Bài làm

a) \(\frac{10^4.81-16.15^2}{\left(-8\right)^4.3^{12}+6^{11}}=\frac{10^4.3^4-4^2.15^2}{8^4.3^{12}+6^{11}}=\frac{30^4-60^2}{2^{12}.3^{12}+6^{11}}=\frac{\left(30^2\right)^2-60^2}{6^{12}+6^{11}}\)

\(=\frac{900^2-60^2}{6^{11}.\left(6+1\right)}=\frac{60^2.\left(15^2-1\right)}{6^{11}.7}=\frac{60^2.224}{6^{11}.7}=\frac{2^9.3^2.5^2.7}{2^{11}.3^{11}.7}=\frac{5^2}{2^2.3^9}=\frac{25}{78732}\)

180 nha

Đặt \(C=\frac{3\left|x\right|+2}{4\left|x\right|-5}\)

\(\Rightarrow\frac{4}{3}C=\frac{4}{3}.\left(\frac{3\left|x\right|+2}{4\left|x\right|-5}\right)=\frac{12\left|x\right|+8}{12\left|x\right|-15}=\frac{12\left|x\right|-15+23}{12\left|x\right|-15}\)

                                                                \(=1+\frac{23}{12\left|x\right|-15}\)

Để C đạt GTLN \(\Leftrightarrow\left(12\left|x\right|-15\right)_{min}\)

Vì \(\left|x\right|\ge0\left(\forall x\right)\Rightarrow12\left|x\right|\ge0\Rightarrow12\left|x\right|-15\ge-15\)

Dấu "=" xảy ra <=> \(\left|x\right|=0\Leftrightarrow x=0\)

Vậy ...

\(\frac{x}{2}=\frac{y}{4}=k\)

=>   \(x=2k;\)\(y=4k\)

Theo bài ra ta có:

\(x^4.y^4=16\)

<=>  \(\left(2k\right)^4.\left(4k\right)^4=16\)

<=> \(4096.k^8=16\)

<=> \(k^8=\frac{1}{256}\)

<=>  \(k=\pm\frac{1}{2}\)

làm nốt phần còn lại

        x/2=y/4

=>   2y=4x

<=>   y=2x

thay vào , ta có

        x⁴ .(2x)⁴ =16

<=> 16x⁸=16

<=>    x⁸ =1

=> x= 1 hoặc x=-1

thay vào ta có 2 cặp (x,y) là ( 1,2) và (-1,-2)

 mn xem giúp em nhanh nhanh dc ko ạ em dag cần gấp

a) \(\frac{21}{x}=\frac{7}{4}\Rightarrow x=\frac{21.4}{7}=12\)

     \(\frac{y}{16}=\frac{7}{4}\Rightarrow y=\frac{16.7}{4}=28\)

      \(\frac{-14}{z}=\frac{7}{4}\Rightarrow x=\frac{\left(-14\right).4}{7}=-8\)

Vậy ...

b) \(\frac{-21}{x}=\frac{-3}{4}\Rightarrow x=\frac{\left(-21\right).4}{-3}=28\)

     \(\frac{y}{-16}=\frac{-3}{4}\Rightarrow y=\frac{\left(-16\right).\left(-3\right)}{4}=12\)

      \(\frac{81}{z}=\frac{-3}{4}\Rightarrow z=\frac{81.4}{-3}=-108\)

\(x^2+5x+6\ge0\)

\(\Leftrightarrow x^2+2x+3x+6\ge0\)

\(\Leftrightarrow x\left(x+2\right)+3\left(x+2\right)\ge0\)

\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x+2\ge0\\x+3\ge0\end{cases}}\) hoặc \(\hept{\begin{cases}x+2\le0\\x+3\le0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge-2\\x\ge-3\end{cases}}\)   hoặc    \(\hept{\begin{cases}x\le-2\\x\le-3\end{cases}}\)

Vậy \(x\ge-2\) hoặc \(x\le-3\)

\(x^2+5x+6\ge0\)

<=>  \(\left(x+2\right)\left(x+3\right)\ge0\)

TH1:  \(\hept{\begin{cases}x+2\ge0\\x+3\ge0\end{cases}}\)<=>   \(\hept{\begin{cases}x\ge-2\\x\ge-3\end{cases}}\)<=>  \(x\ge-2\)

TH2:  \(\hept{\begin{cases}x+2\le0\\x+3\le0\end{cases}}\)<=>   \(\hept{\begin{cases}x\le-2\\x\le-3\end{cases}}\)<=>  \(x\le-3\)

Vậy....