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a)
b) (x – 1)(x + 1)(x2 + 1)
= [x .(x + 1) – 1 .(x + 1)] . (x2 + 1)
= {x.x + x.1 + (-1).x + (-1).1}. (x2 + 1)
= (x2 + x – x – 1) . (x2 + 1)
= (x2 – 1) . (x2 + 1)
= x2 . (x2 +1) – 1.(x2 + 1)
= x2 . x2 + x2 . 1 – (1.x2 + 1.1)
= x4 + x2 – (x2 + 1)
= x4 + x2 – x2 – 1
= x4 – 1
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Này
a: A=x^5-32
Khi x=3 thì A=3^5-32=243-32=211
b: B=x^8-x^7+x^6-x^5+x^4-x^3+x^2-x+x^7-x^6+x^5-x^4+x^3-x^2+x-1
=x^8-1
=2^8-1=255
Bài 1:
a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)
b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)
Bài 2:
\(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)
Bài 3:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)
1) \(5-\left(1+\dfrac{1}{3}\right):\left(1-\dfrac{1}{3}\right)\)
\(=5-\dfrac{4}{3}:\dfrac{2}{3}\)
\(=5-\dfrac{4}{3}\cdot\dfrac{3}{2}\)
\(=5-\dfrac{4}{2}\)
\(=5-2\)
\(=3\)
b) \(\left(1+\dfrac{2}{3}-\dfrac{5}{4}\right)-\left(1-\dfrac{5}{4}\right)+2022-\dfrac{2}{3}\)
\(=1+\dfrac{2}{3}-\dfrac{5}{4}-1+\dfrac{5}{4}++2022-\dfrac{2}{3}\)
\(=\left(1-1\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)+\left(-\dfrac{5}{4}+\dfrac{5}{4}\right)+2022\)
\(=0+0+0+2022\)
\(=2022\)
2) \(0,7^2\cdot x=0,49^2\)
\(\Rightarrow x=\dfrac{0,49^2}{0,7^2}\)
\(\Rightarrow x=\left(\dfrac{0,49}{0,7}\right)^2\)
\(\Rightarrow x=\left(0,7\right)^2\)
\(\Rightarrow x=0,49\)
b) \(x:\left(-0,5\right)^3=\left(0,5\right)^2\)
\(\Rightarrow x=\left(0,5\right)^2\cdot\left(-0,5\right)^3\)
\(\Rightarrow x=\left(-0,5\right)^5\)
\(\Rightarrow x=-\dfrac{1}{32}\)
2:
a: =>x*0,49=0,49^2
=>x=0,49
b: =>x=(0,5)^2*(-1)*(0,5)^3=-(0,5)^5
a) \({x^2} + \dfrac{1}{4}{x^2} - 5{x^2} = (1 + \dfrac{1}{4} - 5){x^2} = - \dfrac{{15}}{4}{x^2}\);
b) \({y^4} + 6{y^4} - \dfrac{2}{5}{y^4} = (1 + 6 - \dfrac{2}{5}){y^4} = \dfrac{{33}}{5}{y^4}\).
c: \(\dfrac{4x^3-8x^2+13x-5}{2x-1}=\dfrac{4x^3-2x^2-6x^2+3x+10x-5}{2x-1}\)
=2x^2-3x+5
\(1,3\left(x+2\right)-x^2+4=0\)
\(\Rightarrow3x+6-x^2+4=0\)
\(\Rightarrow x^2-3x-10=0\)
\(\Rightarrow x^2+2x-5x-10=0\)
\(\Rightarrow x\left(x+2\right)-5\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-2\\x=5\end{cases}}\)
\(2,\left(x+2\right)\left(x-2\right)-x^2\)
\(=x^2-4-x^2=-4\)
(x + 2) (x - 2) - x2
= x2 - 22 - x2
= (x2 - x2) - 22
= 0 - 22 = -22
mik ko chắc nha