`@` `\text {Ans}`

`\downarrow`

`a)`

`3x ( 12x - 4 ) - 9x( 4x - 3 ) = 30`

`=> 3x (12x-4) - 3*3x (4x - 3) = 30`

`=> 3x [12x - 4 - 3(4x-3)] = 30`

`=> 3x (12x - 4 - 12x + 9) = 30`

`=> 3x (-4+9)=30`

`=> 3x*5=30`

`=> 3x=6`

`=> x=2`

Vậy, `x=2`

`b)`

`x( 5 - 2x) + 2x( x - 1)`

`=> x(5-2x) + 2x^2 - 2x=15`

`=> 5x - 2x^2 + 2x^2 - 2x =15`

`=> 3x = 15`

`=> x=5`

Vậy, `x=5.`

a: =>36x^2-12x-36x^2+27x=30

=>15x=30

=>x=2

b: =>5x-2x^2+2x^2-2x=15

=>3x=15

=>x=5

a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)

\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)

\(\Leftrightarrow2x=-8\)

hay x=-4

b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)

\(\Leftrightarrow-10x=-10\)

hay x=1

c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)

\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)

\(\Leftrightarrow-4x=-8\)

hay x=2

a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

a) Ta có: \(\left(2x-5\right)^3=216\)

\(\Leftrightarrow2x-5=6\)

\(\Leftrightarrow2x=11\)

hay \(x=\dfrac{11}{2}\)

b) Ta có: \(2x-3⋮x+4\)

\(\Leftrightarrow-11⋮x+4\)

\(\Leftrightarrow x+4\in\left\{1;-1;11;-11\right\}\)

hay \(x\in\left\{-3;-5;7;-15\right\}\)

Alo, sugeni two wai phem. Si ga no, you woo be the me that nas te, ai gi da

bạn vừa đăng câu này r mà

a: Ta có: \(\left(2x-3\right)^2+6\left(2x-1\right)=7\)

\(\Leftrightarrow\left(2x-3\right)^2+6\left(2x-1\right)-7=0\)

\(\Leftrightarrow4x^2-12x+9+12x-6-7=0\)

\(\Leftrightarrow4x^2=4\)

\(\Leftrightarrow x^2=1\)

hay \(x\in\left\{1;-1\right\}\)

b: Ta có: \(x^2-7x+10=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

bạn học sd máy tính tìm nghiệm chưa?

a) \(\left(2x-3\right)^2+6\left(2x-1\right)=7\\ \Rightarrow4x^2-12x+9+12x-6-7=0\\ \Rightarrow4x^2-4=0\\ \Rightarrow x^2-1=0\\ \Rightarrow x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

b) \(x^2-7x+10=0\\ \Rightarrow\left(x^2-2x\right)-\left(5x-10\right)=0\\ \Rightarrow\left(x-2\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

c) \(-6x^2+13x-5=0\\ \Rightarrow-\left(6x^2-13x+5\right)=0\\ \Rightarrow-\left[\left(6x^2-10x\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left[2x\left(3x-5\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left(2x-1\right)\left(3x-5\right)=0\)

    \(\Rightarrow\left[{}\begin{matrix}-\left(2x-1\right)=0\\3x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\3x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\)

d) \(x^4+7x^2-18=0\\ \Rightarrow\left(x^4-4\right)+\left(7x^2-14\right)=0\\ \Rightarrow\left(x^2-2\right)\left(x^2+2\right)+7\left(x^2-2\right)=0\\ \Rightarrow\left(x^2-2\right)\left(x^2+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+9=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x^2=-9\left(loại\right)\end{matrix}\right.\)

tách đi bạn

a) (2x - 3)(6 - 2x) = 0

=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)

c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)

d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)

e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)

f)\(\dfrac{x-12}{4}=\dfrac{1}{2}=\dfrac{x-12}{4}=\dfrac{2}{4}\)

⇒\(x-12=2\)

   \(x=2+12\)

  x = 14

g)2\(\dfrac{1}{4}.\left(x-7\dfrac{1}{3}\right)=1,5\)

