a) \(x+1^3=2^5-\left(-1^3\right)\)

\(\Rightarrow x+1=33\)

=> x = 32

b) \(3^7-x=1^4-\left(-3^5\right)\)

\(\Rightarrow2187-x=1+243=244\)

=> x = 1943

a) \(\Leftrightarrow x+1=32+1\)

\(\Leftrightarrow x=32\)

Vậy x = 32

b) \(\Leftrightarrow2187-x=1+243\)

\(\Leftrightarrow2187-x=244\)

\(\Leftrightarrow x=1943\)

Vậy x = 1943

a) \(x+1^3=2^5-\left(-1^3\right)\) 

      \(x+1=32-\left(-1\right)\) 

      \(x+1=33\) 

            \(x=33-1\) 

            \(x=32\) 

b) \(3^7-x=1^4-\left(-3^5\right)\) 

\(2187-x=1-\left(-243\right)\) 

\(2187-x=244\) 

            \(x=2187-244\) 

            \(x=1943\)

Bạn ghi đề lại câu a.

a: \(\left(\dfrac{3}{4}x+2\dfrac{1}{2}\right)\cdot\dfrac{-2}{3}=\dfrac{1}{8}\)

=>\(\left(\dfrac{3}{4}x+\dfrac{5}{2}\right)=\dfrac{1}{8}:\dfrac{-2}{3}=\dfrac{-3}{16}\)

=>\(\dfrac{3}{4}x=-\dfrac{3}{16}-\dfrac{5}{2}=-\dfrac{3}{16}-\dfrac{40}{16}=-\dfrac{43}{16}\)

=>\(x=-\dfrac{43}{16}:\dfrac{3}{4}=\dfrac{-43}{16}\cdot\dfrac{4}{3}=\dfrac{-43}{12}\)

b: \(\dfrac{1}{3}\cdot x-0,5x=0,75\)

=>\(x\left(\dfrac{1}{3}-\dfrac{1}{2}\right)=0,75\)

=>\(x\cdot\dfrac{-1}{6}=0,75\)

=>\(x=-0,75\cdot6=-4,5\)

\(a,\left(2x+1\right)^2-4\left(x+2\right)^2=9\\ \Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)-9=0\\ \Leftrightarrow4x^2-4x^2+4x-16x+1-16-9=0\\ \Leftrightarrow-12x=24\\ \Leftrightarrow x=\dfrac{24}{-12}=-2\\ b,\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\\ \Leftrightarrow x^2+6x+9-\left(x^2+4x-32\right)=1\\ \Leftrightarrow x^2-x^2+6x-4x=1-9-32\\ \Leftrightarrow2x=-40\\ \Leftrightarrow x=-20\\ c,3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\\ \Leftrightarrow3\left(x^2+4x+4\right)+\left(4x^2-4x+1\right)-7\left(x^2-9\right)=36\\ \Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\\ \Leftrightarrow3x^2+4x^2-7x^2+12x-4x=36-12-1-63\\ \Leftrightarrow8x=-40\\ \Leftrightarrow x=\dfrac{-40}{8}=-5\)

giúp mk với

Có khác gì đâu bn

khác gì vậy

a) \(\left(2\dfrac{3}{4}-1\dfrac{4}{5}\right)\cdot x=1\)

\(\left(\dfrac{11}{4}-\dfrac{9}{5}\right)\cdot x=1\)

\(\dfrac{19}{20}x=1\)

\(x=\dfrac{20}{19}\)

Vậy \(x=\dfrac{20}{19}\)

b) \(\left(x^2-9\right)\left(3-5x\right)=0\)

TH1:

\(x^2-9=0\)

\(x^2=9\)

\(x^2=3^2=\left(-3\right)^2\)

=>\(x\in\left\{3;-3\right\}\)

TH2:

\(3-5x=0\)

\(5x=3\)

\(x=\dfrac{3}{5}\)

Vậy \(x\in\left\{3;-3;\dfrac{3}{5}\right\}\)

bạn viết lại câu b được không?

2(x⁴+3)-(9)=17

⇒2x⁴+6+9=17

⇒2x⁴+15=17

⇒ 2x⁴=2

⇒ x⁴=1

⇒ x=\(\pm1\)

b) 5x².x+1-3.4²=-47

⇒5x³+1-48=-47

⇒5x³-47=-47

⇒5x³=0

⇒x³=0

⇒x=0

a) \(2\left(x^4+3\right)-\left(-9\right)=17\)

              \(2x^4+6+9=17\)

                           \(2x^4=2\)

                             \(x^4=1\)

                       ⇒     \(x=1\)

a) 2*(x⁴+3)-(-9)=17

=>2x⁴+2*3+9=17

=>2x⁴+6=17-9=8

=>2x⁴=8-6=2

=>x⁴=2/2=1

=>x=1

a) \(x-\dfrac{2}{3}=\dfrac{3}{8}\Rightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{25}{24}\)

b) \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\Rightarrow x-\dfrac{3}{4}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}+\dfrac{3}{4}=1\)

c) \(\dfrac{3}{2}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{4}{5}=\dfrac{7}{10}\)

\(\Rightarrow x=\dfrac{7}{10}-\dfrac{1}{2}=\dfrac{1}{5}\)

d) \(\left|x-2\right|-1=0\Rightarrow\left|x-2\right|=1\)

\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

a: Ta có: \(x-\dfrac{2}{3}=\dfrac{3}{8}\)

\(\Leftrightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{9}{24}+\dfrac{16}{24}=\dfrac{25}{24}\)

b: Ta có: \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\)

\(\Leftrightarrow x-\dfrac{3}{4}=\dfrac{13}{10}\cdot\dfrac{5}{26}=\dfrac{1}{4}\)

hay x=1

a: \(\Leftrightarrow2x+\dfrac{7}{2}=\dfrac{16}{3}:\dfrac{8}{3}=2\)

=>2x=-3/2

hay x=-3/4

b: 2x+3=5

=>2x=2

hay x=1

c: =>3(x-2)=4(5+x)

=>4x+20=3x-6

=>x=-26

a) => (7/2 + 2x) . 8/3 = 16/3
=> 7/2 + 2x = 16/3 : 8/3
=> 7/2 + 2x = 2
=> 2x = 2 - 7/2
=> 2x = -1.5
=> x = -1.5 : 2
=> x = -0.1