a)  2x-3=-x+6

2x - 3 + x - 6 =0

3x -9 = 0

3x = 9

x = 9 : 3

x= 3

c/ \(\left(-12x-4^3\right).8^3=4.8^4\)

\(\left(-12x-64\right).512=16384\)

\(-12x-64=\dfrac{16384}{512}=32\)

\(-12x=32+64=96\)

\(x=\dfrac{96}{\left(-12\right)}=-8\)

a,   2x-3=-x+6

\(\Leftrightarrow\)2x-3+x-6=0

\(\Leftrightarrow\)3x-9=0

\(\Leftrightarrow\)3x=9

\(\Leftrightarrow\)x=3

a) Ta có: \(2x-3=-x+6\)

\(\Leftrightarrow2x+x=6+3\)

\(\Leftrightarrow3x=9\)

hay x=3

Vậy: x=3

b) Ta có: \(\left|2x-7\right|+2=13\)

\(\Leftrightarrow\left|2x-7\right|=11\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=11\\2x-7=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=18\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{9;-2\right\}\)

c) Ta có: \(\left(-12x-4^3\right)\cdot8^3=4\cdot8^4\)

\(\Leftrightarrow\left(-12x-64\right)=\dfrac{4\cdot8^4}{8^3}=32\)

\(\Leftrightarrow-12x=96\)

hay x=-8

Vậy: x=-8

\(a,128-3.\left(x+4\right)=23\\ \Rightarrow3.\left(x+4\right)=105\\ \Rightarrow x+4=35\\ \Rightarrow x=31\\ b,\left[\left(4x+28\right).3+55\right]:5=35\\ \Rightarrow\left(4x+28\right).3+55=175\\ \Rightarrow4x+28.3=120\\ \Rightarrow4x+28=60\\ \Rightarrow4x=32\\ \Rightarrow x=8.\)

c) \(\left(12x-4^3\right).8^3=4.8^4\)

\(12x-64=4.8^4:8^3\)

\(12x-64=32\)

\(12x=32+64\)

\(12x=96\)

\(x=\dfrac{96}{12}\)

\(x=8\)

d) \(720:\left[41-\left(2x-5\right)\right]:5=35\)

\(720:\left(41-2x+5\right):5=35\)

\(720:\left(46-2x\right)=35.5\)

\(720:\left(46-2x\right)=175\)

\(46-2x=720:175\)

\(46-2x=\dfrac{144}{35}\)

\(2x=46-\dfrac{144}{35}\)

\(2x=\dfrac{1466}{35}\)

\(x=\dfrac{1466}{35}:2\)

\(x=\dfrac{733}{35}\)

lm nốt hộ mik 2 ý cuối vs ạ

đơn giản

bn ơi nếu muốn ân smux

thì bn ấn vào chữ X² 

ạ bn viết thế này

mình ko có hiểu

Ta có:

\(A=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.16}\)

\(=\frac{3^{10}.16}{3^9.16}=\frac{3.1}{1.1}=3\)

