a) \(\left|x\right|-\frac{7}{6}=\frac{9}{15}\)

=> \(\left|x\right|=\frac{9}{15}+\frac{7}{6}=\frac{53}{30}\)

=> \(\orbr{\begin{cases}x=\frac{53}{30}\\x=-\frac{53}{30}\end{cases}}\)

b) \(\left|x-\frac{4}{3}\right|=\frac{1}{6}\)

=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{1}{6}\\x-\frac{4}{3}=-\frac{1}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{7}{6}\end{cases}}\)

c) \(\left|x-\frac{4}{3}\right|-\frac{1}{3}=\frac{1}{2}\)

=> \(\left|x-\frac{4}{3}\right|=\frac{1}{2}+\frac{1}{3}\)

=> \(\left|x-\frac{4}{3}\right|=\frac{5}{6}\)

=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{5}{6}\\x-\frac{4}{3}=-\frac{5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=\frac{1}{2}\end{cases}}\)

d) \(\frac{8}{3}-\left|\frac{7}{9}-x\right|=-\frac{1}{5}\)

=> \(\left|\frac{7}{9}-x\right|=\frac{43}{15}\)

=> \(\orbr{\begin{cases}\frac{7}{9}-x=\frac{43}{15}\\\frac{7}{9}-x=-\frac{43}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{94}{45}\\x=\frac{164}{45}\end{cases}}\)

e) \(\left|x-\left(\frac{1}{4}\right)^2\right|-\frac{25}{64}=0\)

=> \(\left|x-\frac{1}{16}\right|=\frac{25}{64}\)

=> \(\orbr{\begin{cases}x-\frac{1}{16}=\frac{25}{64}\\x-\frac{1}{16}=-\frac{25}{64}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{29}{64}\\x=-\frac{21}{64}\end{cases}}\)

f) \(\left(x-\frac{1}{4}\right)^2+\frac{17}{64}=\frac{21}{32}\)

=> \(\left(x-\frac{1}{4}\right)^2=\frac{25}{64}\)

=> \(\left(x-\frac{1}{4}\right)^2=\left(\frac{5}{8}\right)^2\)

=> \(\orbr{\begin{cases}x-\frac{1}{4}=\frac{5}{8}\\x-\frac{1}{4}=-\frac{5}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{3}{8}\end{cases}}\)

\(3\left(x+2\right)^2+\left(2x-3\right)^2-7\left(x-4\right)\left(x+4\right)=64\)

\(\Leftrightarrow3\left(x^2+4x+4\right)+\left(4x^2-12x+9\right)-7\left(x^2-16\right)=64\)

\(\Leftrightarrow3x^2+12x+12+4x^2-12x+9-7x^2+112=64\)

\(\Leftrightarrow12+9+112=64\)(vô lí)

Vậy pt vô nghiệm

 TL:

\(\Leftrightarrow3\left(x^2+4x+4\right)+4x^2-6x+9-7x^2+112=64\)

\(\Leftrightarrow6x+133=64\)  

\(\Leftrightarrow6x=-69\) 

\(\Leftrightarrow x=\frac{-23}{2}\) 

Vậy....

Nguyễn Văn Tuấn AnhSai từ dòng đầu tiên

\(\left(2x-3\right)^2=4x^2-12x+9\ne4x^2-6x+9\)

a: x^3=7^3

=>x^3=343

=>\(x=\sqrt[3]{343}=7\)

b: x^3=27

=>x^3=3^3

=>x=3

c: x^3=125

=>x^3=5^3

=>x=5

d: (x+1)^3=125

=>x+1=5

=>x=4

e: (x-2)^3=2^3

=>x-2=2

=>x=4

f: (x-2)^3=8

=>x-2=2

=>x=4

h: (x+2)^2=64

=>x+2=8 hoặc x+2=-8

=>x=6 hoặc x=-10

j: =>x-3=2 hoặc x-3=-2

=>x=1 hoặc x=5

k:

9x^2=36

=>x^2=36/9

=>x^2=4

=>x=2 hoặc x=-2

l:

(x-1)^4=16

=>(x-1)^2=4(nhận) hoặc (x-1)^2=-4(loại)

=>x-1=2 hoặc x-1=-2

=>x=3 hoặc x=-1

a, 4¹⁰. 2³⁰=2²⁰.2³⁰=2⁵⁰

b,9²⁵.27⁴.81³= 3⁵⁰.3¹².3¹²=3⁷⁴

Tương tự các câu khác....

