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\(a,A=\dfrac{\dfrac{3}{4}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\\ A=\dfrac{\dfrac{405}{572}}{\dfrac{645}{1001}}+\dfrac{\dfrac{5}{12}}{\dfrac{25}{24}}\\ A=\dfrac{189}{172}+\dfrac{2}{5}\\ A=\dfrac{1289}{860}\)
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, \(\frac{2}{3}x=\frac{3}{4}y=\frac{4}{5}z\)
\(\Rightarrow\frac{2x}{3.12}=\frac{3y}{4.12}=\frac{4z}{5.12}\)
\(\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{18+16+15}=\frac{45}{49}\)
Đến đây tự làm tiếp nhé
b, \(2x=3y=5z\Rightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)
=> x = 75, y = 50, z = 30
c, \(\frac{3}{4}x=\frac{5}{7}y=\frac{10}{11}z\)
\(\Rightarrow\frac{3x}{4.30}=\frac{5y}{7.30}=\frac{10z}{11.30}\)
\(\Rightarrow\frac{x}{40}=\frac{y}{42}=\frac{z}{33}\)
\(\Rightarrow\frac{2x}{80}=\frac{3y}{126}=\frac{4z}{132}=\frac{2x-3y+4z}{80-126+132}=\frac{8,6}{86}=\frac{1}{10}\)
=> x=... , y=... , z=...
d, Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow x=2k,y=5k\)
Ta có: xy = 90 => 2k.5k = 90 => 10k2 = 90 => k2 = 9 => k = 3 hoặc -3
Với k = 3 => x = 6, y = 15
Với k = -3 => x = -6, y = -15
Vậy...
e, Tương tự câu d
b) Ta có :\(\text{ 2x = 3y = 5z }=\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=\frac{1}{6}\)
=> \(2x=\frac{1}{6}\Rightarrow x=\frac{1}{12}\)
\(3y=\frac{1}{6}\Rightarrow y=\frac{1}{18}\)
\(5z=\frac{1}{6}\Rightarrow z=\frac{1}{30}\)
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. Đặt \(\dfrac{x}{-3}=\dfrac{y}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=-3k\\y=5k\end{matrix}\right.\)
mà \(x.y=\dfrac{-5}{27}\)
hay \(-3k.5k=\dfrac{-5}{27}\)
\(\Rightarrow-15.k^2=\dfrac{-5}{27}\)
\(\Rightarrow k^2=\dfrac{1}{81}=\left(\pm\dfrac{1}{9}\right)^2\)
Với \(k=\dfrac{1}{9}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{5}{9}\end{matrix}\right.\)
Với \(k=\dfrac{-1}{9}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{-5}{9}\end{matrix}\right.\)
Vậy.......
b. Từ \(\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{3}=\dfrac{z}{5}\end{matrix}\) \(\Rightarrow\begin{matrix}\dfrac{x}{9}=\dfrac{y}{12}\\\dfrac{y}{12}=\dfrac{z}{20}\end{matrix}\) \(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}=\dfrac{x-y+z}{9-12+20}=\dfrac{32}{17}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=\dfrac{32}{17}\Rightarrow x=\dfrac{32.9}{17}=\dfrac{288}{17}\\\dfrac{y}{12}=\dfrac{32}{17}\Rightarrow y=\dfrac{32.12}{17}=\dfrac{384}{17}\\\dfrac{z}{20}=\dfrac{32}{17}\Rightarrow z=\dfrac{32.20}{17}=\dfrac{640}{17}\end{matrix}\right.\)
Vậy.........
1.
\(\left(\dfrac{-2}{3}\right).0,75+1\dfrac{2}{3}:\left(\dfrac{-4}{9}\right)+\left(\dfrac{-1}{2}\right)^2\)
\(=\left(\dfrac{-2}{3}\right).\dfrac{3}{4}+\dfrac{5}{3}.\left(\dfrac{9}{-4}\right)+\dfrac{1}{4}\)
\(=-\dfrac{1}{2}+\dfrac{45}{-12}+\dfrac{1}{4}\)
\(=-\dfrac{6}{12}+\dfrac{-45}{12}+\dfrac{3}{4}\)
\(=\dfrac{-48}{12}\)
\(=-4\)
2.
a) \(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{10}{20}\)
\(\Leftrightarrow x=\dfrac{-11}{20}\)
b) \(\left|x-\dfrac{2}{5}\right|+\dfrac{3}{4}=\dfrac{11}{4}\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=\dfrac{11}{4}-\dfrac{3}{4}\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=-2\Rightarrow x=-2+\dfrac{2}{5}=\dfrac{-8}{5}\\x-\dfrac{2}{5}=2\Rightarrow x=2+\dfrac{2}{5}=\dfrac{12}{5}\end{matrix}\right.\)
3.
a) \(\dfrac{16}{2^n}=2\)
\(\Leftrightarrow2^n=16:2\)
\(\Leftrightarrow2^n=8\)
\(\Leftrightarrow2^n=2^3\)
\(\Leftrightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{81}=-27\)
\(\Leftrightarrow\left(-3\right)^n=\left(-27\right).81\)
\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^3.\left(-3\right)^4\)
\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^7\)
\(\Leftrightarrow n=7\)
4. Ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\) (1)
\(\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\) (2)
Từ (1) và (2) suy ra \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)
Vì \(x-y+x=-49\) ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\)
Vậy \(\left\{{}\begin{matrix}x=\left(-7\right).10=-70\\y=\left(-7\right).15=-105\\z=\left(-7\right).12=-84\end{matrix}\right.\)
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) Ta có: \(6x=4y=3z\Rightarrow\dfrac{6x}{12}=\dfrac{4y}{12}=\dfrac{3z}{12}\Rightarrow\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}=\dfrac{x+2y-3z}{2+6-12}=\dfrac{-2}{-4}=\dfrac{1}{2}.\)
Với: \(\dfrac{x}{2}=\dfrac{1}{2}\Rightarrow x=1.\)
\(\dfrac{2y}{6}=\dfrac{y}{3}=\dfrac{1}{2}\Rightarrow y=\dfrac{1}{2}.3=\dfrac{3}{2}.\)
\(\dfrac{3z}{12}=\dfrac{z}{4}=\dfrac{1}{2}\Rightarrow z=\dfrac{1}{2}.4=\dfrac{4}{2}=2.\)
Vậy: \(x=1;y=\dfrac{3}{2};z=2.\)
\(a,\dfrac{x}{5}=-\dfrac{3}{y}\Rightarrow xy=-15\\ \Rightarrow xy=-1\cdot15=-15\cdot1=-5\cdot3=-3\cdot5\\ \Rightarrow\left(x;y\right)=\left\{\left(-1;-15\right);\left(1;-15\right);\left(15;-1\right);\left(-15;1\right);\left(3;-5\right);\left(-5;3\right);\left(5;-3\right);\left(-3;5\right)\right\}\)\(g,-\dfrac{11}{x}=\dfrac{y}{3}\\ \Rightarrow xy=-33\\ \Rightarrow xy=-3\cdot11=-11\cdot3=-1\cdot33=-33\cdot1\\ \Rightarrow\left(x;y\right)=\left\{\left(-3;11\right);\left(11;-3\right);\left(-11;3\right);\left(3;-11\right);\left(-1;33\right);\left(33;-1\right);\left(-33;1\right);\left(1;-33\right)\right\}\)