Vậy giá trị biểu thức A là 3

de lam

G = 21 mũ 2 . 14 . 125 trên 35 mũ 5 . 6\

giúp mình với

a) ( x - 3 )² - 4 = 0

<=> ( x - 3 )² - 2² = 0

<=> ( x - 3 - 2 )( x - 3 + 2 ) = 0

<=> ( x - 5 )( x - 1 ) = 0

<=> x = 5 hoặc x = 1

b( 2x + 3 )² - ( 2x + 1 )( 2x - 1 ) = 22

<=> 4x² + 12x + 9 - ( 4x² - 1 ) = 22

<=> 4x² + 12x + 9 - 4x² + 1 = 22

<=> 12x + 10 = 22

<=> 12x = 12

<=> x = 1

c) ( 4x + 3 )( 4x - 3 ) - ( 4x - 5 )² = 16

<=> 16x² - 9 - ( 16x² - 40x + 25 ) = 16

<=> 16x² - 9 - 16x² + 40x - 25 = 16

<=> 40x - 34 = 16

<=> 40x = 50

<=> x = 50/40 = 5/4

d) x³ - 9x² + 27x - 27 = -8

<=> ( x - 3 )³ = -8

<=> ( x - 3 )³ = (-2)³

<=> x - 3 = -2

<=> x = 1 

e) ( x + 1 )³ - x²( x + 3 ) = 2

<=> x³ + 3x² + 3x + 1 - x³ - 3x² = 2

<=> 3x + 1 = 2

<=> 3x = 1

<=> x = 1/3

f) ( x - 2 )³ - x( x - 1 )( x + 1 ) + 6x² = 5

<=> x³ - 6x² + 12x - 8 - x( x² - 1 ) + 6x² = 5

<=> x³ + 12x - 8 - x³ + x = 5

<=> 13x - 8 = 5

<=> 13x = 13

<=> x = 1

a) \(\left(x-3\right)^2-4=0\)

=> \(\left(x-3\right)^2-2^2=0\)

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

=> \(\left(x-5\right)\left(x-1\right)=0\)

=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)

=> \(\left(2x+3\right)^2-\left[\left(2x\right)^2-1^2\right]=22\)

=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)

=> \(\left(2x\right)^2+2\cdot2x\cdot3+3^2-4x^2+1=22\)

=> \(4x^2+12x+9-4x^2+1=22\)

=> \(12x+9+1=22\)

=> \(12x+10=22\)

=> 12x = 12

=> x = 1

c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)

=> \(\left(4x\right)^2-3^2-\left[\left(4x\right)^2-2\cdot4x\cdot5+5^2\right]=16\)

=> \(16x^2-9-\left(16x^2-40x+25\right)=16\)

=> \(16x^2-9-16x^2+40x-25=16\)

=> \(-9+40x-25=16\)

=> \(40x=16+25-\left(-9\right)=16+25+9=50\)

=> x = 50/40 = 5/4

d) \(x^3-9x^2+27x-27=-8\)

=> \(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=8\)

=> \(\left(x-3\right)^3=-8\)

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

=> x - 3  = -2 => x = 1

e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)

=> \(x^3+3x^2+3x+1-x^3-3x^2=2\)

=> \(3x+1=2\)

=> \(3x=1\)=> x = 1/3

f) \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x^2=5\)

=> \(x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3-x\left(x^2-1\right)+6x^2=5\)

=> \(x^3-6x^2+12x-8-x^3+x+6x^2=5\)

=> \(\left(12x+x\right)-8=5\)

=> 13x  = 13

=> x = 1

a) (x+3)^2-4=0

=>(x+3)^2 = 4

=>(x+3)^2 = 2^2 = (-2)^2

=>x+3 = 2 hoặc -2

=> x= -1 hoặc -5

Bài 1:

a) 8² . 2² = 64 . 4 = 256

b) 2⁵. 4 = 2⁵. 2²= 2⁵⁺² = 2⁷

c) 9⁴. 243 = 9⁴. 3⁵=(3²)⁴.3⁵=3⁸.3⁵=3⁸⁺⁵=3¹³

Bài 1 :

\(a.8^2.2^2=\left(8.2\right)^2=16^2\)

\(b.2^5.4=2^5.2^2=2^7\)

\(c.9^4.243=\left(3^2\right)^4.3^5=3^8.3^5=3^{13}\)

Bài 2:

a. ( đề sai =.= )

\(b.12.\left(x-1\right):3=43+2^3\)

\(\Leftrightarrow12.\left(x-1\right)=43+2^3.3\)

\(\Leftrightarrow12.\left(x-1\right)=67\)

\(\Leftrightarrow x-1=\frac{67}{12}\)

\(\Leftrightarrow x=\frac{79}{12}\)

chịu

Chịu 

ở bài 1 đầu bài là viết các tich và các thương sau dưới dạng lũy thừa mình viết thiếu 

bạn đăng từng bài một thôi nhé