\(\left(x+\dfrac{3}{5}\right)-\dfrac{1}{2}=\dfrac{1}{3}\\ \Rightarrow x+\dfrac{3}{5}-\dfrac{1}{2}=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{1}{2}\\ \Rightarrow x=\dfrac{10}{30}-\dfrac{18}{30}+\dfrac{15}{30}\\ \Rightarrow x=\dfrac{7}{30}\)

\(\left(x+\dfrac{3}{5}\right)-\dfrac{1}{2}=\dfrac{1}{3}\)

=>\(x=\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{3}{5}=\dfrac{5}{6}-\dfrac{3}{5}=\dfrac{25}{30}-\dfrac{18}{30}=\dfrac{7}{30}\)

\(\dfrac{2}{5}+\left(x-\dfrac{2}{3}\right)=\dfrac{5}{3}\)

=>\(x+\dfrac{2}{5}-\dfrac{2}{3}=\dfrac{5}{3}\)

=>\(x=\dfrac{5}{3}+\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{7}{3}-\dfrac{2}{5}=\dfrac{35}{15}-\dfrac{6}{15}=\dfrac{29}{15}\)

`#3107.101107`

\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\\ =\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^8\cdot2^2\cdot5}\\ =\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^{10}\cdot5}\\ =\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\\ =\dfrac{-2}{6}=-\dfrac{1}{3}\)

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4-2.6^9}{2^8.3^8.2^2+6^8.20}\\ =\dfrac{2^{10}.3^8-2.6^9}{\left(2.3\right)^8.2^2+6^8.20}=\dfrac{2^8.3^8.2^2-2.6^9}{6^8.4+6^8.20}\\ =\dfrac{6^8.4-2.6.6^8}{6^8.\left(4+20\right)}=\dfrac{6^8.\left(4-2.6\right)}{6^8.24}\\ =\dfrac{4-12}{24}=\dfrac{-8}{24}=-\dfrac{1}{3}\)

`#3107.101107`

\(\dfrac{2^8\cdot2^{18}}{8^5\cdot4^6}=\dfrac{2^{8+18}}{\left(2^3\right)^5\cdot\left(2^2\right)^6}=\dfrac{2^{26}}{2^{15}\cdot2^{12}}=\dfrac{2^{26}}{2^{15+12}}=\dfrac{2^{26}}{2^{27}}=\dfrac{1}{2}\)

\(\dfrac{2^8.2^{18}}{8^5.4^6}=\dfrac{2^{8+18}}{\left(2^3\right)^5.\left(2^2\right)^6}\\ =\dfrac{2^{26}}{2^{15}.2^{12}}=\dfrac{2^{26}}{2^{15+12}}\\ =\dfrac{2^{26}}{2^{27}}=\dfrac{1}{2^1}=\dfrac{1}{2}\)

\(\dfrac{5}{11}.\left(\dfrac{-3}{7}\right)+\dfrac{5}{11}.\left(\dfrac{-5}{7}\right)+\left(\dfrac{-8}{7}\right).\dfrac{6}{11}\\ =\dfrac{5}{11}.\left(\dfrac{-3}{7}+\dfrac{-5}{7}\right)+\left(\dfrac{-8}{7}\right).\dfrac{6}{11}\\ =\dfrac{5}{11}.\dfrac{-8}{7}+\dfrac{-8}{7}.\dfrac{6}{11}\\ =\dfrac{-8}{7}.\left(\dfrac{5}{11}+\dfrac{6}{11}\right)\\ =\dfrac{-8}{7}.\dfrac{11}{11}\\ =\dfrac{-8}{7}.1=-\dfrac{8}{7}\)

Ta có: M1 đối đỉnh với M3 

⇒ M1 kề bù với M2 

⇒ \(M1+M2=180^o\) 

Mà: \(M1=3\cdot M2\)

\(\Rightarrow3\cdot M2+M2=180^o\)

\(\Rightarrow4\cdot M2=180^o\)

\(\Rightarrow M2=\dfrac{180^o}{4}=45^o\)

Mà: M4 = M2 = `45^o` 

⇒ M1 = 3.M2 = 3.45 = `135^o` 

Mà: M1 = M3 

⇒ M3 = `135^o`  

`5x^3 : x + 2x - 5x^2 = 1`

`<=> 5x^2 + 2x - 5x^2 = 1`

`<=> 2x = 1`.

`<=> x = 1/2`

Không dùng dấu ⇔ nhé em!

Bài 4:

a. \(x^{10}=x^{7+3}=x^7.x^3\)

b. \(x^{10}=x^{2.5}=\left(x^2\right)^5\)

c. \(x^{10}=x^{12-2}=x^{12}:x^2\)

Bài 5:

a. Đề lỗi rồi bạn.

b. \(5^{x+1}=125\)

\(\Rightarrow5^{x+1}=5^3\)

\(\Rightarrow x+1=3\)

\(\Rightarrow x=3-1=2\)

c. \(\left(3x-1\right)^2=81\)

\(\Rightarrow\left(3x-1\right)^2=\left(\pm9\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}3x-1=9\\3x-1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=10\\3x=-8\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=-\dfrac{8}{3}\end{matrix}\right.\)

$Toru$

a) ĐK: $y\ne 0;x,y\in\mathbb{Q}$

Ta có: $x-y=2(x+y)$

$\Rightarrow x-y-2x-2y=0$

$\Rightarrow -x-3y=0$

$\Rightarrow x=-3y$

Thay $x=-3y$ vào $x-y=\frac{x}{y}$, ta được:

$-3y-y=\frac{-3y}{y}$

$\Rightarrow -4y=-3$

$\Rightarrow y=\frac34(tm)$

Khi đó: $x=-3.\frac34=-\frac94(tm)$

b) ĐK: $y\ne0;x,y\in\mathbb{Q}$

Ta có: $x+y=\frac{x}{y}$

$\Rightarrow y(x+y)=x$

$\Rightarrow x=xy+y^2$

Thay $x=xy+y^2$ vào $x+y=xy$, ta được:

$xy+y^2+y=xy$

$\Rightarrow y^2+y=0$

$\Rightarrow y(y+1)=0$

\(\Rightarrow\left[{}\begin{matrix}y=0\left(loại\right)\\y+1=0\end{matrix}\right.\Rightarrow y=-1\left(tm\right)\)

Khi đó: $x=x.(-1)+(-1)^2$

$\Rightarrow x=-x+1$

$\Rightarrow x+x=1$

$\Rightarrow 2x=1$

$\Rightarrow x=\frac12(tm)$

$Toru$