x5 - 4x3=0
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a.
\(x^3-7x+6=0\)
\(\Leftrightarrow x^3-3x^2+2x+3x^2-9x+6=0\)
\(\Leftrightarrow x\left(x^2-3x+2\right)+3\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^2-x-2x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[x\left(x-1\right)-2\left(x-1\right)\right]\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
f.
\(x^4-4x^3+12x-9=0\)
\(\Leftrightarrow x^4-4x^3+3x^2-3x^2+12x-9=0\)
\(\Leftrightarrow x^2\left(x^2-4x+3\right)-3\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow\left(x^2-4x+3\right)\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x^2-x-3x+3\right)\left(x^2-3\right)=0\)
\(\Leftrightarrow\left[x\left(x-1\right)-3\left(x-1\right)\right]\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\\x=\pm\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow x^5-1=4x^4+4x^3+4x^2+4x+4\)
\(\Leftrightarrow\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)=4\left(x^4+x^3+x^2+x+1\right)\)
\(\Leftrightarrow\left(x-5\right)\left(x^4+x^3+x^2+x+1\right)=0\)
\(\Leftrightarrow x=5\)
Ta có
P ( x ) = 2 x 3 − 3 x + x 5 − 4 x 3 + 4 x − x 5 + x 2 − 2 = x 5 − x 5 + 2 x 3 − 4 x 3 + x 2 + ( 4 x − 3 x ) − 2 = − 2 x 3 + x 2 + x − 2 Và Q ( x ) = x 3 − 2 x 2 + 3 x + 1 + 2 x 2
= x 3 + - 2 x 2 + 2 x 2 + 3 x + 1 = x 3 + 3 x + 1
Khi đó
P ( x ) − Q ( x ) = − 2 x 3 + x 2 + x − 2 − x 3 + 3 x + 1 = − 2 x 3 + x 2 + x − 2 − x 3 − 3 x − 1 = − 2 x 3 − x 3 + x 2 + ( x − 3 x ) − 2 − 1 = − 3 x 3 + x 2 − 2 x − 3
Chọn đáp án B
y’ = (x5 – 4x3 + 2x – 3)’
= (x5)’ – (4x3)’ + (2x)’ – (3)’
= 5x4 – 4.3x2 + 2
= 5x4 – 12x2 + 2.
Ta có: f(x) + g(x) – h(x)
= (x5 – 4x3 + x2 – 2x + 1) + (x5 – 2x4 + x2 – 5x + 3) – (x4 – 3x2 + 2x – 5)
= x5 – 4x3 + x2 – 2x + 1 + x5 – 2x4 + x2 – 5x + 3 – x4 + 3x2 - 2x + 5
= (x5 +x5) – (2x4 + x4) – 4x3 + (x2 + x2 + 3x2)- (2x + 5x + 2x) + (1 + 3 + 5)
= (1 + 1)x5 – (2 + 1)x4 – 4x3 + (1 + 1 + 3)x2 - (2 + 5 + 2)x + (1 + 3 + 5)
= 2x5 – 3x4 – 4x3 + 5x2 – 9x + 9
Ta có
P ( x ) = 2 x 3 − 3 x + x 5 − 4 x 3 + 4 x − x 5 + x 2 − 2 = x 5 − x 5 + 2 x 3 − 4 x 3 + x 2 + ( 4 x − 3 x ) − 2 = − 2 x 3 + x 2 + x − 2 Và Q ( x ) = x 3 − 2 x 2 + 3 x + 1 + 2 x 2 = x 3 + − 2 x 2 + 2 x 2 + 3 x + 1 = x 3 + 3 x + 1
Khi đó
M ( x ) = P ( x ) + Q ( x ) = − 2 x 3 + x 2 + x − 2 + x 3 + 3 x + 1 = − 2 x 3 + x 2 + x − 2 + x 3 + 3 x + 1 = − 2 x 3 + x 3 + x 2 + ( x + 3 x ) − 2 + 1 = − x 3 + x 2 + 4 x − 1
Bậc của M ( x ) = - x 3 + x 2 + 4 x - 1 l à 3
Chọn đáp án C
---> \(x^3\left(x^2-4\right)=0\)
---> \(x^3\left(x-2\right)\left(x+2\right)=0\)
---> \(\hept{\begin{cases}x=0\\x=2\\x=-2\end{cases}}\)BẠN THAY DẤU NGOẶC NHỌN BẰNG DẤU NGOẶC VUÔNG NHÉ BẠN
xin tiick
x5 - 4x3=0
<=> x3(x2-4)=0
<=> x3=0 hoặc x2-4=0
<=> x=0 hoặc x2=4
<=> x=0 hoặc x=2 hoặc x=-2