[(x^8)^3] với x khác 0
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a) \(a^5.a^7:a^{11}\left(a\inℤ\right)\)
\(\Rightarrow a^5.a^7:a^{11}=a^{5+7}:a^{11}=a^{12}:a^{11}=a^1=a\)
b) \(x^6:x^3.x^2\)
\(\Rightarrow x^{6-3}.x^2=x^3.x^2=x^5\)
c) \(\left[\left(x^8\right)^3\right]^0\left(x\ne0\right)\)
\(\Rightarrow\left[\left(x^8\right)^3\right]^0=\left(x^{24}\right)^0=1\)
a, Ta có : \(\frac{y}{x}.\sqrt{\frac{x^2}{y^4}}=\frac{y}{x}.\frac{x}{y^2}=\frac{1}{y}\)
b , Ta có : \(5xy\sqrt{\frac{x^2}{y^6}}=5xy\frac{x}{y^3}=\frac{5x^2}{y^2}\)
c, Ta có : \(0,2x^3y^3\sqrt{\frac{16}{x^4y^8}}=0,2x^3y^3.\frac{4}{x^2y^4}=\frac{0,8x}{y}\)
a) (x^4)^2=x^12/x^5 (x khác 0 )
b) x^10+=25x^8
c) (2x+3)^2=9/121
d) (3x-1)^3=-8/27
Các bạn giúp mình với
a) \(2\dfrac{3}{4}-x=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{11}{4}-x=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{11}{4}-\dfrac{3}{4}=\dfrac{8}{4}=2\)
b) \(x:\dfrac{5}{6}=-\dfrac{3}{5}\)
\(\Rightarrow x=-\dfrac{3}{5}.\dfrac{5}{6}=-\dfrac{15}{30}=-\dfrac{1}{2}\)
c) \(1\dfrac{1}{3}+\dfrac{2}{3}:x=1\)
\(\Rightarrow\dfrac{2}{3}:x=1-1\dfrac{1}{3}\)
\(\Rightarrow\dfrac{2}{3}:x=-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{3}:-\dfrac{1}{3}\)
\(\Rightarrow x=-2\)
d) \(x-\dfrac{1}{9}=\dfrac{8}{3}\)
\(\Rightarrow x=\dfrac{8}{3}+\dfrac{1}{9}\)
\(\Rightarrow x=\dfrac{25}{9}\)
e) \(\dfrac{1}{2}x+650\%x-x=-6\)
\(\Rightarrow\dfrac{1}{2}x+\dfrac{13}{2}x-x=-6\)
\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{13}{2}-1\right)-6\)
\(\Rightarrow6x=-6\)
\(\Rightarrow x=\dfrac{-6}{6}=-1\)
g) \(2\left(x-\dfrac{1}{2}\right)+3\left(-1+\dfrac{x}{3}\right)=x\left(\dfrac{2}{x}-1\right)\) \(\text{Đ}K:x\ne0\)
\(\Rightarrow2x-1-3+x=2-x\)
\(\Rightarrow3x-4=2-x\)
\(\Rightarrow3x+x=2+4\)
\(\Rightarrow4x=6\)
\(\Rightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)
\(\dfrac{1}{x^2-x}+\dfrac{x-3}{x^2-1}\left(x\ne\pm1;x\ne0\right)\)
\(=\dfrac{1}{x\left(x-1\right)}+\dfrac{x-3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+1}{x\left(x-1\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+1+x^2-3x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-2x+1}{\left(x-1\right)\left[x\left(x+1\right)\right]}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+x\right)}\)
\(=\dfrac{x-1}{x^2+x}\)
\([(x^8)^3]=[(x^2)^4)^3]=x^{512}(x\ne0)\)
Chúc bạn học tốt
\(\left[\left(x^8\right)^3\right]=x^{512}\)