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a)\(\frac{x}{5}=\frac{y}{-3}\Rightarrow\frac{x^2+y}{5^2.-3}=\frac{34}{-125}\)
\(\Rightarrow\frac{x}{5}=-\frac{34}{125}\Rightarrow x=-\frac{34}{125}.5=-\frac{34}{25}\)
\(\Rightarrow\frac{y}{-3}=-\frac{34}{125}\Rightarrow y=-\frac{34}{125}.-3=\frac{102}{125}\)
b)\(4x=-5y\Rightarrow\frac{4x}{20}=-\frac{5y}{20}\Rightarrow\frac{x}{5}=\frac{y}{-4}=K\)
\(\frac{x}{5}=K\Rightarrow x=5K;\frac{y}{-4}=K\Rightarrow y=-4K\)
\(x.y=-80\)
\(5K.-4K=-80\)
\(K^2.\left(-4.5\right)=-80\)
\(K^2=-80:\left(-20\right)\)
\(K^2=4\Rightarrow K=2\)
\(\frac{x}{5}=2\Rightarrow x=10\)
\(\frac{y}{-4}=2\Rightarrow y=-8\)
a, Đặt \(\hept{\begin{cases}x=5k\\y=-3k\end{cases}}\)Theo bài ra ta có : \(x^2+y=34\)
\(\left(5k\right)^2-3k=34\Leftrightarrow25k^2-3k=34\Leftrightarrow k\left(25k-3\right)=34\)
\(\Leftrightarrow\orbr{\begin{cases}k=34\\25k-3=34\end{cases}\Leftrightarrow\orbr{\begin{cases}k=34\\k=\frac{37}{25}\end{cases}}}\)
b, Theo bài ra ta có : \(4x=-5y\Leftrightarrow\frac{x}{-5}=\frac{y}{4}\)
Đặt \(\hept{\begin{cases}x=-5k\\y=4k\end{cases}}\)Theo bài ra ta có : \(xy=-80\)
\(\Leftrightarrow-5k.4k=-80\Leftrightarrow-20k^2=-80\Leftrightarrow k^2=4\Leftrightarrow k=\pm2\)
Với k = 2 : \(\hept{\begin{cases}x=-10\\y=8\end{cases}}\)Với k = -2 \(\hept{\begin{cases}x=10\\y=-8\end{cases}}\)
a: \(\dfrac{-0.2}{x}=\dfrac{x}{-0.8}\)
\(\Leftrightarrow x^2=\dfrac{1}{5}\cdot\dfrac{4}{5}=\dfrac{4}{25}\)
=>x=2/5 hoặc x=-2/5
c: \(\dfrac{x-1}{x-2}=\dfrac{-3}{4}\)
=>4(x-1)=-3(x-2)
=>4x-4=-3x+6
=>7x=10
hay x=10/7
d: \(\dfrac{2-x}{5-x}=\dfrac{x+3}{x+2}\)
\(\Leftrightarrow\dfrac{x+3}{x+2}=\dfrac{x-2}{x-5}\)
\(\Leftrightarrow\left(x+3\right)\left(x-5\right)=\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow x^2-2x-15=x^2-4\)
=>-2x=11
hay x=-11/2
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) \(\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}=\dfrac{5x}{15}=\dfrac{3z}{-6}=\dfrac{5x-y+3z}{15-5-6}=\dfrac{-16}{4}=-4\Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-4\\\dfrac{y}{5}=-4\\\dfrac{z}{-2}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-12\\y=-20\\z=8\end{matrix}\right.\)
Bài 1:
1)
\(\dfrac{3x+2}{4}\) = \(\dfrac{5x-3}{3}\)
<=> 3(3x + 2) = 4(5x - 3)
<=> 9x + 6 = 20x - 12
<=> 6 +12 = 20x - 9x
<=> 11x = 18
<=> x = \(\dfrac{18}{11}\)
Vậy: x = \(\dfrac{18}{11}\)
2)
\(\dfrac{x-1}{3x+2}\)= \(\dfrac{1}{5}\)
<=> 5(x - 1) = 3x + 2
<=> 5x - 5 = 3x + 2
<=> 5x - 3x = 2 +5
<=> 2x = 7
<=> x = \(\dfrac{7}{2}\)
Vậy : x = \(\dfrac{7}{2}\)
Bài 1 :
1) Ta có :
\(\dfrac{3x+2}{4}=\dfrac{5x-3}{3}\\ \Leftrightarrow4\cdot\left(5x-3\right)=3\cdot\left(3x+2\right)\\ \Leftrightarrow20x-12=9x+6\\ \Leftrightarrow20x-18=9x\\ \Leftrightarrow20x-9x=18\\ \Leftrightarrow11x=18\\ \Leftrightarrow x=\dfrac{18}{11}\\ Vậy.,...\)
2) Ta có :
\(\dfrac{x-1}{3x+2}=\dfrac{1}{5}\Leftrightarrow5\cdot\left(x-1\right)=3x+2\\ \Leftrightarrow5x-5=3x+2\\ \Leftrightarrow5x-3x-5=2\\ \Leftrightarrow2x-5=2\\ \Leftrightarrow2x=7\\ \Leftrightarrow x=\dfrac{7}{2}\)
Vậy ....
