\(1,x:\left(-\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{3}\right)\\ \Leftrightarrow x=\left(-\dfrac{1}{3}\right)\times\left(-\dfrac{1}{3}\right)^3\\ \Leftrightarrow x=\left(-\dfrac{1}{3}\right)^4=\dfrac{1}{81}\\ 2,\left(\dfrac{4}{5}\right)^5.x=\left(\dfrac{4}{5}\right)^7\\ \Leftrightarrow x=\left(\dfrac{4}{5}\right)^7:\left(\dfrac{4}{5}\right)^5=\left(\dfrac{4}{5}\right)^{7-5}=\left(\dfrac{4}{5}\right)^2=\dfrac{16}{25}\)

\(3,\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

\(4,\left(3x+1\right)^3=-27\\ \Leftrightarrow\left(3x+1\right)^3=\left(-3\right)^3\\ \Leftrightarrow3x+1=-3\\ \Leftrightarrow3x=-4\\ \Leftrightarrow x=-\dfrac{4}{3}\)

\(5,\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\\ \Leftrightarrow x=\left(\dfrac{1}{2}\right)^5:\left(\dfrac{1}{2}\right)^2\\ \Leftrightarrow x=\left(\dfrac{1}{2}\right)^{5-2}=\left(\dfrac{1}{2}\right)^3=\dfrac{1}{8}\)

\(6,\left(-\dfrac{1}{3}\right)^3.x=\dfrac{1}{81}\\ \Leftrightarrow\left(-\dfrac{1}{3}\right)^3.x=\left(-\dfrac{1}{3}\right)^4\\ \Leftrightarrow x=\left(-\dfrac{1}{3}\right)^4:\left(-\dfrac{1}{3}\right)^3=-\dfrac{1}{3}\)

\(7,\left(2x-3\right)^2=16\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

\(8,\left(x-\dfrac{2}{3}\right)^3=\dfrac{1}{27}\\ \Leftrightarrow\left(x-\dfrac{2}{3}\right)^3=\left(\dfrac{1}{3}\right)^3\\ \Leftrightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\)

`@` `\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`

\(2,\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\\ \Leftrightarrow\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\left(\dfrac{1}{3}\right)^5\\ \Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{1}{3}\\ \Leftrightarrow\dfrac{2}{3}x=\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}\\ \Leftrightarrow x=\dfrac{2}{3}:\dfrac{2}{3}=1\)

\(4,\left(2x-1\right)^{10}=49^5\\ \Leftrightarrow\left(2x-1\right)^{10}=\left(7^2\right)^5\\ \Leftrightarrow\left(2x-1\right)^{10}=7^{10}\\ \Leftrightarrow2x-1=7\\ \Leftrightarrow2x=8\\ \Leftrightarrow x=4\)

\(5,x:5^2=\left(\dfrac{3}{5}\right)^2:3^2\\ \Leftrightarrow x:5^2=\dfrac{9}{25}:9\\ \Leftrightarrow x:25-\dfrac{1}{25}\\ \Leftrightarrow x=\dfrac{1}{25}\times25=1\)

\(8,\left(2x-5\right)^4=-81\)

Xem lại đề nhé số mũ chẵn thì phải ra chẵn chứ

Sửa đề \(\left(2x-5\right)^4=81\\ \Leftrightarrow\left(2x-5\right)^4=3^4\\ \Leftrightarrow2x-5=3\\ \Leftrightarrow2x=8\\ \Leftrightarrow x=4\)

Phần bên này mình chỉ làm những câu khác và khó hơn 1 tý thôi còn các câu còn lại bạn có thể tự làm được nhé 

x + y = x . y

⇒ x + y − x . y = 0

⇒ x ( 1 − y ) + y = 0 

⇒ x ( 1 − y ) + ( y − 1 ) = −1 

⇒  ( 1 − y ) ( x − 1 ) = −1 

Ta có bảng sau : 

Vậy (x;y) thuộc (0;0);(2;2)

\(x+y=xy\)\(\Leftrightarrow xy-x-y=0\)

\(\Leftrightarrow x\left(y-1\right)-\left(y-1\right)=1\)( cộng 2 vế với 1 )

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)=1\)

Lập bảng giá trị ta có: 

Vậy \(x=y=0\)hoặc \(x=y=2\)

a: Ta có: \(2\sqrt{28}+3\sqrt{63}-3\sqrt{\dfrac{112}{9}}-\sqrt{\dfrac{196}{7}}\)

\(=4\sqrt{7}+9\sqrt{7}-4\sqrt{7}-2\sqrt{7}\)

\(=7\sqrt{7}\)

b: Ta có: \(\sqrt{8+2\sqrt{7}}-\sqrt{12-\sqrt{140}}-\sqrt{5}\)

\(=\sqrt{7}+1-\sqrt{7}+\sqrt{5}-\sqrt{5}\)

=1

Bài 1:

Ta có: \(A=\dfrac{x}{yz}:\dfrac{y}{zx}=\dfrac{x}{yz}.\dfrac{zx}{y}=\dfrac{x^2}{y^2}=\left(\dfrac{x}{y}\right)^2\)

Mà \(3x=2y\Leftrightarrow\dfrac{x}{y}=\dfrac{2}{3}\Leftrightarrow\left(\dfrac{x}{y}\right)^2=\dfrac{4}{9}\)

   \(\Rightarrow A=\dfrac{4}{9}\)

\(1,\\ 3x=2y\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}\Rightarrow\dfrac{x}{yz}=\dfrac{2}{3z}\\ 3x=2y\Rightarrow\dfrac{y}{x}=\dfrac{3}{2}\Rightarrow\dfrac{y}{zx}=\dfrac{3}{2z}\)

\(2,\\ \dfrac{x}{y^2}=2\Rightarrow x=2y^2\\ \dfrac{x}{y}=16\Rightarrow x=16y\\ \Rightarrow2y^2=16y\Rightarrow2y\left(y-8\right)=0\\ \Rightarrow\left[{}\begin{matrix}y=0\left(ktm.vì.y\ne0\right)\\y=8\end{matrix}\right.\Rightarrow y=8\Rightarrow x=128\)

\(3,\\ \dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a-b}{b}=\dfrac{c-d}{d}\)

Bài 2:

Ta có: \(\dfrac{x}{y^2}:\dfrac{x}{y}=2:16\Leftrightarrow\dfrac{1}{y}=\dfrac{1}{8}\Leftrightarrow y=8\)

Ta lại có \(\dfrac{x}{y}=16\Rightarrow x=128\)

Bài 3:

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{b}-1=\dfrac{c}{d}-1\Leftrightarrow\dfrac{a-b}{b}=\dfrac{c-d}{d}\left(đpcm\right)\)

Dạng 4: Toán lượng dư:

Bài 1:nBr₂=0,05x0,5=0,025 (mol)

PTHH: 2NaI+Br₂→2NaBr+I₂

                  0,025     0,05       (mol)

→ mNaBr=0,05.103= 5,15 (g)

Chúc em học giỏi

Bài 1:nBr₂=0,05x0,5=0,025 (mol)

PTHH: 2NaI+Br₂→2NaBr+I₂

                  0,025     0,05       (mol)

→ mNaBr=0,05.103= 5,15 (g)

k mik nha bn

HT

Ta có:

(165 - 15) : 3 + 1 = 51

Vậy số hạng thứ 51 là 165

165 nha

hok tút 👍

51