x = - 4

320-4x+4³=352
4x+64=320-352
4x+64=-32
4x=-32-64
4x=-96
x=-96:4
x=-24

320-4x+4³=352
4x+64=320-352
4x+64=-32
4x=-32-64
4x=-96
x=-96:4
x=-24

4(x-3)²-320 = 0

=> 4(x-3)² = 320

=> (x-3)² = 320 : 4 = 80 = (8,94427191)² = (-8,94427191)²

TH1:

x - 3 = 8,94427191

=> x = 11,94427191

TH2: 

x - 3 = -8,94427191

=> x = -5,94427191

7(4+x)³-875 = 0

=> 7(4+x)³ = 875

=> (4+x)³ = 875:7 = 125 = 5³

=> 4 + x = 5

=> x = 1

650 - 5(x+4)² = 330

5(x+4)² = 650 - 330 =320

=> (x+4)² = 320 : 5 = 64 = 8²

=> x+4 = 8

=> x = 4

3(5-x)²-15 = 60

=> 3(5-x)² = 75

=> (5-x)² = 25 = 5² =(-5)²

TH1:

5-x =5

=> x = 0

TH2: 5-x = -5

=> x = 10

\(2x\left(x-17\right)+\left(17-x\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-17\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=17\end{cases}}\)

\(9x^2-18x=0\)

\(\Leftrightarrow9x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

b) \(x\left(x-2\right)+5\left(2+x\right)=0\)

\(\Leftrightarrow x^2-2x+10+5x=0\)

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

Dễ thấy phương trình vô nghiệm do vế trái luôn dương

a) 120 x 40 > 120 : 40

b) 280 + 70 < 280 x 70

c) 320 – 80 > 320 : 80

d) 610 + 0 = 610 – 0

a) x = 320 - y
    y = 320 - x
b) 4x - y = 960
      4x   = 960 + y
        x = ( 960 + y ) : 4

x+y=320(*)

4x-y=960(**)

=> x+y+4x-y=320+960

=>5x=1280

=>x=256

Thay x=256 vào (*) suy ra y=64

Thử lại vào (**) thỏa mãn

Vậy x=256

       y=64

x + y = 320          (1)  

\(\Leftrightarrow\)3x + 3y = 960      (*)

4x - y = 960       (2)

thay (*) vào (2) ta được:

     4x - y = 3x + 3y 

\(\Leftrightarrow\)4x - 3x = 3y + y

\(\Leftrightarrow\)x = 4y (**)

thay (**) vào (1) ta được:

   4y + y = 320

\(\Leftrightarrow\)5y = 320

\(\Leftrightarrow\)y = 320 : 5 = 64

\(\Rightarrow\)x = 64 * 4 = 256

Vậy x = 256;   y = 64

Em lớp 8 thui nên chỉ biết lam cách đấy 

Chị k cho em nha///

a) \(^{x^3}\) - 7x+6=0

\(\Leftrightarrow\) \(^{x^3}\) - x-6x+6=0

\(\Leftrightarrow\) \(\left(x^3-x\right)\) - \(\left(6x-6\right)\) =0

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

\(\Leftrightarrow\) x\(\left(x+1\right)\)\(\left(x-1\right)\) - 6\(\left(x-1\right)\) =0

\(\Leftrightarrow\) \(\left(x-1\right)\) \(\left[x-6\left(x+1\right)\right]\) =0

\(\Leftrightarrow\) \(\left(x-1\right)\) \(\left(6-5x\right)\) =0

\(\Leftrightarrow\) \(\left[\begin{matrix}x-1=0\\6-5x=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[\begin{matrix}x=1\\5x=-6\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[\begin{matrix}x=1\\x=-\frac{6}{5}\end{matrix}\right.\)

Những câu sau dùng phương pháp phân tích đa thức thành nhân tử nhé!

x⁴- 4x³+3x²+4x-4= 0

(x-1)(x+1)(x-2)²=0

x=1 ;x=-1;x=2

a)x^3 - 7x - 6

= x^3 + x^2 - x^2 - 6x - x - 6

= (x^3 + x^2) - (x^2 + x) - (6x + 6)
= x^2(x + 1) - x(x + 1) - 6(x + 1)

= (x + 1)(x^2 - x - 6)

= (x + 1)(x^2 - 3x + 2x - 6)
= (x + 1){(x^2 - 3x) + (2x - 6)}

= (x + 1){(x(x - 3) + 2(x - 3)}
= (x + 1)(x - 3)(x + 2)

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

\(\left[\left(2x\right)^2+2.2x.1+1\right]-4=0\)

\(\left(2x+1\right)^2-2^2=0\)

\(\left(2x+1-2\right).\left(2x+1+2\right)=0\) 

\(\left(2x-1\right).\left(2x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}\)

\(x^4-3x^3-x+3=0\)

\(x^3.\left(x-3\right)-\left(x-3\right)=0\)

\(\left(x-3\right).\left(x^3-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^3-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

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

\(x^2.\left(x-1\right)-\left[\left(2x\right)^2-2.2x.2+2^2\right]=0\)

\(x^2.\left(x-1\right)-\left(2x-2\right)^2=0\)

\(x^2.\left(x-1\right)-4.\left(x-1\right)^2=0\)

\(\left(x-1\right).\left[x^2-4.\left(x-1\right)\right]=0\)

\(\left(x-1\right).\left[x^2-2.x.2+2^2\right]=0\)

\(\left(x-1\right).\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)

Vậy \(\begin{cases}x=1\\x=2\end{cases}\)

Tham khảo nhé~

$a)(x-25)-175=0$

$\to x-25=0+175=175$

$\to x=175+25$

$\to x=200$

$b)485-(6.x+60)=5$

$\to 6.x+60=485-5=480$

$\to 6.x=480-60=420$

$\to x=420:6=70$

c) ( x - 31 ) - 68 = 0

x - 31 - 68 = 0

x - 99 = 0

x - 99 +99 = 0 + 99

x = 0 + 99

x = 99