lớp  3 đã học x4 = x nhân 4 rooif à

hay đề sai

a)x=6

b)x=khong biet

a, x4 + 5x=120
=> 10*x+4+50+x=120
=>11*x+54=120
=>11*x=120-54=66
=>x=66/11
=>x=6
b,1x3+4x+15=180
=>103 + 10*x + 40 +x +15 = 180
=>158 +11*x = 180
=>11*x = 180-158=22
=>x=22/11
=>x=2
(dấu * là dấu nhân; dấu / là dấu chia nha =))

Ta có \(\frac{2.3.4+2.3.4.2018+2.3.4.2019-2.3.4.2020}{5.6.7+5.6.7.2018+5.6.7.2019-5.6.7.2020}\)

\(=\frac{2.3.4\left(1+2018+2019-2020\right)}{5.6.7.\left(1+2018+2019-2020\right)}=\frac{2.3.4}{5.6.7}=\frac{4}{35}\)

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

\(\Leftrightarrow x^4-5x^3+10x+4=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3-6x^2+6x+4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\\x^2-4x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=2\pm\sqrt{6}\end{matrix}\right.\)

\(X^4+4=5X\left(X^2-2\right)\\ \Leftrightarrow X^4-5X^3+10X+4=0\\ \Leftrightarrow\left(X+1\right)\left(X-2\right)\left(X^2-4X-2\right)\)

\(\left[{}\begin{matrix}X+1=0\\X-2=0\\X^2-4X-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}X=-1\\X=2\\X=2+\sqrt{6}\\X=2-\sqrt{6}\end{matrix}\right.\)

VẬY......

\(P=x^4+x^3-x^2-x^4+3x^2-5x-1.\)

\(\Leftrightarrow P=\left(x^4-x^4\right)+x^3+\left(3x^2-x^2\right)-5x-1\)

\(\Leftrightarrow P=x^3+2x^2-5x-1\)

Vậy chọn  D

a. =>  5x= 56 

x4 = 120 - 56 = 64 nhé 

b => 4x=42 

1x3 = 180 - ( 42 + 15 ) =133

a) 10x+4+50+x=120

11x+54=120

11x=120-54=66

x=6

b) Làm tương tự nha

a: \(x^2-y^2-x-y\)

\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-1\right)\)

f: \(x^3-5x^2-5x+1\)

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

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