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a, 5 x 4 + 9 x 5 - 5 x 3 - 10
= 5 x ( 4 + 9 - 3) - 10
= 5 x ( 13 - 3) - 10
= 5 x 10 - 10
= 40
b, 4 x 7 + 4 x 6 - 4 x 3 - 4
= 4 x 7 + 4 x 6 - 4 x 3 - 4 x1
= 4 x ( 7 + 6 - 3 -1)
= 4 x 9
= 36
a) \(\frac{2}{9}+\frac{1}{5}+\frac{7}{9}+\frac{4}{5}=\left(\frac{2}{9}+\frac{7}{9}\right)+\left(\frac{1}{5}+\frac{4}{5}\right)= 1+1=2 \)
b) \(\frac{19}{37}+\left(1+\frac{19}{37}\right)=\left(\frac{19}{37}-\frac{19}{37}\right)+1=0+1=1\)
c) \(\frac{9}{8}-\left(\frac{17}{7}-\frac{3}{7}\right)+\frac{7}{8}=\frac{9}{8}+\frac{7}{8}-\left(\frac{17}{7}-\frac{3}{7}\right)=2-2=0\)
d) \(\frac{3}{5}\times\frac{8}{27}\times\frac{5}{3}=\frac{3}{5}\times\frac{5}{3}\times\frac{8}{27}=1\times\frac{8}{27}=\frac{8}{27}\)
e) \(\frac{7}{19}\times\frac{1}{3}+\frac{7}{19}\times\frac{2}{3}=\frac{7}{19}\times\left(\frac{1}{3}+\frac{2}{3}\right)=\frac{7}{19}\times1=\frac{7}{19}\)
f) \(\frac{12}{5}\times4-4\times\frac{7}{5}=4\times\left(\frac{12}{5}-\frac{7}{5}\right)=4\times1=4\)
x=4
=>x+1=5
A=(x+1)x^5 -(x+1)x^4+(x+1)x^3-(x+1)x^2+(x+1)x-1
=x^6+x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2-x+1
=x^6-x-1
=4^6-4-1
=4091
\(a,A=5\cdot4^5-5\cdot4^4+5\cdot4^3-5\cdot4^2+5\cdot4+1\\ A=4^4\left(20-5\right)+4^2\left(20-5\right)+\left(20-5\right)\\ A=15\left(4^4+4^2+1\right)=15\cdot273=4095\)
\(b,x=7\Leftrightarrow x+1=8\\ \Leftrightarrow B=x^{2006}-\left(x+1\right)x^{2005}+\left(x+1\right)x^{2004}-...+\left(x+1\right)x^2-\left(x+1\right)x-5\\ B=x^{2006}-x^{2006}-x^{2005}+x^{2005}+x^{2004}-...+x^3+x^2-x^2-x-5\\ B=-x-5=-12\)
a)
A(x) = \(x^6-2x^5+9x^4+5x^3-4x+7\)
B(x) = \(-x^6-2x^5+9x^4-5x^3+2x-7\)
b)
A(x)+B(x) = \(-4x^5+18x^4-2x\)
A(x)-B(x) = \(2x^6+10x^3-6x+14\)
3, đk : x =< 3/5
TH1 : \(x-2=3-5x\Leftrightarrow6x=5\Leftrightarrow x=\dfrac{5}{6}\)(ktm)
TH2 : \(x-2=5x-3\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\)(tm)
4, \(\Leftrightarrow8x-14=3x+21\Leftrightarrow5x=35\Leftrightarrow x=7\)
Bài 3:
\(\Leftrightarrow x-2=3-5x\\ \Leftrightarrow x+5x=3+2\\ \Leftrightarrow6x=5\\ \Leftrightarrow x=\dfrac{5}{6}\)
Vậy \(x=\dfrac{5}{6}\)
Bài 4:
\(\Leftrightarrow8x-14=3x+3+18\)
\(\Leftrightarrow8x-3x=3+18+14\\ \Leftrightarrow5x=35\\ \Leftrightarrow x=\dfrac{35}{5}=7\)
Vậy x = 7
Thay x = 4 vào A ta được:
5.4⁵ - 5.4⁴ + 5.4³ - 5.4² + 5.4 - 1
= 5.1024 - 5.256 + 5.64 - 5.16 + 5.4 - 1
= 5120 - 1280 + 320 - 80 + 20 - 1
= 4099
15{240-[4x(25-5x4)+80]:4}
=15.{240-[4x5+80]:4}
=15.{240-100:4}
=15.215
=3225
Lời giải:
a. PT $\Leftrightarrow (3-2x-3-2x)(3-2x+3+2x)=8$
$\Leftrightarrow -4x.6=8$
$\Leftrightarrow -24x=8\Leftrightarrow x=\frac{-1}{3}$
b.
$9x^5-72x^2=0$
$\Leftrightarrow 9x^2(x^3-8)=0$
$\Leftrightarrow x^2=0$ hoặc $x^3=8$
$\Leftrightarrow x=0$ hoặc $x=2$
c.
$5x^4-8x^2-4=0$
$\Leftrightarrow 5x^4-10x^2+2x^2-4=0$
$\Leftrightarrow 5x^2(x^2-2)+2(x^2-2)=0$
$\Leftrightarrow (5x^2+2)(x^2-2)=0$
$\Leftrightarrow 5x^2+2=0$ (loại) hoặc $x^2-2=0$ (chọn)
$\Leftrightarrow x=\pm \sqrt{2}$
d.
PT $\Leftrightarrow [x^2(x+1)-4(x+1)]:(x-2)=0$
$\Leftrightarrow (x^2-4)(x+1):(x-2)=0$
$\Leftrightarrow (x-2)(x+2)(x+1):(x-2)=0$
$\Leftrightarrow (x+2)(x+1)=0$
$\Leftrightarrow x+2=0$ hoặc $x+1=0$
$\Leftrightarrow x=-2$ hoặc $x=-1$
a: Ta có: \(\left(3-2x\right)^2-\left(3+2x\right)^2=8\)
\(\Leftrightarrow9-12x+4x^2-9-12x-4x^2=8\)
\(\Leftrightarrow-24x=8\)
hay \(x=-\dfrac{1}{3}\)
b: Ta có: \(9x^5-72x^2=0\)
\(\Leftrightarrow9x^2\left(x^3-8\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
4x : 5x= 4x/5x=4/5
4x : 5x = 4x/5x = 4/5 x = 4/5