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Bài 1: Tìm x,y,z
a) Đặt \(\frac{x}{6}=\frac{y}{5}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=6k\\y=5k\end{matrix}\right.\)
Ta có: \(xy=192\)
\(\Leftrightarrow6k\cdot5k=192\)
\(\Leftrightarrow30k^2=192\)
\(\Leftrightarrow k^2=6.4\)
\(\Leftrightarrow k=\pm\frac{4\sqrt{10}}{5}\)
Ta có: \(\left\{{}\begin{matrix}x=6k\\y=5k\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\cdot\frac{\pm4\sqrt{10}}{5}=\pm\frac{24\sqrt{10}}{5}\\y=5\cdot\pm\frac{4\sqrt{10}}{5}=\pm4\sqrt{10}\end{matrix}\right.\)
Vậy: \(\left(x,y\right)=\left\{\left(\frac{24\sqrt{10}}{5};4\sqrt{10}\right);\left(\frac{-24\sqrt{10}}{5};-4\sqrt{10}\right)\right\}\)
b) Đặt \(\frac{x}{-3}=\frac{y}{7}=a\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3a\\y=7a\end{matrix}\right.\)
Ta có: \(x^2-y^2=-360\)
\(\Leftrightarrow\left(-3a\right)^2-\left(7a\right)^2=-360\)
\(\Leftrightarrow9a^2-49a^2+360=0\)
\(\Leftrightarrow360-40a^2=0\)
\(\Leftrightarrow40a^2=360\)
\(\Leftrightarrow a^2=9\)
hay \(a=\pm3\)
Trường hợp 1: a=3
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\cdot3\\y=7\cdot3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-9\\y=21\end{matrix}\right.\)
Trường hợp 2: a=-3
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\cdot\left(-3\right)\\y=7\cdot\left(-3\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=-21\end{matrix}\right.\)
Vậy: (x,y)={(-9;21);(9;-21)}
c) Ta có: \(\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{4}\)
\(\Leftrightarrow\frac{x-1}{2}=\frac{2y+4}{6}=\frac{3z-9}{12}\)
mà x-2y+3z=46
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{x-1}{2}=\frac{2y+4}{6}=\frac{3z-9}{12}=\frac{x-1-2y-4+3z-9}{2-6+12}=\frac{46-14}{8}=\frac{32}{8}=4\)
Do đó:
\(\left\{{}\begin{matrix}x-1=4\cdot2=8\\2y+4=4\cdot6=24\\3z-9=4\cdot12=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\2y=20\\3z=57\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=10\\z=19\end{matrix}\right.\)
Vậy: (x,y,z)=(9;10;19)
Bài 1:
a, \(x^2\) +2\(x\) = 0
\(x.\left(x+2\right)\) = 0
\(\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
\(x\) \(\in\) {-2; 0}
b, (-2.\(x\)).(-4\(x\)) + 28 = 100
8\(x^2\) + 28 = 100
8\(x^2\) = 100 - 28
8\(x^2\) = 72
\(x^2\) = 72 : 8
\(x^2\) = 9
\(x^2\) = 32
|\(x\)| = 3
\(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Vậy \(\in\) {-3; 3}
c, 5.\(x\) (-\(x^2\)) + 1 = 6
- 5.\(x^3\) + 1 = 6
5\(x^3\) = 1 - 6
5\(x^3\) = - 5
\(x^3\) = -1
\(x\) = - 1
a)
Ta có
\(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{3x}{6}=\frac{y}{5}\)
Áp dụng tc của dãy tỉ só bằng nhau
\(\Rightarrow\frac{3x}{6}=\frac{y}{5}=\frac{3x-y}{6-5}=\frac{10}{1}=10\)
=> x=2.10=20
y=5.10=50
Ta có
\(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{x^2}{4}=\frac{y^2}{25}=\frac{xy}{10}=\frac{30}{10}=3\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\sqrt{12}\\x=-\sqrt{12}\end{array}\right.\)
\(\left[\begin{array}{nghiempt}y=\sqrt{75}\\y=-\sqrt{75}\end{array}\right.\)
Mà 2;5 cùng dấu
=> x; y cùng dấu
Vậy \(\left(x;y\right)=\left(\sqrt{12};\sqrt{75}\right);\left(-\sqrt{12};-\sqrt{75}\right)\)
Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
Bài 2:
a) \(\frac{4^2\cdot25^2+16\cdot125}{2^3\cdot5^2}=\frac{2^4\cdot5^4+2^4\cdot5^3}{2^3\cdot5^2}=\frac{2^4\cdot5^3\cdot\left(5+1\right)}{2^3\cdot5^2}=2\cdot5\cdot6=60\)
b) \(\frac{6^8\cdot2^4-4^5\cdot18^4}{27^3\cdot8^4-3^9\cdot2^{13}}=\frac{2^8\cdot3^8\cdot2^4-2^{10}\cdot2^4\cdot3^8}{3^9\cdot2^{12}-3^9\cdot2^{13}}=\frac{2^{12}\cdot3^8-2^{14}\cdot3^8}{3^9\cdot2^{12}\cdot\left(1-2\right)}=\frac{2^{12}\cdot3^8\cdot\left(1-2^2\right)}{3^9\cdot2^{12}\cdot\left(-1\right)}\)
\(=\frac{2^2-1}{3}=\frac{3}{3}=1\)
Thanks bạn .