\(110-3\times\left(8+x\right)=1\)

\(3\times\left(8+x\right)=110-1\)

\(3\times\left(8+x\right)=109\)

\(8+x=\dfrac{109}{3}\)

\(x=\dfrac{109}{3}-8\)

\(x=\dfrac{85}{3}\)

Chúc bạn học tốt

$110-3\times \left(8+x\right)=1$

$3\times \left(8+x\right)=110-1$

$3\times \left(8+x\right)=109$

$8+x=\genfrac{}{}{0ex}{}{109}{3}$

$x=\genfrac{}{}{0ex}{}{109}{3}-8$

$x=\genfrac{}{}{0ex}{}{85}{3}$
 

Ta có:

`x : y : z = 3 : 8 : 5`

`\Rightarrow `\(\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}\)

`\Rightarrow `\(\dfrac{3x}{9}=\dfrac{y}{8}=\dfrac{2z}{10}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{3x}{9}=\dfrac{y}{8}=\dfrac{2z}{10}=\dfrac{3x+y-2z}{9+8-10}=\dfrac{14}{7}=2\)

`\Rightarrow`\(\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}=2\)

`\Rightarrow`

`x = 3.2 = 6`

`y = 8.2 = 16`

`z = 5.2 = 10`

Vậy, `x = 6; y = 16; z = 10.`

\(\sqrt[]{5-x^6}+\sqrt[]{3x^4-2}=1\left(1\right)\)

Điều kiện \(\left\{{}\begin{matrix}5-x^6\ge0\\3x^4-2\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^6\le5\\x^4\ge\dfrac{2}{3}\end{matrix}\right.\)  \(\) \(\Rightarrow\left\{{}\begin{matrix}-\sqrt[6]{5}\le x\le\sqrt[6]{5}\\\left[{}\begin{matrix}x\le-\sqrt[4]{\dfrac{2}{3}}\\x\ge\sqrt[4]{\dfrac{2}{3}}\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-\sqrt[6]{5}\le x\le-\sqrt[4]{\dfrac{2}{3}}\\\sqrt[4]{\dfrac{2}{3}}\le x\le\sqrt[6]{5}\end{matrix}\right.\) \(\left(2\right)\)

\(\Rightarrow\left(1\right)\) thỏa \(\Leftrightarrow\left\{{}\begin{matrix}5-x^6\le1\\3x^4-2\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^6\le4\\x^4\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le\sqrt[3]{2}\\0\le x\le1\end{matrix}\right.\) \(\Leftrightarrow0\le x\le1\left(3\right)\)

\(\left(2\right),\left(3\right)\Rightarrow\sqrt[4]{\dfrac{2}{3}}\le x\le1\) \(\Rightarrow\sqrt[4]{\dfrac{2}{3}}< x< 1\)

thanks

\(\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}\)

\(=\dfrac{67}{9}+\dfrac{51}{11}-\dfrac{31}{9}\)

\(=\dfrac{67}{9}-\dfrac{31}{9}+\dfrac{51}{11}\)

\(=4+\dfrac{51}{11}\)

\(=\dfrac{95}{11}\)

Chúc bạn học tốt

`@` `\text {Ans}`

`\downarrow`

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

`\Rightarrow`\(\left(x-\dfrac{1}{3}\right)^2-\left(x-\dfrac{1}{3}\right)^1=0\)

`\Rightarrow`\(\left(x-\dfrac{1}{3}\right)\left(x-\dfrac{1}{3}-1\right)=0\)

`\Rightarrow`\(\left[{}\begin{matrix}x-\dfrac{1}{3}=0\\x-\dfrac{4}{3}=0\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy, `x \in`\(\left\{\dfrac{1}{3};\dfrac{4}{3}\right\}.\)

`@` `\text {Ans}`

`\downarrow`

`a)`

\(5^x+5^{x+2}=650\)

`\Rightarrow 5^x + 5^x . 5^2 = 650`

`\Rightarrow 5^x . (1 + 5^2) = 650`

`\Rightarrow 5^x . 26 = 650`

`\Rightarrow 5^x = 650 \div 26`

`\Rightarrow 5^x = 25`

`\Rightarrow 5^x = 5^2`

`\Rightarrow x = 2`

Vậy, `x = 2`

`b)`

`(4x + 1)^2 = 25.9`

`\Rightarrow (4x + 1)^2 = 225`

`\Rightarrow (4x + 1)^2 = (+-15^2)`

`\Rightarrow`\(\left[{}\begin{matrix}4x-1=15\\4x-1=-15\end{matrix}\right.\)

`\Rightarrow `\(\left[{}\begin{matrix}4x=16\\4x=-14\end{matrix}\right.\)

`\Rightarrow `\(\left[{}\begin{matrix}x=4\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy, `x \in`\(\left\{-\dfrac{7}{2};4\right\}\)

`c)`

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

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

`\Rightarrow 2^x . (1 + 2^3) = 144`

`\Rightarrow 2^x . 9 = 144`

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

`\Rightarrow 2^x = 16`

`\Rightarrow 2^x = 2^4`

`\Rightarrow x = 4`

Vậy, `x = 4`

`d)`

\(3^{2x+2}=9^{x+3}\)

`\Rightarrow `\(3^{2x+2}=\left(3^2\right)^{x+3}\)

`\Rightarrow `\(3^{2x+2}=3^{2x+6}\)

`\Rightarrow 2x + 2 = 2x + 6`

`\Rightarrow 2x - 2x = 6 - 2`

`\Rightarrow 0 = 4 (\text {vô lý})`

Vậy, `x` không có giá trị nào thỏa mãn.