Bài 2 ;
1) Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{x+y}{3+4}=\dfrac{21}{7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=3\cdot3=9\\y=3\cdot4=12\end{matrix}\right.\\ Vậy...\)
2) Ta có : \(3x=5y\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-y}{5-3}=\dfrac{-16}{2}=-8\\ \Rightarrow\left\{{}\begin{matrix}x=-8\cdot5=-40\\y=-8\cdot3=-24\end{matrix}\right.\\ Vậy....\)
3) Ta có : \(4x=7y\Leftrightarrow\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{x^2}{7^2}=\dfrac{y^2}{4^2}=\dfrac{x\cdot y}{7\cdot4}\\ \Leftrightarrow\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{112}{28}=4\\ \Rightarrow\left\{{}\begin{matrix}x=4\cdot7=28\\y=4\cdot4=16\end{matrix}\right.\\ Vậy...\)
a)Xét \(x=\dfrac{y}{2}=\dfrac{z}{3}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=k\\y=2k\\z=3k\end{matrix}\right.\) (1)
Thay (1) vào 4x - 3y + 2z = 36
\(\Rightarrow4.k-3.2k+2.3k=36\)
\(\Rightarrow4k-6k+6k=36\Rightarrow4k=36\)
\(\Rightarrow k=\dfrac{36}{4}=9\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=2.4=8\\z=3.4=12\end{matrix}\right.\)
Vậy...............................................................
b) Xét \(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{7}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\\z=7k\end{matrix}\right.\) (2)
Thay (2) vào 2x - 3z = 44
\(\Rightarrow2.5k-3.7k=44\)
\(\Rightarrow-11k=44\Rightarrow k=-4\)
\(\Rightarrow\left\{{}\begin{matrix}x=5.\left(-4\right)=-20\\y=4.\left(-4\right)=-16\\z=7.\left(-4\right)=-28\end{matrix}\right.\)
Vậy,................................................
c) Xét \(\dfrac{-x}{7}=\dfrac{y}{11}=\dfrac{-z}{5}=\dfrac{x}{-7}=\dfrac{z}{-5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=-7k\\y=11k\\z=-5k\end{matrix}\right.\) (3)
Thay (3) vào -3z - 2y - x = -88
\(\Rightarrow-3.\left(-5k\right)-2.11k-\left(-7k\right)=-88\)
\(\Rightarrow15k-22k+7k=-88\Rightarrow0k=88\)
\(\Rightarrow k\in\varnothing\)
Suy ra: Không có cặp ( x; y; z) thỏa mãn
Vậy.................................................................
d) Xét \(\dfrac{y}{12}=\dfrac{x}{-5}=\dfrac{z}{11}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=-5k\\y=12k\\z=11k\end{matrix}\right.\) (4)
Thay (4) vào 5y - 2z = 114
\(\Rightarrow6.12k-2.11k=114\)
\(\Rightarrow50k=114\Rightarrow k=2,28\)
\(\Rightarrow\left\{{}\begin{matrix}x=-5.2,28=-11,4\\y=12.2,28=27,36\\z=25,08\end{matrix}\right.\)
Vậy..............................................
