-32:(-2)^n=4 giúp mình với ạ
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\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)
\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)
= \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{1}{2}\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{6}{5}\)
= \(\dfrac{1}{4}-\dfrac{6}{5}\)
= \(-\dfrac{19}{20}\)
\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)
\(a,\left(\dfrac{-1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)
\(=\left(-\dfrac{1}{2}\right)^2\left(4-2\right)\)
\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)
\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)
\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)
\(=\dfrac{1}{3}+\left(-\dfrac{1}{5}\right)=\dfrac{2}{15}\)
\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)
\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)
\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)
\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)
\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)
\(=-\dfrac{5}{11}\)
\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)
\(=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=1\)
Tổng của hai đáy:
\(60:8\times2=15\left(m\right)\)
Tổng số phần bằng nhau:
\(1+2=3\) (phần)
Đáy lớn là:
\(15:3\times2=10\left(m\right)\)
Đáy bé là:
\(15-10=5\left(m\right)\)
\(\left|x\right|+x=\dfrac{1}{3}\)
\(\Rightarrow\left|x\right|=\dfrac{1}{3}-x\)
\(\left|x\right|=\left\{{}\begin{matrix}xkhix\ge0\\-xkhix< 0\end{matrix}\right.\)
Với \(x\ge0\Rightarrow x=\dfrac{1}{3}-x\Rightarrow2x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{6}\left(tm\right)\)
Với \(x< 0\Rightarrow-x=\dfrac{1}{3}-x\Rightarrow-x+x=\dfrac{1}{3}\Rightarrow0=\dfrac{1}{3}\left(VL\right)\)
Vậy \(x=\dfrac{1}{6}\)
Gọi số ban đầu tìm là x
Khi chuyển dấu phẩy sang phải một chữ số thì số đó gấp 10 lần số ban đầu là 10x
Ta có:
10x - x = 435,96
9x = 435,96
x = 435,96 : 9
x = 48,44
Vậy số thập phân ban đầu cần tìm là 48,44
Gọi số cần tìm là a
Khi dời dấu phẩy của một chữ số thì số đó tăng 10 lần so với số ban đầu là 10a. 10a - a = 435,96 <=> 9a = 435,96 <=> a = 435,96 : 9 <=> a = 48,44 Vậy số thập phân ban đầu là : 48,44\(2.16\ge2^n>4\)
\(2.2^4\ge2^n>2^2\)
\(2^5\ge2^n>2^2\)
=> \(n\in\left\{3,4,5\right\}\)
Vậy: \(n\in\left\{3,4,5\right\}\)
a) Để A là phân số thì : \(n-2\ne0=>n\ne2\)
b) Để A nhận giá trị nguyên âm lớn nhất
\(=>A=-1\\ =>\dfrac{n-6}{n-2}=-1\\ =>n-6=-\left(n-2\right)\\ =>n-6=-n+2\\ =>n+n=6+2\\ =>2n=8\\ =>n=4\left(TMDK\right)\)
c) \(A=\dfrac{n-6}{n-2}=\dfrac{n-2-4}{n-2}=1-\dfrac{4}{n-2}\)
Để A nhận gt số nguyên thì : \(\dfrac{4}{n-2}\in Z=>4⋮\left(n-2\right)\\ =>n-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\\ =>n\in\left\{3;1;4;0;6;-2\right\}\)
Đến đây bạn lập bảng giá trị rồi thay từng gt n vào bt A, giá trị nào cho A là STN thì bạn nhận gt đó ạ.
d) Mình nghĩ bạn thiếu đề ạ
\(1,x:\left(-\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{3}\right)\\ \Leftrightarrow x=\left(-\dfrac{1}{3}\right)\times\left(-\dfrac{1}{3}\right)^3\\ \Leftrightarrow x=\left(-\dfrac{1}{3}\right)^4=\dfrac{1}{81}\\ 2,\left(\dfrac{4}{5}\right)^5.x=\left(\dfrac{4}{5}\right)^7\\ \Leftrightarrow x=\left(\dfrac{4}{5}\right)^7:\left(\dfrac{4}{5}\right)^5=\left(\dfrac{4}{5}\right)^{7-5}=\left(\dfrac{4}{5}\right)^2=\dfrac{16}{25}\)
\(3,\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(4,\left(3x+1\right)^3=-27\\ \Leftrightarrow\left(3x+1\right)^3=\left(-3\right)^3\\ \Leftrightarrow3x+1=-3\\ \Leftrightarrow3x=-4\\ \Leftrightarrow x=-\dfrac{4}{3}\)
\(5,\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\\ \Leftrightarrow x=\left(\dfrac{1}{2}\right)^5:\left(\dfrac{1}{2}\right)^2\\ \Leftrightarrow x=\left(\dfrac{1}{2}\right)^{5-2}=\left(\dfrac{1}{2}\right)^3=\dfrac{1}{8}\)
\(6,\left(-\dfrac{1}{3}\right)^3.x=\dfrac{1}{81}\\ \Leftrightarrow\left(-\dfrac{1}{3}\right)^3.x=\left(-\dfrac{1}{3}\right)^4\\ \Leftrightarrow x=\left(-\dfrac{1}{3}\right)^4:\left(-\dfrac{1}{3}\right)^3=-\dfrac{1}{3}\)
\(7,\left(2x-3\right)^2=16\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(8,\left(x-\dfrac{2}{3}\right)^3=\dfrac{1}{27}\\ \Leftrightarrow\left(x-\dfrac{2}{3}\right)^3=\left(\dfrac{1}{3}\right)^3\\ \Leftrightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\)
`@` `\text {Ans}`
`\downarrow`
(Vế 1)
`1.`
`x \div(-1/3)^3 =-1/3`
`=> x= (-1/3) \times (-1/3)^3`
`=> x= (-1/3)^4`
`2.`
`(4/5)^5 *x = (4/5)^7`
`=> x = (4/5)^7 \div (4/5)^5`
`=> x=(4/5)^2`
`3.`
`(x+1/2)^2 =1/16`
`=> (x+1/2)^2 = (+-1/4)^2`
`=>`\(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{4}-\dfrac{1}{2}\\x=-\dfrac{1}{4}-\dfrac{1}{2}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
`4.`
`(3x+1)^3 = -27`
`=> (3x+1)^3 = (-3)^3`
`=> 3x+1=-3`
`=> 3x=-3-1`
`=> 3x =-4`
`=> x=-4/3`
`5.`
`(1/2)^2*x=(1/2)^5`
`=> x=(1/2)^5 \div (1/2)^2`
`=> x=(1/2)^3`
`6.`
`(-1/3)^3*x=1/81`
`=> (-1/3)^3*x = (1/3)^4`
`=> x= (1/3)^4 \div (-1/3)^3`
`=> x=(-1/3)`
`7.`
`(2x-3)^2 = 16`
`=> (2x-3)^2 = (+-4)^2`
`=>`\(\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=7\\2x=-1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
`8.`
`(x-2/3)^3 = 1/27`
`=> (x-2/3)^3 = (1/3)^3`
`=> x-2/3=1/3`
`=> x=1/3 + 2/3`
`=> x=1`
\(-32:\left(-2\right)^n=4\\ =>\left(-2\right)^n=\left(-32\right):4=-8=\left(-2\right)^3\\ =>n=3\)