`e)`

\(x^{15}=x^2\)

`\Rightarrow `\(x^{15}-x^2=0\)

`\Rightarrow `\(x^2\cdot\left(x^{13}-1\right)=0\)

`\Rightarrow `\(\left[{}\begin{matrix}x^2=0\\x^{13}-1=0\end{matrix}\right.\)

`\Rightarrow `\(\left[{}\begin{matrix}x=0\\x^{13}=1\end{matrix}\right.\)

`\Rightarrow `\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy, `x \in`\(\left\{0;1\right\}.\)

nhanh giúp mik với mọi ng 

chỉ cần kết quả thui

`@` `\text {Ans}`

`\downarrow`

`63 \div (3x) = 3`

`3x = 63 \div 3`

`3x = 21`

`x = 21 \div 3`

`x = 7`

Vậy, `x = 7.`

3x=63:3

3x=21

x=21:3

x=7

vậy x=7

\(100000=56743+x\)

\(x=100000-56743\)

\(x=43257\)

= 43257 bạn nhé

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\dfrac{7}{5}\cdot\dfrac{8}{19}+\dfrac{7}{5}\cdot\dfrac{12}{19}-\dfrac{7}{5}\cdot\dfrac{1}{19}\)

`=`\(\dfrac{7}{5}\cdot\left(\dfrac{8}{19}+\dfrac{12}{19}-\dfrac{1}{19}\right)\)

`=`\(\dfrac{7}{5}\cdot\dfrac{19}{19}=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)

`b)`

\(-\dfrac{3}{5}\cdot\dfrac{5}{7}+\left(-\dfrac{3}{5}\right)\cdot\dfrac{3}{7}+\left(-\dfrac{3}{5}\right)\cdot\dfrac{6}{7}\)

`=`\(-\dfrac{3}{5}\cdot\left(\dfrac{5}{7}+\dfrac{3}{7}+\dfrac{6}{7}\right)\)

`=`\(-\dfrac{3}{5}\cdot\dfrac{14}{7}\)

`=`\(-\dfrac{3}{5}\cdot2=-\dfrac{6}{5}\)

`c)`

\(10\dfrac{2}{9}+\left(2\dfrac{2}{5}-7\dfrac{2}{9}\right)\)

`=`\(10\dfrac{2}{9}+2\dfrac{2}{5}-7\dfrac{2}{9}\)

`=`\(\left(10\dfrac{2}{9}-7\dfrac{2}{9}\right)+2\dfrac{2}{5}\)

`=`\(3+2\dfrac{2}{5}=\dfrac{27}{5}\)

`d)`

\(6\dfrac{3}{10}-\left(3\dfrac{4}{7}+2\dfrac{3}{10}\right)\)

`=`\(6\dfrac{3}{10}-3\dfrac{4}{7}-2\dfrac{3}{10}\)

`=`\(\left(6\dfrac{3}{10}-2\dfrac{3}{10}\right)-3\dfrac{4}{7}\)

`=`\(4-3\dfrac{4}{7}=\dfrac{3}{7}\)

a) \(\dfrac{7}{5}.\dfrac{8}{19}+\dfrac{7}{5}.\dfrac{12}{19}-\dfrac{7}{5}.\dfrac{1}{19}\)

\(=\dfrac{7}{5}.\left(\dfrac{8}{19}+\dfrac{12}{19}-\dfrac{1}{19}\right)\)

\(=\dfrac{7}{5}.1\)

\(=\dfrac{7}{5}\)

b) \(\dfrac{-3}{5}.\dfrac{5}{7}+\dfrac{-3}{5}.\dfrac{3}{7}+\dfrac{-3}{5}.\dfrac{6}{7}\)

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

\(=\dfrac{-3}{5}.2\)

\(=\dfrac{-6}{5}\)

c) \(10\dfrac{2}{9}+\left(2\dfrac{2}{5}-7\dfrac{2}{9}\right)\)

\(=\dfrac{92}{9}+\dfrac{12}{5}-\dfrac{65}{9}\)

\(=\dfrac{92}{9}-\dfrac{65}{9}+\dfrac{12}{5}\)

\(=3+\dfrac{12}{5}\)

\(=\dfrac{15}{5}+\dfrac{12}{5}\)

\(=\dfrac{27}{5}\)

d) \(6\dfrac{3}{10}-\left(3\dfrac{4}{7}+2\dfrac{3}{10}\right)\)

\(=\dfrac{63}{10}-\dfrac{25}{7}-\dfrac{23}{10}\)

\(=\dfrac{63}{10}-\dfrac{23}{10}-\dfrac{25}{7}\)

\(=4-\dfrac{25}{7}\)

\(=\dfrac{28}{7}-\dfrac{25}{7}\)

\(=\dfrac{3}{7}\)

Chúc bạn học tốt