e) Xét \(\dfrac{x}{25}=\dfrac{y}{17}=\dfrac{z}{32}=k\)
\(\left\{{}\begin{matrix}x=25k\\y=17k\\z=32k\end{matrix}\right.\) (5)
Thay (5) vào -2z + 3y - 4x = -452
\(\Rightarrow\left(-2\right).32k+3.17k-4.25k=-452\)
\(\Rightarrow-113k=-452\Rightarrow k=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=25.5=100\\y=17.4=68\\z=32.4=128\end{matrix}\right.\)
Vậy.......................................................
a) Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(x=\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\\ \Rightarrow\dfrac{4x}{4}-\dfrac{3y}{6}+\dfrac{2z}{6}=\dfrac{4x-3y+2z}{4-6+6}=\dfrac{36}{4}=9\)
+) \(\dfrac{x}{1}=9\Rightarrow x=9\)
+) \(\dfrac{y}{2}=9\Rightarrow y=18\)
+) \(\dfrac{z}{3}=9\Rightarrow z=27\)
Vậy x = 9; y = 18; z = 27.
tương tự
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!!!
Bài 1:
a: \(\Leftrightarrow\dfrac{x+2}{2}=x-5\)
=>2x-10=x+2
=>x=12
b: \(\Leftrightarrow\left(x+2\right)^2=100\)
=>x+2=10 hoặc x+2=-10
=>x=-12 hoặc x=8
c: \(\Leftrightarrow\left(2x-5\right)^3=27\)
=>2x-5=3
=>2x=8
=>x=4
b: Ta có: x/y=7/9
nên x/7=y/9
=>x/49=y/63
Ta có: y/z=7/3
nên y/7=z/3
=>y/63=z/27
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{49}=\dfrac{y}{63}=\dfrac{z}{27}=\dfrac{x-y+z}{49-63+27}=\dfrac{-15}{13}\)
Do đó: x=-735/13; y=-945/13; z=-405/13
c: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}=\dfrac{2x+5y-2z}{2\cdot7+5\cdot20-2\cdot32}=\dfrac{100}{50}=2\)
Do đó: x=14; y=40; z=64
d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{2}=\dfrac{x-y-z}{8-5-2}=3\)
Do đó: x=24; y=15; z=6
a)\(\dfrac{x}{5}=\dfrac{y}{-3}va\) \(x^2+y=34\)
Từ \(\dfrac{x}{5}=\dfrac{y}{-3}\)=\(\dfrac{x^2}{25}=\dfrac{y}{-3}\)
Áp dụng t/c của dãy tỉ số bằng nhau,ta có:
\(\dfrac{x}{5}=\dfrac{y}{-3}\)=\(\dfrac{x^2}{25}=\dfrac{y}{-3}\)=\(\dfrac{x^2+y}{25-3}=\dfrac{34}{22}\)=\(\dfrac{17}{11}\)
=>x=\(\dfrac{85}{11}\);y=\(-\dfrac{51}{11}\)
b)\(4x=-5y\)=>\(\dfrac{x}{-5}=\dfrac{y}{4}\)
Đặt \(\dfrac{x}{-5}=\dfrac{y}{4}\)=k=>x=-5k;y=4k
Ta có:xy=-80
=>\(\left(-5k\right)\left(4k\right)=-80\)
=>\(-20k^2=-80\)
=>\(k^2=4\)
=>k=-2 hoặc k=2
=>x=-5k=-5.-2=10
y=4k=4.-2=-8
Và x=-5k=-5.2=-10
y=4k=4.2=8
Vậy
b, Ta có:
\(4x=-5y\Rightarrow\dfrac{x}{-5}=\dfrac{y}{4}\)
Đặt \(\dfrac{x}{-5}=\dfrac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=-5k\\y=4k\end{matrix}\right.\)
Thay vào \(xy=-80\) ta được:
\(-5k.4k=-80\Rightarrow k^2=4\Rightarrow k=\pm2\)
+, Xét \(k=2\Rightarrow\left\{{}\begin{matrix}x=-5.2=-10\\y=4.2=8\end{matrix}\right.\)
+, Xét \(k=-2\Rightarrow\left\{{}\begin{matrix}x=-5.-2=10\\y=4.-2=-8\end{matrix}\right.\)
Vậy...................
Chúc bạn học tốt!!!