\(\dfrac{9}{4}.\left(x-\dfrac{22}{3}\right)=1,5\) 

      \(\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}:\dfrac{9}{4}\)

       \(x-\dfrac{22}{3}=\dfrac{2}{3}\)

       \(x=\dfrac{2}{3}+\dfrac{22}{3}\) 

      \(x=8\)

\(a,\left(3x-7\right)^2=\left(2-2x\right)^2\)

a,\(=>\left(3x-7\right)^2-\left(2-2x\right)^2=0\)

\(< =>\left(3x-7+2-2x\right)\left(3x-7-2+2x\right)=0\)

\(< =>\left(x-5\right)\left(5x-9\right)=0=>\left[{}\begin{matrix}x=5\\x=1,8\end{matrix}\right.\)

b, \(x^2-8x+6=0< =>x^2-2.4x+16-10=0\)

\(< =>\left(x-4\right)^2-\sqrt{10}^2=0\)

\(=>\left(x-4+\sqrt{10}\right)\left(x-4-\sqrt{10}\right)=0\)

\(=>\left[{}\begin{matrix}x=4-\sqrt{10}\\x=4+\sqrt{10}\end{matrix}\right.\)

c, \(4x^2-2x-1=0\)

\(< =>\left(2x\right)^2-2.2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{5}{4}=0\)

\(=>\left(2x-\dfrac{1}{2}\right)^2-\left(\dfrac{\sqrt{5}}{2}\right)^2=0\)

\(=>\left(2x+\dfrac{-1+\sqrt{5}}{2}\right)\left(2x-\dfrac{1+\sqrt{5}}{2}\right)=0\)

\(=>\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{4}\\x=\dfrac{1+\sqrt{5}}{4}\end{matrix}\right.\)

d,\(x^4-4x^2-32=0\)

đặt \(t=x^2\left(t\ge0\right)=>t^2-4t-32=0\)

\(< =>t^2-2.2t+4-6^2=0\)

\(=>\left(t-2\right)^2-6^2=0=>\left(t-8\right)\left(t+4\right)=0\)

\(=>\left[{}\begin{matrix}t=8\left(tm\right)\\t=-4\left(loai\right)\end{matrix}\right.\)\(=>x=\pm\sqrt{8}\)

a) \(2^x:4=16\\ \Rightarrow2^x=64\\ \Rightarrow2^x=2^6\\ \Rightarrow x=6\)

b) \(4^{x-3}=256\\ \Rightarrow4^{x-3}=4^4\\ \Rightarrow x-3=4\\ \Rightarrow x=7\)

c) \(\left(2x+1\right)^3=343\\ \Rightarrow\left(2x+1\right)^3=7^3\\ \Rightarrow2x+1=7\\ \Rightarrow x=3\)

d) \(10+2x=4^5:4^3\\ \Rightarrow10+2x=16\\ \Rightarrow x=3\)

a,2^x:4=16            

2^x=16.4=64

2^x=2^6

=>x=6

b,4^x-3=256

4^x-3=4^4

=>x-3=4

x=4+3=7

c,(2x+1)^3=343

(2x+1)^3=7^3

=>2x+1=7

2x=7-1=6

x=6:2=3

d,10+2x=4^5:4^3

10+2x=4^2=16

2x=16-10=6

x=6:2=3

vc đăng bài giống nhau ạ

a, 2^x : 4 = 16

⇒ 2^x : 2² = 2⁴

⇒ x - 2 = 4

⇒ x       = 6

b, 4^x-3 = 256 

⇒ 4^{x - 3} =  4⁴

⇒ x - 3 = 4

⇒ x       = 7

c, (2x + 1)³ = 343

⇒ (2x + 1)³ = 7³

⇒  2x + 1 = 7

⇒ 2x        = 6

⇒ x          = 3

d, 10 + 2x = 4⁵ : 4³

⇒ 10 + 2x = 16

⇒          2x = 6

⇒             x = 3