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\(\Rightarrow\)\(\text{Cái đề của bn hơi bị sai chắc là z hả?}\)
\(x^3+12x^2+48x+10=-54\)
\(\text{Ta có thể thấy:}\)\(x^3+12x^2+48x+10\)\(\text{là 1 HĐT}\)
\(\Rightarrow\)\(\text{Nhưng k có số nào mũ 3 =10}\)
\(x^3+12x^2+48x+10+54\)\(\text{(để ra HĐT)}\)
\(\text{Về trái +54 thì phế phải cx phải cộng 54}\)
\(x^3+12x^2+48x+10+54=-54+54\)
\(x^3+12x^2+48x+10+54=0\)
\(x^3+12x^2+48x+64=0\)
\(\left(x+4\right)^3=0\)
\(x+4=0\)
\(\Rightarrow x=-4\)
\(\text{Hok tốt nhé}\)
a: 78x(x-97)-x+97=0
=>(x-97)(78x-1)=0
=>\(\left[{}\begin{matrix}x=97\\x=\dfrac{1}{78}\end{matrix}\right.\)
b: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
=>\(x\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
c: \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
=>\(x^2+4x+4-x^2+4=0\)
=>4x+8=0
=>x+2=0
=>x=-2
\(a,78x\left(x-97\right)-x+97=0\)
\(\Leftrightarrow78x\left(x-97\right)-\left(x-97\right)=0\)
\(\Leftrightarrow\left(x-97\right)\left(78x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-97=0\\78x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=97\\x=\dfrac{1}{78}\end{matrix}\right.\)
\(b,\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=0\\x-2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
\(c,\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[\left(x+2\right)-\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+2-x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\cdot4=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Ta có
x 3 – 12 x 2 + 48 x – 64 = 0 ⇔ x 3 – 3 . x 2 . 4 + 3 . x . 4 2 – 4 3 = 0 ⇔ ( x – 4 ) 3 = 0
ó x – 4 = 0 ó x = 4
Vậy x = 4
Đáp án cần chọn là: B
a, 2x(x-5) - x ( 3 + 2x ) = 26
=> 2x^2 - 10x - 3x - 2x ^ 2 = 26
=> - 13 x = 26
=> x = -2
a, \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
\(\Leftrightarrow-13x=26\)
\(\Leftrightarrow x=-2\)
Vậy x = -2
b, \(3x^3-48x=0\)
\(\Leftrightarrow3x\left(x^2-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4;x=-4\end{cases}}\)
Vậy x = 0 hoặc x = 4 hoặc x = -4
x4+4x3-4x2-48x-48=0
=> x4+4(x3-x2) - 48x = 48
=> x4 + 4[x2(x-1)] - 48x = 48
\(x^4+4x^3-4x^2-48x-48=0\)
\(\Leftrightarrow\)\(x^4-2x^3-4x^2+6x^3-12x^2-24x+12x^2-24x-48=0\)
\(\Leftrightarrow\)\(x^2\left(x^2-2x-4\right)+6x\left(x^2-2x-4\right)+12\left(x^2-2x-4\right)=0\)
\(\Leftrightarrow\)\(\left(x^2-2x-4\right)\left(x^2+6x+12\right)\)
\(\Leftrightarrow\)\(\left[\left(x-1\right)^2-5\right]\left(x^2+6x+12\right)=0\)
\(\Leftrightarrow\)\(\left(x-1-\sqrt{5}\right)\left(x-1+\sqrt{5}\right)\left(x^2+6x+12\right)=0\)
Ta có: \(x^2+6x+12=\left(x+3\right)^2+3>0\)
\(\Rightarrow\)\(\orbr{\begin{cases}x-1-\sqrt{5}=0\\x-1+\sqrt{5}=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1+\sqrt{5}\\x=1-\sqrt{5}\end{cases}}\)
Vậy...
c: Ta có: \(x^3-12x^2+48x-64=0\)
\(\Leftrightarrow x-4=0\)
hay x=4
c: Ta có: \(x^3-12x^2+48x-64=0\)
\(\Leftrightarrow x-4=0\)
hay x=4
X=1 vì
78 x X - 48 x X =30
( 78-48) x X =30
30 x X =30
x= 30:30
x=1