f(x)4x3 - x2+2x-5
g(x)=4x3+2x2-x-5
Tìm x sao cho f(x)=g(x)
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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
f(x) = x5 + 3x2 − 5x3 − x7 + x3 + 2x2 + x5 − 4x2 + x7
= (x5 + x5) + (3x2 + 2x2 – 4x2) + (-5x3 + x3) + (-x7 + x7)
= 2x5 + x2 – 4x3.
= 2x5 - 4x3 + x2
Đa thức có bậc là 5
g(x) = x4 + 4x3 – 5x8 – x7 + x3 + x2 – 2x7 + x4 – 4x2 – x8
= (x4 + x4) + (4x3 + x3) – (5x8 + x8) – (x7 + 2x7) + (x2 – 4x2)
= 2x4 + 5x3 – 6x8 – 3x7 – 3x2
= -6x8 - 3x7 + 2x4 + 5x3 - 3x2.
Đa thức có bậc là 8.
a) \(\Leftrightarrow x^2-x-x^2+2x=5\)
\(\Leftrightarrow x=5\)
b) \(\Leftrightarrow4x\left(x^2-9\right)=0\)
\(\Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0
\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}4x=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy x = 0 , x = 3 hoặc x = -3
\(a,\Leftrightarrow x^2-x-x^2+2x=5\\ \Leftrightarrow x=5\\ b,\Leftrightarrow4x\left(x^2-9\right)=0\\ \Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(2x-x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\ d,\Leftrightarrow\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72=0\\ \Leftrightarrow\left(x^2-9x+17\right)^2-3^2-72=0\\ \Leftrightarrow\left(x^2-9x+17\right)^2-81=0\\ \Leftrightarrow\left(x^2-9x+17-9\right)\left(x^2-9x+17+9\right)=0\\ \Leftrightarrow\left(x-8\right)\left(x-1\right)\left(x^2-9x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=1\\\left(x-\dfrac{9}{2}\right)^2+\dfrac{23}{4}=0\left(vô.n_0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)
+) Ta có
2 g ( x ) = 2 − x 4 + 2 x 3 − 3 x 2 + 4 x + 5 = − 2 x 4 + 4 x 3 − 6 x 2 + 8 x + 10 Ta có f ( x ) − 2 ⋅ g ( x ) = 5 x 4 + 4 x 3 − 3 x 2 + 2 x − 1 − − 2 x 4 + 4 x 3 − 6 x 2 + 8 x + 10 = 5 x 4 + 4 x 3 − 3 x 2 + 2 x − 1 + 2 x 4 − 4 x 3 + 6 x 2 − 8 x − 10 = 5 x 4 + 2 x 4 + 4 x 3 − 4 x 3 + − 3 x 2 + 6 x 2 + ( 2 x − 8 x ) − 1 − 1 = 7 x 4 + 3 x 2 − 6 x − 11
Hệ số cần tìm là -11
Chọn đáp án C
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