tìm x biết (1/16)^x+2=(1/32)^6
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3/2+5/4+9/8/+17/16+33/32-6+x-1/x+1=31/32-2/2015
=(1+1/2)+(1+1/4)+(1+1/8)+(1+1/16)+(1+1/32-6+x-1/x+1=31/32-2/2015
=(1/2+1/4+1/8+1/16+1/32)+(1+1+1+1+1)-6+x-1/x+1=31/32-2/2015
=31/32+5-6+x-1/x+1=31/32-2/2015
=5-6+x-1/x+1=31/32-2/2015-31/32
=-1+x-1/x+1=-2/2015
=x-1/x+1=-2/2015- -1
=x-1/x+1=2013/2015
=>x=2014
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a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
![](https://rs.olm.vn/images/avt/0.png?1311)
x- 32:16=48
=>x-2=48
=>x=48+2
=>x=50
(x-32):16=48
=>x-32= 48x 16
=>x-32= 768
=>x=768+32
=>x=800
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1 a x=4
b x=-4
c x=-7
d x=3
e x=10
g x=60
h x=36
i x=16
2a 1,2,3,4,5,6,7,8,9
b 1,2,3,4,5,6,7,8,9.........
c rỗng
3a 0
b 0
c10
![](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 )
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
x + 5 x 2 - ( 32 + 16 x 3 : 6 - 15 ) = 0
x + 5 x 2 - 25 = 0
x + 5 x 2 = 0 + 25
x + 5 x 2 = 25
x + 10 = 25
x = 25 - 10
x = 15
\(\left(\dfrac{1}{16}\right)^{x+2}=\left(\dfrac{1}{32}\right)^6\)
\(\Leftrightarrow\dfrac{1}{16^{x+2}}=\dfrac{1}{32^6}\)
\(\Leftrightarrow\dfrac{1}{2^{4x+8}}=\dfrac{1}{2^{30}}\)
\(\Leftrightarrow4x+8=30\Leftrightarrow4x=22\Leftrightarrow x=\dfrac{11}{2}\)