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Ta đặt:
\(\frac{y}{4}=\frac{x}{2}=k\)
<=>y=4k;x=2k
Thay y=4k;x=2k vào x.y=32 ta được:
x.y=32
<=>4k.2k=32
8.k^2=32
<=>k^2=32:8
k^2=4
<=>k=2 hoặc -2
Nếu k=2:
y=2.4=8
x=2.2=4
Nếu k=-2:
y=-2.4=-8
x=-2.2=-4
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(\frac{x}{y}=\frac{3}{2};5x=7z\Rightarrow\frac{x}{3}=\frac{y}{2};\frac{x}{7}=\frac{z}{5}\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{x}{10}=\frac{2y}{28}\)
Ap dụng tính chất DTSBN
\(\frac{x}{21}=\frac{2y}{28}=\frac{z}{10}=\frac{x-2y+z}{21-28+10}=\frac{32}{3}\)
\(\hept{\begin{cases}\frac{x}{21}=\frac{32}{3}\Rightarrow x=224\\\frac{y}{14}=\frac{32}{3}\Rightarrow x=\frac{448}{3}\\\frac{z}{10}=\frac{32}{3}\Rightarrow x=\frac{320}{3}\end{cases}}\)
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a)\(\left|\frac{1}{3}x+\frac{5}{4}\right|-\frac{1}{8}=0\)
\(\Leftrightarrow\left|\frac{1}{3}x+\frac{5}{4}\right|=\frac{1}{8}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}x+\frac{5}{4}=\frac{1}{8}\\\frac{1}{3}x+\frac{5}{4}=-\frac{1}{8}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{27}{8}\\x=-\frac{33}{8}\end{cases}}\)
Vậy x=-27/8 và x=-33/8
b) \(\frac{x-2}{32}=\frac{2}{x-2}\)
\(\Leftrightarrow\left(x-2\right)^2=64\)
\(\Leftrightarrow\left(x-2\right)^2=8^2\)
\(\Leftrightarrow\hept{\begin{cases}x-2=8\\x-2=-8\end{cases}}\Leftrightarrow\hept{\begin{cases}x=10\\x=-6\end{cases}}\)
vậy x=10 hoặc x=-6
![](https://rs.olm.vn/images/avt/0.png?1311)
b)\(2^{x-1}+5\cdot2^{x-2}=\frac{7}{32}\)
\(2^x:2+5\cdot2^x:2^2=\frac{7}{32}\)
\(2^x:2+2^x:\frac{4}{5}=\frac{7}{32}\)
\(2^x\cdot\left(\frac{1}{2}+\frac{5}{4}\right)=\frac{7}{32}\)
\(2^x\cdot\frac{7}{4}=\frac{7}{32}\)
\(2^x=\frac{7}{32}:\frac{7}{4}=\frac{1}{8}\)
\(2^x=\frac{2^0}{2^3}=2^{-3}\)
\(\Rightarrow x=-3\)
a) \(4^x+4^{x+3}=4160\)
\(\Rightarrow4^x+4^x.4^3=4160\)
\(\Rightarrow4^x.\left(1+4^3\right)=4160\)
\(\Rightarrow4^x.65=4160\)
\(\Rightarrow4^x=64\)
\(\Rightarrow4^x=4^4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
b) \(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
\(\Rightarrow2^x.\frac{1}{2}+5.2^x.\frac{1}{4}=\frac{7}{32}\)
\(\Rightarrow2^x.\left(\frac{1}{2}+5.\frac{1}{4}\right)=\frac{7}{32}\)
\(\Rightarrow2^x.\frac{7}{4}=\frac{7}{32}\)
\(\Rightarrow2^x=\frac{7}{32}:\frac{7}{4}\)
