x/12=y/13=z/13 và x+Y+Z=160
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Áp dụng t/c dtsbn:
\(\dfrac{x}{12}=\dfrac{y}{13}=\dfrac{z}{15}=\dfrac{x+y+z}{12+13+15}=\dfrac{160}{40}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.12=48\\y=4.13=52\\z=4.15=60\end{matrix}\right.\)
a) ADTCDTSBN
có: \(\frac{x}{12}=\frac{y}{13}=\frac{z}{15}=\frac{x+y+z}{12+13+15}=\frac{160}{40}=4\)
=> x/12 = 4 => x = 48
...
b) ta có: \(x=\frac{y}{6}=\frac{z}{3}=\frac{2x}{2}=\frac{3y}{18}=\frac{4z}{12}\)
ADTCDTSBN
có: \(\frac{2x}{2}=\frac{3y}{18}=\frac{4z}{12}=\frac{2x-3y+4z}{2-18+12}=\frac{16}{-4}=-4\)
=>...
c) ta có: \(\frac{x}{2}=\frac{y}{-3}=\frac{z}{3}=\frac{2x}{4}=\frac{3y}{-9}=\frac{2z}{8}\)
ADTCTDBN
có: \(\frac{2x}{4}=\frac{3y}{-9}=\frac{2z}{8}=\frac{2x+3y+2z}{4-9+8}=\frac{1}{3}\)
=>...
7) 5x=4y ⇒\(\dfrac{x}{4}=\dfrac{y}{5}\)
Nhân cả hai vế với \(\dfrac{x}{4}\), ta có: \(\left(\dfrac{x}{4}\right)^2=\dfrac{x}{4}.\dfrac{y}{5}=\dfrac{xy}{20}=\dfrac{20}{20}=1\)
\(\left(\dfrac{x}{4}\right)^2=1\Rightarrow\left[{}\begin{matrix}\dfrac{x}{4}=1\\\dfrac{x}{4}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}y=5\\y=-5\end{matrix}\right.\)
4) áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{0,5}=\dfrac{y}{0,3}=\dfrac{z}{0,2}=\dfrac{z-y+x}{0,2-0,3+0,5}=\dfrac{1}{\dfrac{2}{5}}=\dfrac{5}{2}\)
\(\dfrac{x}{0,5}=\dfrac{5}{2}\Rightarrow x=\dfrac{5}{4}\)
\(\dfrac{y}{0,3}=\dfrac{5}{2}\Rightarrow y=\dfrac{3}{4}\)
\(\dfrac{z}{0,2}=\dfrac{5}{2}\Rightarrow z=\dfrac{1}{2}\)
6) áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x+11}{13}=\dfrac{y+12}{14}=\dfrac{z+13}{15}=\dfrac{x+11+y+12+z+13}{13+14+15}=\dfrac{42}{42}=1\)
\(\dfrac{x+11}{13}=1\Rightarrow x=2\)
\(\dfrac{y+12}{13}=1\Rightarrow y=1\)
\(\dfrac{z+13}{15}=1\Rightarrow z=2\)
7) \(5x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{5}=k\)
\(\Rightarrow x=4k,y=5k\)
\(x.y=20\\ \Rightarrow4k.5k=20\\ \Rightarrow20k^2=20\\ \Rightarrow k^2=1\\ \Rightarrow\left[{}\begin{matrix}k=-1\\k=1\end{matrix}\right.\)
\(x=4k\Rightarrow\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)
\(y=5k\Rightarrow\left[{}\begin{matrix}y=-5\\y=5\end{matrix}\right.\)
Vậy \(\left(x,y\right)=\left\{\left(-4;-5\right);\left(4;5\right)\right\}\)
\(\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{15}.\)
\(\Rightarrow\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{15}=\frac{x+11+y+12+z+13}{13+14+15}=\frac{\left(x+y+z\right)+\left(11+12+13\right)}{42}\)
\(=\frac{6+36}{42}=\frac{42}{42}=1\) ( Áp dụng tính chất dãy tỉ số bằng nhau )
\(\Rightarrow\hept{\begin{cases}\frac{x+11}{13}=1\\\frac{y+12}{14}=1\\\frac{z+13}{15}=1\end{cases}}\Rightarrow\hept{\begin{cases}x+11=13\\y+12=14\\z+13=15\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y=2\\z=2\end{cases}}\)
Vậy \(x=y=z=2\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{15}=\frac{x+11+y+12+z+13}{13+14+15}\)
\(=\frac{\left(x+y+z\right)+\left(11+12+13\right)}{13+14+15}=\frac{16+36}{42}=\frac{42}{42}=1\)
\(\Rightarrow\frac{x+11}{13}=1\Rightarrow x+11=13\Rightarrow x=13-11=2\)
\(\Rightarrow\frac{y+12}{14}=1\Rightarrow y+12=14\Rightarrow y=14-12=2\)
