a: Ta có: \(\left[\left(6x-39\right):3\right]\cdot28=5628\)

\(\Leftrightarrow\left(6x-39\right):3=201\)

\(\Leftrightarrow6x-39=603\)

\(\Leftrightarrow6x=642\)

hay x=107

b: Ta có: \(4x^3+12=120\)

\(\Leftrightarrow4x^3=108\)

\(\Leftrightarrow x^3=27\)

hay x=3

c: Ta có: \(1500:\left[\left(30x+40\right):x\right]=30\)

\(\Leftrightarrow\left(30x+40\right):x=50\)

\(\Leftrightarrow30x+40=50x\)

hay x=-2

\(a,223-6\left(x+5\right)=49\\ 6\left(x+5\right)=223-49\\ 6\left(x+5\right)=174\\ x+5=174:6\\ x+5=29\\ x=29-5\\ x=24\\ ----\\ b,4x^3+12=120\\ 4x^3=120-12\\ 4x^3=108\\ x^3=108:4\\ x^3=27\\ Mà:3^3=27\\ Nên:x^3=3^3\\ Vậy:x=3\)

a. 223 - 6. (x + 5) = 49

6. (x + 5) = 223 - 49

6. (x + 5) = 174

x + 5 = 174 : 6

x + 5 = 29

x = 29 - 5

x = 24

b. 4x³ + 12 = 120

4x³ = 120 - 12 = 108

x³ = 108 : 4 = 27

x³ = 3³

x= 3

3.2x—3=45

3.2x=45+3

3.2x=48

2x=48:3

2x=16

x=16:2

x=8

4x .3+12=120

4x.3=120–12

4x.3=108

4x=108:3

4x=36

x=36:4

x=9

\(3.2x-3=45\)

\(3.2x=45+3\)

\(3.2x=48\)

\(2x=\frac{48}{3}\)

\(2x=16\)

\(x=8\)

Vậy \(x=8\)

\(1500:\left[\left(36x+40\right):x\right]=30\)

\(\left(36x+40\right):x=1500:30\)

\(36x+40=50x\)

\(50x-36x=40\)

\(14x=40\)

\(x=\frac{40}{14}\)

\(x=\frac{20}{7}\)

Vậy \(x=\frac{20}{7}\)

a)3.2x-3 = 45

3.(2x-1) = 45

2x-1 = 45 : 3

2x-1 = 15

2x = 16

x = 8

các bài cn lại bn dựa vào phần a mak lm nha!

\(a,x^4-4x^3-19x^2+106x-120=0\\ \Rightarrow\left(x-4\right)\left(x^3-19x+30\right)=0\Rightarrow\left(x-4\right)\left(x+5\right)\left(x^2-5x+6\right)=0\\ \Rightarrow\left(x-4\right)\left(x+5\right)\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-5\\x=2\\x=3\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{-5;2;3;4\right\}\)

\(b,4x^4+12x^3+5x^2-6x-15=0\\ \Rightarrow\left(x-1\right)\left(4x^3+16x^2+21x+15\right)=0\\ \Rightarrow\left(x-1\right)\left[\left(4x^3+10x^2\right)+\left(6x^2+15x\right)+\left(6x+15\right)\right]=0\\ \Rightarrow\left(x-1\right)\left[2x^2\left(2x+5\right)+3x\left(2x+5\right)+3\left(2x+5\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(2x+5\right)\left(2x^2+3x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{2}\\2x^2+3x+3=0\left(vô.lí\right)\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{1;-\dfrac{5}{2}\right\}\)

a: \(4x^3+12=120\)

=>\(4x^3=108\)

=>\(x^3=27=3^3\)

=>x=3

b: \(\left(x-4\right)^2=64\)

=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)

c: (x+1)^3-2=5^2

=>\(\left(x+1\right)^3=25+2=27\)

=>x+1=3

=>x=2

d: 136-(x+5)^2=100

=>(x+5)^2=36

=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)

e: \(4^x=16\)

=>\(4^x=4^2\)

=>x=2

f: \(7^x\cdot3-147=0\)

=>\(3\cdot7^x=147\)

=>\(7^x=49\)

=>x=2

g: \(2^{x+3}-15=17\)

=>\(2^{x+3}=32\)

=>x+3=5

=>x=2

h: \(5^{2x-4}\cdot4=10^2\)

=>\(5^{2x-4}=\dfrac{100}{4}=25\)

=>2x-4=2

=>2x=6

=>x=3

i: (32-4x)(7-x)=0

=>(4x-32)(x-7)=0

=>4(x-8)*(x-7)=0

=>(x-8)(x-7)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

k: (8-x)(10-2x)=0

=>(x-8)(x-5)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)

m: \(3^x+3^{x+1}=108\)

=>\(3^x+3^x\cdot3=108\)

=>\(4\cdot3^x=108\)

=>\(3^x=27\)

=>x=3

n: \(5^{x+2}+5^{x+1}=750\)

=>\(5^x\cdot25+5^x\cdot5=750\)

=>\(5^x\cdot30=750\)

=>\(5^x=25\)

=>x=2

= ( 21 + 12 - 3 + 44 + 55 ) x 0 

= 0 

( 21 + 12 - 3 + 44 + 55) x ( 2 x 6 -4x3) = (21+12-3+44+55)x( 12-12)

=(21+12-3+44+55)x0 = 0

đúng

809703

= 10+1-12-3 = 11 -12 - 3

                     = -1 - 3

                     = -4

# AHT

= âm 4

là -4 nha

\(4^0\cdot9^3-6^9\cdot\dfrac{120}{8^4\cdot3^{12}+6^{11}}\) hay \(4^0\cdot9^3-6^9\cdot\dfrac{120}{8^4}\cdot3^{12}+6^{11}\)?

Các pạn giải hộ mk nha . Sáng mai mk phải nộp bài rùi huhu