bạn câu đầu tiền là 0,2 hay 02 vậy ạ

bạn ơi câu đầu tiên là 0 nhé ! mình viết liền quá ! Xin lỗi bạn 

các bạn giúp mình với đi !!!

cảm ơn xong chẳng có ai :)))

a:  2².x+3².x=39                                     b:   4².x-3².x=49

=> (4+9).x=39                                         =>(16-9).x=49

=>13.x=39                                               =>7.x=49                     NHỚ CHO ANH NHA:)    <3

=>x=39:13                                               =>x=49:7

=>x=3                                                       =>x=7

x/2+x+x/3+x+x+x/4=23/4

⇒ 6x/12+12x/12+4x/12+12x/12+12x/12+3x/12=23/4

⇒ (6x+12x+4x+12x+12x+3x)/12=23/4

⇒ 49x/12=23/4

⇒ 49x=23/4.12

⇒ 49x=69

⇒ x=69/49

Giải:

\(\dfrac{x}{2}+x+\dfrac{x}{3}+x+x+\dfrac{x}{4}=\dfrac{23}{4}\)

\(x.\left(\dfrac{1}{2}+1+\dfrac{1}{3}+1+1+\dfrac{1}{4}\right)=\dfrac{23}{4}\) 

                               \(x.\dfrac{49}{12}=\dfrac{23}{4}\)

                                      \(x=\dfrac{23}{4}:\dfrac{49}{12}\) 

                                      \(x=\dfrac{69}{49}\)

b) \(x^3-6x^2+9x=0\)

\(\Leftrightarrow x.\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow x.\left(x-3\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Vậy \(x=0\)hoặc \(x=3\)

a. ( x - 1 )³ + 1 + 3x ( x - 4 ) = 0

<=> x³ - 3x² + 3x - 1 + 1 + 3x² - 12x = 0

<=> x³ - 9x = 0

<=> x ( x² - 9 ) = 0

<=> \(\orbr{\begin{cases}x=0\\x^2-9=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}\)

b. x³ - 6x² + 9x = 0

<=> x ( x² - 6x + 9 ) = 0

<=> x ( x - 3 )² = 0

<=> \(\orbr{\begin{cases}x=0\\\left(x-3\right)^2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

a) ( x - 1 )³ + 1 + 3x( x - 4 ) = 0

⇔ x³ - 3x² + 3x - 1 - 1 + 3x² - 12x = 0

⇔ x³ - 9x = 0

⇔ x( x² - 9 ) = 0

⇔ \(\orbr{\begin{cases}x=0\\x^2-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}\)

b) x³ - 6x² + 9x = 0

⇔ x( x² - 6x + 9 ) = 0

⇔ x( x - 3 )² = 0

⇔ \(\orbr{\begin{cases}x=0\\\left(x-3\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

\(a,78-\left(x-3\right)=16\)

\(78-x+3=16\)

\(51-x=16\)

\(x=51-16=35\)

\(b,\left(2x+3\right):4=60\)

\(2x+3=60\cdot4=240\)

\(2x=240-3=237\)

\(x=\frac{237}{2}\)

\(c,\left(2x-8\right)=\left(2x-8\right)^7\)

\(\left(2x-8\right)-\left(2x-8\right)^7=0\)

Để mik nghĩ thêm tí r làm tiép

Để tui làm nốt ý C cho !!!

( 2x - 8 ) = ( 2x - 8 )⁷

=> ( 2x - 8 ) . 1 - ( 2x - 8 ) . ( 2x - 8 )⁶ = 0

=> ( 2x - 8 ) . [ 1 - ( 2x - 8 )⁶ ] = 0

=> 1 - ( 2x - 8 )⁶ = 0

=> ( 2x - 8 )⁶ = 1

=> ( 2x - 8 )⁶ = ( 1⁶ )

\(\Rightarrow\orbr{\begin{cases}2x-8=1\\2x-8=-1\end{cases}\Rightarrow\orbr{\begin{cases}2x=9\\2x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{9}{2}\\x=\frac{7}{2}\end{cases}}}\)

Vậy ........

c) \(\left(2x-8\right)=\left(2x-8\right)^7\)

\(\Leftrightarrow\left(2x-8\right)^7-\left(2x-8\right)=0\)

\(\Leftrightarrow\left(2x-8\right)\left[\left(2x-8\right)^6-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-8=0\\\left(2x-8\right)^6-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\\left(2x-8\right)^6=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x\in\left\{\frac{7}{2};\frac{9}{2}\right\}\end{cases}}\)

Vậy \(x\in\left\{4;\frac{7}{2};\frac{9}{2}\right\}\)

                          Bài làm :

\(a\text{)}...\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}\)

\(b\text{)}...\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-4=0\\x-\frac{1}{4}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=4\\x=0+\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=\frac{1}{4}\end{cases}}\)

\(c\text{)}...\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}-x=0\\\frac{1}{2}+1\div x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}-0\\1\div x=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}\)

\(d\text{)}...\Leftrightarrow\left(x+3\right)\left(x-4+2\right)=0\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

Bài làm :

\(a,\left(x-3\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}\)

\(b,\left(\frac{1}{2}x-4\right)\left(x-\frac{1}{4}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x-4=0\\x-\frac{1}{4}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=4\\x=\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=\frac{1}{4}\end{cases}}\)

\(c,\left(\frac{1}{3}-x\right).\left(\frac{1}{2}+1:x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{3}-x=0\\\frac{1}{2}+1:x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}\)

\(d,\left(x+3\right)\left(x-4\right)+2\left(x+3\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x-4+2\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

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