{x^5} - {x^5}} \right) - \dfrac{1}{2}{x^3} + 6{x^2} + \dfrac{3}{4}x - 2\\ = - \dfrac{1}{2}{x^3} + 6{x^2} + \dfrac{3}{4}x - 2\end{array}\)
\(\begin{array}{l}a)\frac{x}{{ - 3}} = \frac{7}{{0,75}}\\ \Rightarrow x.0,75 = ( - 3).7\\ \Rightarrow x = \frac{{( - 3).7}}{{0,75}} = - 28\end{array}\)
Vậy x = 28
\(\begin{array}{l}b) - 0,52:x = \sqrt {1,96} :( - 1,5)\\ - 0,52:x = 1,4:( - 1,5)\\ x = \dfrac{(-0,52).(-1,5)}{1,4}\\x = \frac{39}{{70}}\end{array}\)
Vậy x = \(\frac{39}{{70}}\)
\(\begin{array}{l}c)x:\sqrt 5 = \sqrt 5 :x\\ \Leftrightarrow \frac{x}{{\sqrt 5 }} = \frac{{\sqrt 5 }}{x}\\ \Rightarrow x.x = \sqrt 5 .\sqrt 5 \\ \Leftrightarrow {x^2} = 5\\ \Leftrightarrow \left[ {_{x = - \sqrt 5 }^{x = \sqrt 5 }} \right.\end{array}\)
Vậy x \( \in \{ \sqrt 5 ; - \sqrt 5 \} \)
Chú ý:
Nếu \({x^2} = a(a > 0)\) thì x = \(\sqrt a \) hoặc x = -\(\sqrt a \)
a: \(\dfrac{x}{-3}=\dfrac{7}{0.75}=\dfrac{28}{3}\)
=>\(x=\dfrac{28\left(-3\right)}{3}=-28\)
b: \(-\dfrac{0.52}{x}=\dfrac{\sqrt{1.96}}{-1.5}=\dfrac{1.4}{-1.5}\)
=>\(x=0.52\cdot\dfrac{1.5}{1.4}=\dfrac{39}{70}\)
c: \(\dfrac{x}{\sqrt{5}}=\dfrac{\sqrt{5}}{x}\)
=>\(x^2=5\)
=>\(x=\pm\sqrt{5}\)
Theo cột dọc:
Theo hàng ngang:
\(\begin{array}{l}P(x) - Q(x) = 6{x^3} + 8{x^2} + 5x - 2 - ( - 9{x^3} + 6{x^2} + 2x + 3)\\ = 6{x^3} + 8{x^2} + 5x - 2 + 9{x^3} - 6{x^2} - 2x - 3\\ = (6 + 9){x^3} + (8 - 6){x^2} + (5 - 2)x + ( - 2 - 3)\\ = 15{x^3} + 2{x^2} + 3x - 5\end{array}\)