a, 4¹⁰.2³⁰ = (2²)¹⁰.2³⁰ = 2²⁰.2³⁰ = 2⁵⁰
b, 9²⁵.27⁴.81³ = (3²)²⁵.(3³)⁴.(3⁴)³ = 3⁵⁰.3¹².3¹² = 3⁷⁴
c, 25⁵⁰.125⁵ = (5²)⁵⁰.(5³)⁵ = 5¹⁰⁰.5¹⁵ = 5¹¹⁵
d, 64³.4⁸.16⁴ = (2⁶)³.(2²)⁸.(2⁴)⁴ = 2¹⁸.2¹⁶.2¹⁶ = 2⁵⁰
e, 3⁸ : 3⁶ = 3²
2¹⁰ : 8³ = 2¹⁰ : (2³)³ = 2¹⁰ : 2⁹ = 2
12⁷ : 6⁷ = (12 : 6)⁷ = 2⁷
@Dương Tuyết Mai

`a,`\(2^x -15= 2^4+1\)

`-> 2^x-15=17`

`-> 2^x=17+15`

`-> 2^x=32`

`-> 2^x=2^5`

`-> x=5`

`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?

`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)

`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)

`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)

`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)

Mà `1/65+1/64-1/63-1/62 \ne 0`

`-> x+66=0`

`-> x=-66`

a: =>2^x=2^4+16=32

=>x=5

b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)

=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)

=>x+66=0

=>x=-66

Bài 4:

a, \(x^3+12x^2+48x+64=x^3+4x^2+8x^2+32x+16x+64\)

\(=x^2.\left(x+4\right)+8x.\left(x+4\right)+16.\left(x+4\right)\)

\(=\left(x+4\right).\left(x^2+8x+16\right)=\left(x+4\right).\left(x^2+4x+4x+16\right)\)

\(=\left(x+4\right).\left(x+4\right)^2=\left(x+4\right)^3\)(1)

Thay \(x=6\) vào (1) ta được:

\(\left(6+4\right)^3=10^3=1000\)

Vậy...........

b, \(x^3-6x^2+12x-8=x^3-2x^2-4x^2+8x+4x-8\)

\(=x^2.\left(x-2\right)-4x.\left(x-2\right)+4.\left(x-2\right)\)

\(=\left(x-2\right).\left(x^2-4x+4\right)=\left(x-2\right).\left(x^2-2x-2x+4\right)\)

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

Thay \(x=22\) vào (2) ta được:

\(\left(22-2\right)^3=20^3=8000\)

Vậy.............

Chúc bạn học tốt!!!

Bài 2:

a, \(\left(x+9\right)^3=27=3^3\)

\(\Rightarrow x+9=3\Rightarrow x=-6\)

Vậy.........

b, \(8-12x-x^3+6x^2=-64\)

\(\Rightarrow-\left(x^3-6x^2+12x-8\right)=-64\)

\(\Rightarrow x^3-2x^2-4x^2+8x+4x-8=64\)

\(\Rightarrow x^2.\left(x-2\right)-4x.\left(x-2\right)+4.\left(x-2\right)=64\)

\(\Rightarrow\left(x-2\right).\left(x^2-4x+4\right)=64\)

\(\Rightarrow\left(x-2\right).\left(x^2-2x-2x+4\right)=64\)

\(\Rightarrow\left(x-2\right).\left(x-2\right)^2=64\)

\(\Rightarrow\left(x-2\right)^3=4^3\Rightarrow x-2=4\Rightarrow x=6\)

Vậy............

Chúc bạn học tốt!!!

4. Tính giá trị biểu thức

a) x³ + 12x² + 48x + 64 khi x = 6

Ta có:

x³ + 12x² + 48x + 64 =

= (x³ + 64) + (12x² + 48x)

= (x³ + 4³) + 12x(x + 4)

= (x + 4)(x² - 4x + 4²) + 12x(x + 4)

= (x + 4)(x² - 4x + 16 +12x)

= (x + 4)(x² + 8x + 16)

= (x + 4)(x + 4)²

= (x + 4)³

Thế x = 6 vào biểu thức vừa tìm, ta được:

(x + 4)³ = (6 + 4)³ = 10³ = 1000

Vậy 1000 là giá trị của biểu thức x³ + 12x² + 48x + 64 khi x = 6.

b) x³ - 6x² + 12x - 8 khi x = 22

Ta có:

x³ - 6x² + 12x - 8 =

= (x³ - 8) - (6x² - 12x)

= (x³ - 2³) - 6x(x - 2)

= (x - 2)(x² + 2x + 2²) - 6x(x - 2)

= (x - 2)(x² + 2x + 4 - 6x)

= (x - 2)(x² - 4x + 4)

= (x - 2)(x - 2)²

= (x - 2)³

Thế x = 22 vào biểu thức vừa tìm, ta được:

(x - 2)³ = (22 - 2)³ = 20³ = 8000

Vậy 8000 là giá trị của biểu thức x³ - 6x² + 12x - 8 khi x = 22.