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a, \(\left[x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\right]x^2-1\)
\(=\left[x\left(x^2-16\right)-\left(x^2+1\right)\right]x^2-1\)
\(=\left[x^3-16x-x^2-1\right]x^2-1\)
\(=x^5-16x^3-x^4-x^2-1\)
b, \(\left(y-3\right)y+3y^2+9-y^2+2\left(y^2-2\right)\)
\(=y^2-3y+3y^2+9-y^2+2y^2-4\)
\(=5y^2-3y+5\)
c, \(\left(x+y\right)\left(x^2x^2-xy+y^2\right)\)
\(=x^5-x^2y+xy^2+x^4y-xy^2+y^3\)
d, \(\left(\dfrac{1}{2}xy+\dfrac{3}{4}y\right).\dfrac{1}{2}xy-\dfrac{3}{4}y\)
\(=\dfrac{1}{4}x^2y^2+\dfrac{3}{8}xy^2-\dfrac{3}{4}y\)
\(=\dfrac{1}{4}y.\left(x^2y+\dfrac{3}{2}xy-3\right)\)
Chúc bạn học tốt!!!
Đặt \(\dfrac{x}{3}=\dfrac{y}{2}=k\left(k\in Z\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=2k\end{matrix}\right.\)
Mà \(x.y^2=96\)
\(\Rightarrow3k.4k^2=96\)
\(\Rightarrow12k^3=96\)
\(\Rightarrow k^3=8\)
\(\Rightarrow k=2\)
a. Đặt \(\dfrac{x}{3}=\dfrac{y}{2}=k\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=2k\end{matrix}\right.\)
mà x . y2 = 96
hay \(3k.\left(2k\right)^2=96\)
\(\Rightarrow3k.4.k^2=96\)
\(\Rightarrow12.k^3=96\)
\(\Rightarrow k^3=8=2^3\)
\(\Rightarrow k=2\)
Với k = 2 \(\Rightarrow\begin{matrix}x=3.2=6\\y=2.2=4\end{matrix}\)
Vậy........
b. Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x}{6}=\dfrac{4y}{20}=\dfrac{2x+4y}{6+20}=\dfrac{28}{26}=\dfrac{14}{13}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{14}{13}\Rightarrow x=\dfrac{3.14}{13}=\dfrac{42}{13}\\\dfrac{y}{5}=\dfrac{14}{13}\Rightarrow y=\dfrac{5.14}{13}=\dfrac{70}{13}\end{matrix}\right.\)
Vậy...........
e, Đặt \(\dfrac{x}{4}=\dfrac{y}{5}=k\left(k\in Z\right)\)
\(\Leftrightarrow x=4k,y=5k\) (1)
Theo bài ra ta có: xy = 80
Từ (1) \(\Rightarrow4k.5k=80\Rightarrow20.k^2=80\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k^2=2^2\\k^2=\left(-2\right)^2\end{matrix}\right.\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
+ Với k = 2 \(\Rightarrow\left\{{}\begin{matrix}x=8\\y=10\end{matrix}\right.\)
+ Với k = -2 \(\Rightarrow\left\{{}\begin{matrix}x=-8\\y=-10\end{matrix}\right.\)
Vậy \(\left(x,y\right)\in\left\{\left(8,10\right);\left(-8,-10\right)\right\}\)
a) \(\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}=\dfrac{5x}{15}=\dfrac{3z}{-6}=\dfrac{5x-y+3z}{15-5-6}=\dfrac{-16}{4}=-4\Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-4\\\dfrac{y}{5}=-4\\\dfrac{z}{-2}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-12\\y=-20\\z=8\end{matrix}\right.\)
1. Áp dụng tc dãy TSBN, ta có:
\(\dfrac{x}{6}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{x+y-z}{6+5-3}=\dfrac{54}{8}=\dfrac{27}{4}\)
+\(\dfrac{x}{6}=\dfrac{27}{4}\Rightarrow x=\dfrac{27.6}{4}=\dfrac{81}{2}\)
+\(\dfrac{y}{5}=\dfrac{27}{4}\Rightarrow y=\dfrac{27.5}{4}=\dfrac{135}{4}\)
+\(\dfrac{z}{3}=\dfrac{27}{4}\Rightarrow z=\dfrac{27.3}{4}=\dfrac{81}{4}\)
Vậy \(x=\dfrac{81}{2};y=\dfrac{135}{4};z=\dfrac{81}{4}\)
2,Áp dụng tc dãy TSBN, ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{c}{4}=\dfrac{x+2y-3c}{2+2.3+3.4}=\dfrac{-20}{20}=-1\)
+\(\dfrac{x}{2}=-1\Rightarrow x=-1.2=-2\)
+\(\dfrac{y}{3}=-1\Rightarrow y=-1.3=-3\)
+\(\dfrac{c}{4}=-1\Rightarrow c=-1.4=-4\)
Vậy \(x=-2;y=-3;c=-4\)
Theo đề bài ta có :
\(\dfrac{x-y}{3}=\dfrac{x+y}{13}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x-y}{3}=\dfrac{x+y}{13}=\dfrac{x-y+x+y}{3+13}=\dfrac{2x}{16}=\dfrac{x}{8}\)
\(\dfrac{x}{8}=\dfrac{xy}{200}\Leftrightarrow\) \(\dfrac{x}{xy}=\dfrac{8}{200}\Rightarrow\) \(\dfrac{1}{y}=\dfrac{1}{25}\) \(\Rightarrow y=25\)
Thay y = 25 vào biểu thức ta có :
\(\dfrac{x-25}{3}=\dfrac{x+25}{13}\)
\(\Leftrightarrow\) \(13x-325=3x+75\)
\(\Leftrightarrow13x-3x=75+325\)
\(\Leftrightarrow10x=400\)
\(\Rightarrow x=40\)
Vậy \(x=40\) ; \(y=25\)
b: 2x^3-1=15
=>2x^3=16
=>x=2
\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)
=>\(\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)
=>y-25=32; z+9=50
=>y=57; z=41
d: 3/5x=2/3y
=>9x=10y
=>x/10=y/9=k
=>x=10k; y=9k
x^2-y^2=38
=>100k^2-81k^2=38
=>19k^2=38
=>k^2=2
TH1: k=căn 2
=>\(x=10\sqrt{2};y=9\sqrt{2}\)
TH2: k=-căn 2
=>\(x=-10\sqrt{2};y=-9\sqrt{2}\)
a) \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)
Từ \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Rightarrow\dfrac{x^3}{2^3}=\dfrac{y^3}{4^3}=\dfrac{z^3}{6^3}\)
\(\Leftrightarrow\dfrac{x^2}{2^2}=\dfrac{y^2}{4^2}=\dfrac{z^2}{6^2}\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{1}{4}\Rightarrow x^2=\dfrac{1}{4}\cdot4\Rightarrow x^2=1\Rightarrow x=1\)
\(\dfrac{y^2}{16}=\dfrac{1}{4}\Rightarrow y^2=\dfrac{1}{4}\cdot16\Rightarrow y^2=4\Rightarrow y=2\)
\(\dfrac{z^2}{36}=\dfrac{1}{4}\Rightarrow z^2=\dfrac{1}{4}\cdot36\Rightarrow z^2=9\Rightarrow z^2=3\)
Xin lỗi mình chỉ làm được câu a)
xy = 96 => x = 96/y => 2/x = y/48
=> y/48 = 3/y => y = 12 hoặc -12
=> x = 8 hoặc -8
\(\dfrac{2}{x}=\dfrac{3}{y}\) và x.y =96
\(=>\dfrac{x}{2}=\dfrac{y}{3}=k\)
=> x = 2k và y = 3k
Thay vào x.y = 96
(2k . 3k) = 96
\(6k^2=96\)
\(k^2=96:6\)
\(k^2=16\)
\(k=-4\) hoặc \(+4\)
Với k = - 4 => x = 2 . ( - 4 ) = - 8
y = 3 . ( - 4) = - 12
Với k = 4 => x = 2 . 4 = 8
y = 3 . 4 = 12