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 \(x=\dfrac{1}{6}\)

\(\left|x\right|+x=\dfrac{1}{3}\left(1\right)\)

TH1 : \(x\ge0\)

\(\left(1\right)=>x+x=\dfrac{1}{3}\\ =>2x=\dfrac{1}{3}\\ =>x=\dfrac{1}{3}:2=\dfrac{1}{6}\left(TMDK\right)\)

\(TH2:x< 0\)

\(\left(1\right)=>-x+x=\dfrac{1}{3}\\ =>0=\dfrac{1}{3}\)( Vô lí )

Vậy `x=1/6`

tiếp đi bạn

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

\(2.16\ge2^n>4\)

\(2.2^4\ge2^n>2^2\)

\(2^5\ge2^n>2^2\)

=> \(n\in\left\{3,4,5\right\}\)

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`

\(9.27\le3n\le243\\ =>9.27:3\le3n:3\le243:3\\=>81\le n\le81\\ =>n=81\)

\(9.27\le3^n\le243\)

\(3.3^3\le3^n\le3^5\)

\(3^4\le3^n\le3^5\)

\(n\in\left\{4,5\right\}\)

Vậy: \(n\in\left\{4,5\right\}\)

\(\dfrac{1}{2}.2^{n+4}.2^n=2^5\\ =>2^{n+4+n}=2^5:\dfrac{1}{2}\\ =>2^{2n+4}=2^5.2\\ =>2^{2n+4}=2^6\\ =>2n+4=6\\ =>2n=2=>n=1\)

`@` `\text {Ans}`

`\downarrow`

\(\dfrac{1}{2}\cdot2^{n+4}\cdot2^n=2^5\)

`\Rightarrow `\(\dfrac{1}{2}\cdot2^n\cdot2^4\cdot2^n=2^5\)

`\Rightarrow `\(2^{n\cdot2+4}=2^5\div\dfrac{1}{2}\)

`\Rightarrow `\(2^{n\cdot2+4}=2^6\)

`\Rightarrow `\(n\cdot2+4=6\)

`\Rightarrow `\(2n=2\)

`\Rightarrow n=1`

Sửa đề : \(2^x+2^{x+3}=144\\ =>2^x.\left(1+2^3\right)=144\\ =>2^x=\dfrac{144}{9}=16=2^4\\ =>x=4\)

`@` `\text {Ans}`

\(2^x+2^{x+3}=144\)

`\Rightarrow 2^x + 2^x + 2^3 = 144`

`\Rightarrow 2^x (8+1)=144`

`\Rightarrow 2^x*9=144`

`\Rightarrow 2^x=144 \div 9`

`\Rightarrow 2^x = 16`

`\Rightarrow 2^x = 2^4`

`\Rightarrow x=4`