\(\Rightarrow\frac{z+13}{15}=1\Rightarrow z+13=15\Rightarrow z=15-13=2\)
Vậy \(x=y=z=2\)
C1:
\(\text{Ta có: }\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{15};x+y+z=6\left(1\right)\)
\(\text{Áp dụng tính dãy tỉ số bằng nhau: }\)
\(\Rightarrow\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{15}=\frac{\left(x+11\right)+\left(y+12\right)+\left(z+13\right)}{13+14+15}=\frac{x+11+y+12+z+13}{42}=\frac{\left(x+y+z\right)+\left(11+12+13\right)}{42}\left(2\right)\)
\(\text{Thay (1) và (2), ta được: }\)
\(\frac{6+36}{42}=\frac{42}{42}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x+11}{13}=1\\\frac{y+12}{14}=1\\\frac{z+13}{15}=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+11=13\\y+12=14\\z+13=15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=2\\z=2\end{matrix}\right.\)
\(\text{Vậy }x=2;y=2;z=2\)
C2:
\(x+y+z=6\left(1\right)\)
\(\text{Ta có: }\frac{x+11}{13}=\frac{y+12}{14}\Rightarrow x+11=\frac{y+12}{14}.13\Rightarrow x=\frac{13.\left(y+12\right)}{14}-11\)
\(\frac{y+12}{14}=\frac{z+13}{15}\Rightarrow z+13=\frac{y+12}{14}.15\Rightarrow z=\frac{15.\left(y+12\right)}{14}-13\)
\(\text{Thay }x=\frac{13.\left(y+12\right)}{14}-11;z=\frac{15.\left(y+12\right)}{14}-13\text{ vào }\left(1\right)\)
\(\Rightarrow\frac{13.\left(y+12\right)}{14}-11+y+\frac{15.\left(y+12\right)}{14}-13=6\)
\(\Leftrightarrow\frac{13.\left(y+12\right)}{14}+y+\frac{15.\left(y+12\right)}{14}-\left(11+13\right)=6\)
\(\Leftrightarrow\frac{13.\left(y+12\right)}{14}+\frac{14y}{14}+\frac{15.\left(y+12\right)}{14}-24=6\)
\(\Leftrightarrow\frac{13.\left(y+12\right)+14y+15.\left(y+12\right)}{14}=6+24\)
\(\Leftrightarrow\frac{\left(y+12\right).\left(13+15\right)+14y}{14}=30\)
\(\Leftrightarrow\left(y+12\right).28+14y=30.14\)
\(\Leftrightarrow14.\left[\left(y+12\right).2+y\right]=420\)
\(\Leftrightarrow2y+12.2+y=420:14\)
\(\Leftrightarrow3y+24=30\)
\(\Leftrightarrow3y=30-24\)
\(\Leftrightarrow3y=6\)
\(\Leftrightarrow y=6:3\)
\(\Leftrightarrow y=2\)
\(\text{Khi đó: }x=\frac{13.\left(2+12\right)}{14}-11\left(\text{Do y=2}\right)\)
\(\Leftrightarrow x=\frac{13.14}{14}-11\)
\(\Leftrightarrow x=13-11\)
\(\Leftrightarrow x=2\)
\(\text{Khi đó: }z=\frac{15.\left(2+12\right)}{14}-13\left(\text{Do y=2}\right)\)
\(\Leftrightarrow z=\frac{15.14}{14}-13\)
\(\Leftrightarrow z=15-13\)
\(\Leftrightarrow z=2\)
\(\text{Vậy }x=2;y=2;z=2\)
Theo đề bài :
\(\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{13}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{13}=\frac{x+y+z+36}{40}=\frac{21}{20}\)
=> \(\frac{x+11}{13}=\frac{21}{20}\)
=> \(\frac{y+12}{14}=\frac{21}{20}\)
=> \(\frac{z+13}{13}=\frac{21}{20}\)
Rồi đến đây bạn tự làm nốt đi ha !!!
Bài 3:
a: A=-5-7-8=-20
b: =-3+7+2*(-8)=-16+4=-12
c: =2-3*15-8=-55
Bài 4:
a: x-y+2=-12-25+2=-37+2=-35
b: x+y+z-y-3=x+z-3=-12-15-3=-30
c: =20-12-(25-15)+12=20-10=10
d: =25-(-12+25+15)+3
=25+12-25-15-3
=-3-3=-6
Áp dụng tính chất của dãy tỉ số bằng nhau ta được:
\(\dfrac{x}{12}=\dfrac{y}{13}=\dfrac{z}{13}=\dfrac{x+y+z}{12+13+13}=\dfrac{160}{38}=\dfrac{80}{19}\)
Do đó: \(\left\{{}\begin{matrix}x=\dfrac{960}{19}\\y=z=\dfrac{1040}{19}\end{matrix}\right.\)
giúp ạ