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![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
a) Đặt \(\frac{x}{10}=\frac{y}{6}=k\)
\(\Rightarrow x=10k,y=6k\)
Mà \(xy=60\)
\(\Rightarrow10k6k=60\)
\(\Rightarrow60k^2=60\)
\(\Rightarrow k^2=1\)
\(\Rightarrow k=\pm1\)
+) \(k=1\Rightarrow x=10;y=6\)
+) \(k=-1\Rightarrow x=-10;y=-6\)
Vậy cặp số \(\left(x;y\right)\) là \(\left(10;6\right);\left(-10;-6\right)\)
b) Hình như đề sai !!!
c) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x^2}{9}=\frac{y^2}{16}=\frac{x^2+y^2}{9+16}=\frac{100}{25}=4\)
+) \(\frac{x^2}{9}=4\Rightarrow x^2=36\Rightarrow x=\pm6\)
+) \(\frac{y^2}{16}=4\Rightarrow y^2=64\Rightarrow y=\pm8\)
( x, y cùng dấu )
Vậy cặp số ( x; y ) là ( 6; 8 ) ; ( -6; -8 )
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a) 32.x+2=1342176728
32.x=134217728-2
32.x=134217726
x=134217726:32
x=4194303,938
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm :
\(\left(\frac{1}{16}\right)^{x+2}=\left(\frac{1}{32}\right)^6\)
\(\Rightarrow\left(\frac{1}{2^4}\right)^{x+2}=\left(\frac{1}{2^5}\right)^6\)
\(\Rightarrow\left(\frac{1}{2}\right)^{4x+8}=\left(\frac{1}{2}\right)^{30}\)
\(\Rightarrow4x+8=30\)
\(\Rightarrow4x=22\)
\(\Rightarrow x=\frac{11}{2}\)
Học tốt nhé
\(\left(\frac{1}{16}\right)^{x+2}=\left(\frac{1}{32}\right)^6=>\left[\left(\frac{1}{2}\right)^4\right]^{x+2}=\left[\left(\frac{1}{2}\right)^4\right]^6\)
\(\left(\frac{1}{2}\right)^{4.\left(x+2\right)}=\left(\frac{1}{2}\right)^{30}\)
4.(x+2)=30
(x+2)=30:4
(x+2)=7,5
x=7,5+2
x=9,5
vậy x=9,5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2^{4-x}}{16^5}=32^6\)
\(\Rightarrow\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)
\(\Rightarrow\frac{2^{4-x}}{2^{20}}=2^{30}\)
\(\Rightarrow2^{4-x}=2^{30}.2^{20}\)
\(\Rightarrow2^{4-x}=2^{50}\)
\(\Rightarrow4-x=50\)
\(\Rightarrow x=-46\)
\(\left(\dfrac{1}{16}\right)^{x+2}=\left(\dfrac{1}{32}\right)^6\)
\(\Leftrightarrow\dfrac{1}{16^{x+2}}=\dfrac{1}{32^6}\)
\(\Leftrightarrow\dfrac{1}{2^{4x+8}}=\dfrac{1}{2^{30}}\)
\(\Leftrightarrow4x+8=30\Leftrightarrow4x=22\Leftrightarrow x=\dfrac{11}{2}\)