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T giải thử thôi nhé :w
a) \(1\frac{1}{4}x^2y\left(\frac{-5}{6}xy\right)^0.\left(-2\frac{1}{3}xy\right)\)
\(=\frac{5}{4}x^2y\left(\frac{-5}{6}xy\right)^0.\left(-\frac{5}{2}xy\right)\)
\(=1.\frac{5}{4}x^2y\left(-\frac{5}{2}xy\right)\)
\(=-\frac{5}{4}x^2y.1.\frac{5}{2}xy\)
\(=-1.\frac{5}{4}.\frac{5}{2}x^3y^2\)
\(=-1.\frac{25x^3y^2}{8}\)
\(=-\frac{25x^3y^2}{8}\)
a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)
=>x+1=0
hay x=-1
b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
\(A=x^3.\left(-\dfrac{5}{4}x^2y\right).\left(\dfrac{2}{5}x^3y^4\right).\\ A=-\dfrac{1}{2}x^8y^5.\)
- Bậc: 8.
- Hệ số: \(-\dfrac{1}{2}.\)
- Biến: \(x;y.\)
\(B=\left(-\dfrac{3}{4}x^5y^4\right).\left(xy^2\right).\left(-\dfrac{8}{9}x^2y^3\right).\\ B=\dfrac{2}{3}x^8y^9.\)
- Bậc: 9.
- Hệ số: \(\dfrac{2}{3}.\)
- Biến: \(x;y.\)
1.
a)\(\left(\dfrac{1}{2}\cdot\left(-2\right)\cdot\dfrac{-1}{3}\right)\cdot\left(x^2\cdot x^2\cdot x^2\right)\cdot\left(y^2\cdot y^3\right)\cdot z\)
\(\dfrac{1}{3}x^6y^5z\)
Deg=12
a) \(A=\left(\dfrac{1}{2^3}.3.\dfrac{13}{3}\right)\left(a^{3+2+1}\right)\left(x^{1+3}\right)\left(y^{1+2}\right)=\dfrac{13}{8}.a^6.x^4.y^3\)
\(B=\left[2^k.\left(-\dfrac{1}{2}\right)^2\right]\left(x^{2k+2}\right)\left(y^{3k+2.2}\right)\left(z^{4k+}\right)=2^{k-2}.x^{2\left(k+1\right)}.y^{3k+4}.z^{4k}\)
Bài 1:
Ta có:
\(\left(\dfrac{ab}{2}-\dfrac{6ab}{7}\right):\dfrac{5b^2}{14}=\left(\dfrac{7ab}{14}-\dfrac{12ab}{14}\right).\dfrac{14}{5b^2}\)
\(=\dfrac{-5ab}{14}.\dfrac{14}{5b^2}=\dfrac{-a}{b}\)(1)
Thay \(a=\dfrac{2007}{2010};b=\dfrac{2011}{2010}\) vào (1) ta được:
\(\dfrac{-\dfrac{2007}{2010}}{\dfrac{2011}{2010}}=-\dfrac{2007}{2011}\)
Vậy......................
Chúc bạn học tốt!!!
Bài 2:
\(\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right).\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}< \dfrac{x}{8}< -\dfrac{1}{2}.\dfrac{3}{4}:\dfrac{1}{8}+1\)
\(\Rightarrow2.\left(-\dfrac{9}{2}\right)-\dfrac{1}{4}< \dfrac{x}{8}< -3+1\)
\(\Rightarrow-\dfrac{37}{4}< \dfrac{x}{8}< -2\)
\(\Rightarrow\dfrac{-74}{8}< \dfrac{x}{8}< -\dfrac{16}{8}\)
\(\Rightarrow-74< x< -16\)
\(\Rightarrow x\in\left\{-73;-72;-71;....;-18;-17\right\}\)
Vậy..............................
Chúc bạn học tốt!!!
Lời giải:
Gọi đa thức là $T$
\(T=(\frac{-2007}{3})^8(xy)^8.(\frac{-6}{2007})^8(x^2y)^8\)
\(=\frac{2007^8}{3^8}.x^8y^8.\frac{6^8}{2007^8}.x^{16}.y^8\)
\(=\frac{6^8}{3^8}.x^{8+16}.y^{8+8}=2^8.x^{24}y^{16}\)
Ta có: \(\left(-\dfrac{2007}{3}xy\right)^8\cdot\left(-\dfrac{6}{2007}x^2y\right)^8\)
\(=\left(\dfrac{2007}{3}\cdot\dfrac{6}{2007}\right)^8\cdot x^8\cdot x^{16}\cdot y^8\cdot y^8\)
\(=256x^{24}y^{16}\)