\(\Rightarrow2^x=\frac{1}{8}\)
\(\Rightarrow2^x=2^{-3}\)
\(\Rightarrow x=-3\)
Vậy \(x=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. <=> |x-2| - 5 > 8
hoặc|x - 2| - 5 < -8
<=> |x - 2| > 13
hoặc|x - 2| < -3 (loại)
=> x - 2 > 13
hoặc x - 2 < -13
=> x> 15
hoặc x < -11
vậy x > 15 hoặc x < -11
b. <=> -12 < 25 - |x + 3| < 12
<=>-12 + (-25) < -|x +3| < 12 - 25
<=> -37 < -|x + 3| < -13
<=> 37 > |x + 3 > 13
<=> 13 < x + 3 < 37
hoặc -37 < x + 3 < -13
<=> 10 < x < 34
hoặc -40 < x < -16
vậy 10 < x < 34 hoặc -40 < x < -16
c. <=> 41 - |x + 5| > 32
hoặc 41 - |x + 5| < -32
<=> 41 - 32 > | x + 5|
hoặc 41 + 32 < | x + 5|
<=> 9> |x+5|
hoặc 73< |x + 5|
<=> -9 < x+5 < 9
hoặc x + 5 > 73
hoặc x + 5 < -73
<=> -14 < x < 14
hoặc x > 68
hoặc x < -78
vậy -14 < x < 14 hoặc x > 68 hoặc x < -78
![](https://rs.olm.vn/images/avt/0.png?1311)
a. 2x-1+ 5.2x-1:2=7/32
=> 2x+1.(1+5/2)=7/32
=>2x+1.7/2=7/32
=> 2x+1=1/16=1/24
=> x+1=-4=>x=-5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2^{4-x}}{16^5}=32^6\)
=> \(\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)
=> \(\frac{2^{4-x}}{2^{20}}=2^{30}\)
=> \(2^{4-x}=2^{30}.2^{20}\)
=> \(2^{4-x}=2^{50}\)
=> 4 - x = 50
=> x = 4 - 50 = -46
\(\frac{3^{2x+3}}{9^3}=9^{14}\)
=> \(\frac{3^{2x+3}}{\left(3^2\right)^3}=\left(3^2\right)^{14}\)
=> \(\frac{3^{2x+3}}{3^6}=3^{28}\)
=> \(3^{2x+3}=3^{28}.3^6\)
=> \(3^{2x+3}=3^{34}\)
=> 2x + 3 = 34
=> 2x = 34 - 3
=> 2x = 31
=> x = 31/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{32}{2^x}=2\)
\(\Rightarrow\frac{2^5}{2^x}=2^1\)
\(\Rightarrow5-x=1\)
\(\Rightarrow x=5-1\)
\(\Rightarrow x=4\)
\(\frac{32}{2^x}=2\)
=> 2x = 32 : 2
=> 2x = 16
=> 2x = 24
=> x = 4
Năm mới vui vẻ nha!
Hk tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
\(\left(\frac{1}{3}\right)^2\times27=3^x\)
\(\frac{1^2}{3^2}\times3^3=3^x\)
\(3^1=3^x\)
\(x=1\)
b.
\(\frac{64}{\left(-2\right)^x}=-32\)
\(\frac{\left(-2\right)^6}{\left(-2\right)^x}=\left(-2\right)^5\)
\(\left(-2\right)^x=\frac{\left(-2\right)^6}{\left(-2\right)^5}\)
\(\left(-2\right)^x=-2\)
\(x=1\)
c.
\(3x^2-\frac{1}{2}x=0\)
\(x\times\left(3x-\frac{1}{2}\right)=0\)
TH1:
\(x=0\)
TH2:
\(3x-\frac{1}{2}=0\)
\(3x=\frac{1}{2}\)
\(x=\frac{1}{2}\div3\)
\(x=\frac{1}{2}\times\frac{1}{3}\)
\(x=\frac{1}{6}\)
Vậy x = 0 hoặc x = 1/6
2x = 32
2x = 25
\(\Rightarrow\)x = 5
Hk tốt
bang 3