\(2^x+2^{x+4}=544\)
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a) ADTCDTSGNTC
\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{a+b-c}{2+3-4}=\frac{5}{1}=5\)
\(\Rightarrow a=10\)
\(b=15\)
\(c=20\)
b) ADTCDTSBNTC
\(\frac{x}{-7}=\frac{y}{4}=\frac{2x}{-14}=\frac{3y}{12}=\frac{2x-3y}{-14-12}=\frac{-78}{-26}=3\)
\(\Rightarrow x=-21\)
\(y=12\)
c) ADTCDTSBNTC
\(\frac{a}{2}=\frac{b}{3}=\frac{2b}{6}=\frac{c}{4}=\frac{3c}{12}=\frac{a+2b-3c}{2+6-12}=\frac{-20}{-4}=5\)
\(\Rightarrow a=10\)
\(b=15\)
\(c=20\)
k mk nhe
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\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{12}\)
\(\frac{y}{4}=\frac{2}{5}\Rightarrow\frac{y}{12}=\frac{2}{15}\)
\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{2}{15}\)
\(ADTCDTSBNTC\)
\(\frac{x}{8}=\frac{y}{12}=\frac{2}{15}=\frac{x+y-2}{8+12-15}=\frac{10}{5}=2\)
\(\Rightarrow x=16\)
\(y=24\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)
Khi đó xy + yz + zx = 31
,<=> 2k.3k +3k.5k + 2k.5k = 31
=> 6k2 + 15k2 + 10k2 = 31
=> 31k2 = 31
=> k2 = 1
=> k = \(\pm\)1
Khi k = 1 => x = 2 ; y = 3 ; z = 5
Khi k = -1 => x = - 2 ; y = -3 ; z = -5
Vậy các cặp số (x;y;z) thỏa mãn là (2;3;5) ; (-2;-3;-5)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{3}\left(x-3\right)-\left(x-\frac{1}{2}\right)=\frac{1}{2}\)
\(\Leftrightarrow\frac{2}{3}x-2-x+\frac{1}{2}=\frac{1}{2}\)
\(\Leftrightarrow x\left(\frac{2}{3}-1\right)-\left(2-\frac{1}{2}\right)=\frac{1}{2}\)
\(\Leftrightarrow-\frac{1}{3}x-\frac{3}{2}=\frac{1}{2}\)
\(\Leftrightarrow-\frac{1}{3}x=2\)
\(\Leftrightarrow x=-6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(N=4a+4b+4c=4\left(a+b+c\right)\)
Thế a + b + c = 28 vào biểu thức N ta được :
\(4.28=112\)
Vậy N có giá trị là 112
Ta có : \(a+b+c=28\Rightarrow4\left(a+b+c\right)=4.28\Rightarrow4a+4b+4c=112\)
Mà \(4a+4b+4c=N\Rightarrow N=112\)
Vậy \(N=112\)
![](https://rs.olm.vn/images/avt/0.png?1311)
0,25 x 40 - 2 : 0,25
= 0,25 x 40 - 32 x 0,25
= 0,25 x (40 - 32)
= 0,25 x 8
= 2
#Học tốt!!!
~NTTH~
\(2^x+2^{x+4}=544\)
\(2^x\left(1+2^4\right)=544\)
\(2^x\left(1+16\right)=544\)
\(2^x\cdot17=544\)
\(2^x=544:17\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
2x+2x.24=544
(2x)2.16=544
2^2x.16=544
2^2x=34
2x=34
x=34:2
x=17