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Bài 13 : So sánh
2225 và 3150
Ta có :
2225 = ( 23 )75 = 875
3150 = ( 32 )75 = 975
Vì 875 < 975 ( 8 < 9 )
Nên 2225 < 3150
a) a:b:c:d=2:3:4:5
<=>a/2=b/3=c/4=d/5=a+b+c+d/2+3+4+5=-42/14=-3
a/2=-3<=>a=-6
b/3=-3<=>b=-9
c/4=-3<=>c=-12
d/5=-3<=>d=-15
b)a/b=7/20<=>a/7=b/20
b/c=5/8 <=>b/5=c/8<=>b/20=c/32
<=>a/7=b/20=c/32=2a/14=5b/100=2c/64=2a+5b-2c/14+100+64=100/178=50/89
minh ko chac dung dau nha
a) Ta có : (1/16)10 = [(1/2)4]10 = (1/2)40
Vì (1/2)40 < (1/2)50 nên (1/16)10 < (1/2)50
b) Ta có : 430 = ( 2 . 2)30 = 230 . 230 = (22)15 . (23)10 > 315 . 810 > 3 . 310 .810 = 3 . (3 . 8)10 = 3 .2410
Vậy nên 230 + 330 + 430 > 2410 . 3
Mình chỉ giải thế thôi, còn đâu bn tự làm tiếp
b) (5/2-3x)=25/9
3x = 5/2-25/9
3x =-5/18
x =-5/18:3
x=-5/54
\(e.\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=\left(-2\right)^5\)
\(x-1=-2\)
\(x\) \(=-2+1\)
\(x\) \(=-1\)
Vậy \(x=-1\)
bài 12 :
a,\(\left(x-\frac{1}{2}\right)^2=0\)
Mà: 02=0
=> \(\left(x-\frac{1}{2}\right)^2=0^2\)
\(\Rightarrow x-\frac{1}{2}=0\)
\(\Rightarrow x=\frac{1}{2}\)
b, \(\left(x-2\right)^2=1\)
Mà : 1=12
\(\Rightarrow\left(x-2\right)^2=1^2\)
=> x - 2 = 1
=> x = 3
c, \(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)=-2\)
Vì -8 =-23
nên ...
=> 2x =-1
=> x=0.5
d.\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
cái này cũng như mấy cái trên thôi
Bài 12:
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(x-2=\pm1\)
- Nếu \(x-2=1\)
\(x=3\)
- Nếu \(x-2=-1\)
\(x=1\)
c) \(\left(2x-1\right)^3=-8\)
\(\Rightarrow2x-1=-2\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(x+\frac{1}{12}=\pm\frac{1}{4}\)
- Nếu \(x+\frac{1}{12}=\frac{1}{4}\)
\(x=\frac{1}{6}\)
- Nếu \(x+\frac{1}{12}=-\frac{1}{4}\)
\(x=-\frac{1}{3}\)
Bài 13: có người làm rồi
Bài 14:
a) \(25^3\div5^2\)
\(=\left(5^2\right)^3\div5^2\)
\(=5^6\div5^2=5^4\)
b) \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
c) \(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
\(=3-1+\frac{1}{4}:2\)
\(=2+\frac{1}{8}=2\frac{1}{8}\)
Bài 1:
a: \(\left(-1\dfrac{2}{3}-x\right)\left(1+x^3\right)=0\)
=>\(\left[{}\begin{matrix}x^3+1=0\\-x-1\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^3=-1\\x+\dfrac{5}{3}=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
b: \(\left(0,5-3x\right)^4+\dfrac{8}{81}=\dfrac{8}{9}\)
=>\(\left(3x-\dfrac{1}{2}\right)^4=\dfrac{8}{9}-\dfrac{8}{81}=\dfrac{64}{81}\)
=>\(\left[{}\begin{matrix}3x-\dfrac{1}{2}=\dfrac{2}{3}\\3x-\dfrac{1}{2}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{7}{6}\\3x=\dfrac{1}{2}-\dfrac{2}{3}=-\dfrac{1}{6}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{7}{18}\\x=-\dfrac{1}{18}\end{matrix}\right.\)
Bài 2:
a: \(\dfrac{2024}{3333}>0;-\dfrac{5324}{9191}< 0\)
Do đó: \(\dfrac{2024}{3333}>-\dfrac{5324}{9191}\)
b: \(\dfrac{20}{33}=\dfrac{20\cdot2}{33\cdot2}=\dfrac{40}{66}< \dfrac{53}{66}\)