1: Tìm x

a) Ta có: \(2\cdot3^x=3^{12}\cdot34+20\cdot27^4\)

\(\Leftrightarrow2\cdot3^x=3^{12}\cdot34+20\cdot3^{12}\)

\(\Leftrightarrow2\cdot3^x=3^{12}\cdot\left(34+20\right)\)

\(\Leftrightarrow2\cdot3^x-3^{12}\cdot54=0\)

\(\Leftrightarrow2\cdot3^x=3^{12}\cdot2\cdot27\)

\(\Leftrightarrow3^x=3^{12}\cdot3^3\)

\(\Leftrightarrow3^x=3^{15}\)

hay x=15

Vậy: x=15

b) Ta có: \(\left(2^x+1\right)^2+3\left(2^2+1\right)=2^2\cdot10\)

\(\Leftrightarrow\left(2^x+1\right)^2=40-3\cdot5=25\)

\(\Leftrightarrow\left[{}\begin{matrix}2^x+1=5\\2^x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2^x=4\\2^x=-6\left(loại\right)\end{matrix}\right.\Leftrightarrow x=2\)

Vậy: x=2

\(\text{a, 3(x+1)+4x=10}\)

\(\Rightarrow3x+3+4x=10\)

\(\Rightarrow7x+3=10\)

\(\Rightarrow7x=10-3=7\)

\(\Rightarrow x=1\)

c, x+1/10+x+2/9=x+3/8+x+4/7

=> (x+1/10 +1) +(x+2/9 +1)= ( x+3/8 +1) +(x+4/7 +1)

=> x+11/10 + x+11/9 = x+11/8 + x+11/7

...............

a) \(3\left(x+1\right)+4x=10\)

\(\Rightarrow3x+3+4x=10\)

\(\Rightarrow3x+4x=10-3\)

\(\Rightarrow7x=7\)

\(\Rightarrow x=7\)

a) \(5\times\left(3+7\times x\right)=400\)

\(3+7\times x=80\)

\(7\times x=77\)

\(x=11\)

b) \(x\times37+x\times63=1200\)

\(x\times\left(37+63\right)=1200\)

\(x\times100=1200\)

\(x=12\)

c) \(x\times6+12:3=40\)

\(x\times6+4=40\)

\(x\times6=36\)

\(x=6\)

d) \(4+6\times\left(x+1\right)=70\)

\(6\times\left(x+1\right)=66\)

\(x+1=11\)

\(x=10\)

e) \(163:x+34:x=10\)

\(\left(163+34\right):x=10\)

\(197:x=10\)

\(x=19,7\)

\(a.\left(x+1\right)\left(x^2-x+1\right)-x\left(x^2-5\right)=71\)

\(\Leftrightarrow x^3+1-x^3+5x=71\)

\(\Leftrightarrow5x=71-1\)

\(\Leftrightarrow5x=70\)

\(\Leftrightarrow x=70:5=14\)

\(b.\left(2x-3\right)^3-8x\left(x-1\right)^2+4x\left(4x+1\right)+27=0\)

\(\Leftrightarrow8x^3-12x^2+18x-27-8x\left(x^2-2x+1\right)+16x^2+4x+27=0\)

\(\Leftrightarrow8x^3-12x^2+18x-27-8x^3+16x^2-8x+16x^2+4x+27=0\)

\(\Leftrightarrow20x^2+14x=0\)

\(\Leftrightarrow x\left(20x+14\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\20x+14=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{7}{10}\end{cases}}}\)

a) ta có: (x+1)(x^2 -x+1) -x(x^2 -5)=71

          <=>x^3 +1 -x^3 +5x=71

         <=>5x=70

         <=>x=14

b) ta có:(2x-3)^3 -8x(x-1)^2 +4x(4x+1)+27=0

        <=>[ (2x-3)^3  +27)]  -  [ 8x(x-1)^2  -4x(4x+1)]=0

       <=> (2x-3+3)[ (2x-3)^2 - (2x-3).3  +3^2]   - 2x [ 4(x^2 -2x +1) -2(4x+1)]=0

       <=>2x( 4.x^2 - 12x +9 - 6x +9 +9)   -  2x( 4.x^2 -8x+4 -8x -2)=0

       <=>2x(4.x^2  -18x +27)  -  2x(4.x^2 -16x +2)=0

      <=>2x(4.x^2 -18x+27 -4.x^2 +16x-2)=0

     <=>2x(25-2x)=0

<=>x=0 hoặc 25-2x=0 <=> x=0 hoặc x=25/2

=> 3( x - 1 -4x^2 + 4x) + 4( 3x^2 + 3ax + 2x + 2a) = -27

=>  3x - 3 - 12x^2 + 12x + 12x^2 + 12ax + 8x + 8a = -27

=>  23x + 12ax + 8a - 3 = -27 

=> x ( 23 + 12a)               = -27 + 3 -8a

=> x                                   =\(\frac{-8a-24}{23+12a}\)

BẠn nhân hết ra sau rút gon bạn sẽ mất hết x^2 chỉ còn x bài này chắc chắn là tím x theo a

4.x=29/12;-29/12

Câu cuối ban viết rõ lên nhé