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1) Ta có : \(\frac{x}{5}=\frac{y}{4}=\frac{2x}{10}=\frac{2x+y}{10+4}=\frac{28}{14}=2\)
Nên : \(\frac{x}{5}=2\Rightarrow x=10\)
\(\frac{y}{4}=2\Rightarrow y=8\)
a) Ta có: \(\frac{a}{3}=\frac{b}{4}.\)
=> \(\frac{a}{3}=\frac{b}{4}\) và \(a.b=48.\)
Đặt \(\frac{a}{3}=\frac{b}{4}=k\Rightarrow\left\{{}\begin{matrix}a=3k\\b=4k\end{matrix}\right.\)
Có: \(a.b=48\)
=> \(3k.4k=48\)
=> \(12k^2=48\)
=> \(k^2=48:12\)
=> \(k^2=4\)
=> \(k=\pm2.\)
TH1: \(k=2.\)
\(\Rightarrow\left\{{}\begin{matrix}a=2.3=6\\b=2.4=8\end{matrix}\right.\)
TH2: \(k=-2.\)
\(\Rightarrow\left\{{}\begin{matrix}a=\left(-2\right).3=-6\\b=\left(-2\right).4=-8\end{matrix}\right.\)
Vậy \(\left(a;b\right)=\left(6;8\right),\left(-6;-8\right).\)
Chúc bạn học tốt!
Bài 1: Làm:
a,
- x - 2/3 = - 6/7
<=> - x = - 6/7 + 2/3 = -18/21 + 14/21
<=> - x = - 4/21
<=> x = 4/21.
Vậy x = 4/21.
b,
x/- 27 = - 3 / x
<=> x^2 = - 27 . (- 3)
<=> x^2 = 81
<=> x thuộc {9;- 9}
Vậy x thuộc {9;- 9}.
c,
x / y = 2 / 5
<=> x / 2 = y / 5 = 2x - y / 2.2 - 5 = 3 / -1 = - 3.
(T/c dãy tỷ số bằng nhau)
=> x / 2 = - 3 <=> x = - 6.
y / 5 = - 3 <=> y = - 15.
Vậy x = - 6 ; y = - 15.
Bài 2: Làm:
1/2 a = 2/3 b = 3/4 c
<=> a/2 = 2b/3 = 3c/4
<=> a/2.6 = 2b/3.6 = 3c/4.6 (mỗi vế nhân với 1/6)
<=> a/12 = 2b/18 = 3c/24
<=> a/12 = b/9 = c/8 (Rút gọn)
Áp dụng tính chất dãy tỷ số bằng nhau ta có:
a/12 = b/9 = c/8 = a - b/ 12 - 9 = 15 / 3 = 5 (Theo đề bài)
=> a/12 = 3 <=>a = 36
b/9 = 3 <=> b = 27
c/8 = 3 <=> c = 24
Vậy a = 36 ; b = 27 ; c = 24.
Học tốt !
a) \(\left(\frac{5}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right)\cdot2\frac{2}{17}\right]\)
\(=\left(\frac{1}{5}-\frac{126}{125}\right):\frac{4}{7}:\left[\left(\frac{13}{4}-\frac{59}{9}\right)\cdot\frac{36}{17}\right]\)
\(=\left(\frac{25}{125}-\frac{126}{125}\right):\frac{4}{7}:\left[-\frac{119}{36}\cdot\frac{36}{17}\right]\)
\(=-\frac{101}{125}:\frac{4}{7}:\left(-7\right)=-\frac{101}{125}\cdot\frac{7}{4}\cdot\left(-\frac{1}{7}\right)=\frac{101}{500}\)
b) \(\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):\left(-2\right)\)
\(=\left(-\frac{1}{2}-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right)\cdot\left(-\frac{1}{2}\right)\)
\(=-\frac{11}{10}:\left(-3\right)+\frac{1}{3}-\frac{1}{12}\)
\(=\frac{11}{30}+\frac{1}{3}-\frac{1}{12}=\frac{37}{60}\)
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
a) \(x:\left(\frac{3}{4}\right)^3=\left(\frac{3}{4}\right)^2\)
\(x=\left(\frac{3}{4}\right)^2.\left(\frac{3}{4}\right)^3\)
\(x=\left(\frac{3}{4}\right)^5\)
\(x=\frac{243}{1024}\)
vay \(x=\frac{243}{1024}\)
b) \(\left(\frac{2}{5}\right)^5.x=\left(\frac{2}{5}\right)^8\)
\(x=\left(\frac{2}{5}\right)^8:\left(\frac{2}{5}\right)^5\)
\(x=\left(\frac{2}{5}\right)^3\)
\(x=\frac{8}{125}\)
vay \(x=\frac{8}{125}\)
4) \(\left(0,36\right)^8=\left(0,6^2\right)^8=\left(0,6\right)^{16}\)
\(\left(0,216\right)^4=\left(0,6^3\right)^4=\left(0,6\right)^{12}\)
5) a) \(\left(3,5\right)^3=42,875\)
b) \(\left(-\frac{4}{11}\right)^2=\frac{16}{121}\)
c) \(\left(0,5\right)^4.6^4=3^4=81\)
d) \(\left(-\frac{1}{3}\right)^5:\left(\frac{1}{6}\right)^5=\left(-2\right)^5=-32\)
a)\(\frac{1}{4}-\frac{1}{3}x=\frac{2}{5}-\frac{3}{2}x\)
\(\Leftrightarrow\)\(\frac{15-20x}{60}=\frac{24-90x}{60}\)
\(\Leftrightarrow15-20x=24-90x\)
\(\Leftrightarrow-20x+90x=24-15\)
\(\Leftrightarrow70x=9\)
\(\Leftrightarrow x=\frac{9}{70}\)
c) (1/2-1/6)*3^x+4-4*3^x=3^16-4*3^13
=1/3*3^x*3^4-4*3^x=3^13*3^3-4*3^13
=27*3^x-4*3^x=3^13*(27-4)
=3^x*(27-4)=3^13*(27-4)
=>x=13