\(\left(\frac12\right)^3X\left(\frac14\right)^2:B\frac{120^3}{40^3}:C\frac{390^4}{130^4}\)
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SS
26 tháng 6 2019
1. sai dấu nhé
2.a, \(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(3^2.5\right)^{10}.5^{20}}{\left(5^2.3\right)^{15}}=\frac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)
b, \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(\frac{4}{5}\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\cdot2\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\right)^5\cdot2^5}{\left(\frac{2}{5}\right)^5\cdot\frac{2}{5}}=2^5\div\frac{2}{5}=32\cdot\frac{5}{2}=80\)
c, \(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{2^{15}}=3^2=9\)
DH
1
G
6 tháng 8 2017
Giải:
a) \(\dfrac{120^3}{40^3}=\left(\dfrac{120}{40}\right)^3=30^3=2700\)
b) \(\dfrac{390^4}{130^4}=\left(\dfrac{390}{130}\right)^4=30^4=810000\)
c) \(\dfrac{3^2}{\left(0,375\right)^2}=\left(\dfrac{3}{0,375}\right)^2=8^2=64\)
Đáp số: a) 2700; b) 810000; c) 64.
Chúc bạn học tốt!!!
LT
3
23 tháng 9 2016
\(\frac{120^3}{40^3}\)= (120:40)3 = 33 = 27
\(\frac{390^4}{130^4}\)=(390 : 130)4=34 = 81
\(\frac{3^2}{\left(0,375\right)^2}\)= (3:0,375)2 = 82 =64
MM
1
13 tháng 8 2017
a) $\dfrac{120^3}{40^3}=(\dfrac{120}{40})^3=3^3=27$
b) $\dfrac{390^4}{130^4}=(\dfrac{390}{130})^4=3^4=81$
c) $\dfrac{3^2}{(0,375)^2}=(3:0,375)^2=(3:\dfrac{3}{8})^2=8^2=64$
14 tháng 8 2017
a) \(\left(\dfrac{1}{5}\right)^5.5^5=\left(\dfrac{1}{5}.5\right)^5=1^5=1\)
b) \(\left(0,125\right)^3.512=\left(0,512\right)^3.8^3=\left(0,512.8\right)^3=1^3=1\)
c) \(\left(0,25\right)^4.1024=\left[\left(0,25\right)^2\right]^2.32^2=\left(\dfrac{1}{6}\right)^2.32^2=\left(\dfrac{1}{6}.32\right)^2=2^2=4\)
d) \(\dfrac{120^3}{40^3}=\left(\dfrac{120}{40}\right)^3=3^3=64\)
e) \(\dfrac{390^4}{130^4}=\left(\dfrac{390}{130}\right)^4=3^4=81\)
g) \(\dfrac{3^2}{\left(0,375\right)^2}=\left(\dfrac{3}{0,375}\right)^3=8^3=512\)
7 tháng 5 2020
Bài 1:
a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)
\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)
\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)
Suy ra: \(12x-45-12x^2+45x=0\)
\(\Leftrightarrow-12x^2+57x-45=0\)
\(\Leftrightarrow-12x^2+12x+45x-45=0\)
\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)
\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)
mà \(-3\ne0\)
nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)
b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
\(\Leftrightarrow\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}+\frac{\left(x-3\right)^2}{4}=0\)
\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)
Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)
\(\Leftrightarrow-x^2+16x-39=0\)
\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)
\(\Leftrightarrow x^2-13x-3x+39=0\)
\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)
\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)
Vậy: Tập nghiệm S={3;13}
c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)
\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)
\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)
\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)
Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)
\(\Leftrightarrow-21x^2+26x+11=0\)
\(\Leftrightarrow-21x^2-7x+33x+11=0\)
\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)
B và C tượng trưng cho j zậy?
hai y khac nhau a b c y