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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{x}{42}=\frac{15}{21}=\frac{5}{7}\Rightarrow7x=42.5\)
\(\Rightarrow7x=210\)
\(\Rightarrow x=30\)
Tương tự: \(\frac{45}{y}=\frac{5}{7}\Rightarrow5y=45.7\)
\(\Rightarrow5y=315\)
\(\Rightarrow y=63\)
\(\frac{120}{z}=\frac{5}{7}\Rightarrow5z=120.7\)
\(\Rightarrow5z=840\)
\(\Rightarrow z=168\)
Vậy x = 30; y = 63 và z = 168
Ta có : \(\frac{15}{21}=\frac{5}{7}\rightarrow\frac{x}{42}=\frac{45}{y}=\frac{120}{z}=\frac{5}{7}\)
Mà : \(\frac{x}{42}=\frac{5}{7}\rightarrow x=\frac{42\cdot5}{7}=30\)
\(\frac{45}{y}=\frac{5}{7}\rightarrow y=\frac{45\cdot7}{5}=63\)
\(\frac{120}{z}=\frac{5}{7}\rightarrow z=\frac{120.7}{5}=168\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a,Ta có: \(\frac{33}{a}=\frac{45}{-120}=\frac{-y}{8}=\frac{z}{160}=\frac{3}{-8}\)
Do: \(\frac{33}{a}=\frac{3}{-8}\Rightarrow-8.33=3.a\Leftrightarrow-264=3.a\Leftrightarrow a=-88\)
\(\frac{-y}{8}=\frac{3}{-8}\Rightarrow8.y=3.8\Leftrightarrow y=3\)
\(\frac{z}{160}=\frac{3}{-8}\Rightarrow-8.z=3.160\Leftrightarrow-8.z=480\Leftrightarrow z=-60\)
Vậy: \(a=-88\) ; \(y=3\) ; \(z=-60\)
b, Ta có: \(\frac{x+1}{5}=\frac{y}{20}=\frac{6}{10}=\frac{3}{5}\)
Do: \(\frac{x+1}{5}=\frac{3}{5}\Rightarrow\left(x+1\right)5=3.5\Leftrightarrow x+1=3\Leftrightarrow x=2\)
\(\frac{y}{20}=\frac{3}{5}\Rightarrow y.5=3.20\Leftrightarrow y.5=60\Leftrightarrow y=12\)
Vậy: \(x=2\) ; \(y=12\)
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Tick cho mình nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
a; \(\dfrac{-x}{4}\) = \(\dfrac{-2}{x}\)
-\(x.x\) = -2.4
-\(x^2\) = -8
\(x^2\) = 8
\(\left[{}\begin{matrix}x=-\sqrt{8}\\x=\sqrt{8}\end{matrix}\right.\)
Vậy \(x\in\) {-\(\sqrt{8}\); \(\sqrt{8}\)}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{x-5}=-\frac{4}{x+2}\)
\(\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)
\(\frac{x}{-2}=-\frac{8}{x}\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x=\pm4\)
\(-\frac{2}{x}=\frac{y}{3}\)
\(\Leftrightarrow xy=-6\)
\(\Leftrightarrow x;y\inƯ\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
Xét bảng
x | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -1 |
y | -6 | 6 | -3 | 3 | -2 | 2 | -1 | 6 |
Vậy.................
\(\frac{2x-9}{240}=\frac{39}{80}\)
\(\Leftrightarrow2x-9=\frac{240.39}{80}\)
\(\Leftrightarrow2x-9=117\)
\(\Leftrightarrow2x=126\)
\(\Leftrightarrow x=63\)
Tìm số x,y,z,t biết : \(\frac{x}{140}=\frac{-18}{y}=\frac{z}{-21}=-\frac{135}{t}=\frac{2679}{6251}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x}{140}=\frac{-18}{y}=\frac{z}{-21}=\frac{-135}{t}=\frac{2679}{6251}\)
\(\Rightarrow x=\frac{140.2679}{6251}=60\)
\(y=\frac{-18.6251}{2679}=-42\)
\(z=\frac{-21.2679}{6251}=-9\)
\(t=\frac{-135.6251}{2679}=315\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{-13}{15}< \frac{y}{5}< \frac{-13}{21}\)
\(\frac{-13}{15}=\frac{-13.7}{15.7}=\frac{-91}{105}\)
\(\frac{y}{5}=\frac{21.y}{5.21}=\frac{21y}{105}\)
\(\frac{-13}{21}=\frac{-13.5}{21.5}=\frac{-65}{105}\)
=>\(\frac{-91}{105}< \frac{21y}{105}< \frac{-65}{105}\)
=> \(-91< 21y< -65\)
=> \(y=-4\)
Vì 21.(-4)=-84 (-91<-84<-65)
=>\(y=-4\)
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Chữ I là giá trị tuyệt đối nhé!