Chia các đa thức:
a, \(\left(x^3-7x+6\right)\div\left(x+3\right)\)
b, \(\left(x^3-9x^2+6x+10\right)\div\left(x+1\right)\)
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2x3 + 3x2 + 6x + 5 = 02
<=> 2x3 + x2 + 5x + 2x2 + x + 5 = 0
<=> x(2x2 + x + 5) + (2x2 + x + 5) = 0
<=> (2x2 + x + 5)(x + 1) = 0
<=> x + 1 = 0 (vì 2x2 + x + 5 \(\ge\) 4,875 > 0 \(\forall\) x)
<=> x = - 1
Vậy tập nghiệm của pt là \(S=\left\{-1\right\}\)
b) 4x4 + 12x3 + 5x2 - 6x - 15 = 0
<=> 4x4 + 10x3 + 2x3 + 5x2 - 6x - 15 = 0
<=> 2x3(2x + 5) + x2(2x + 5) - 3(2x + 5) = 0
<=> (2x + 5)(2x3 + x2 - 3) = 0
<=> (2x + 5)(2x3 - 2x2 + 3x2 - 3) = 0
<=> (2x + 5)(x - 1)(2x2 + 3x + 3) = 0
<=> (2x + 5)(x - 1)[x2 + (x + 3/2)2 + 3/4]= 0
Mà x2 + (x + 3/2)2 + 3/4 > 0\(\forall x\)
\(\Rightarrow\left[\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=-\frac{5}{2}\\x=1\end{matrix}\right.\)
Vậy ...
\(A=4x^2+6x=2x\left(2x+3\right)\)
\(B=\left(2x+3\right)^2-x\left(2x+3\right)=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\)
\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)
\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)
\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)
\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)
a/ \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\\3+2^{x+1}=24-\left[16-\left(4-1\right)\right]\)
\(3+2^{x+1}=24-\left(16-3\right)\\ 3+2^{x-1}=24-13\\ 3+2^{x-1}=11\\ 2^{x+1}=11-3\\ 2^{x-1}=8\)
\(2^{x-1}=2^3\\ \Rightarrow x-1=3\\x=3+1\\ x=4\)
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=205550\)
\(\left(x.100\right)+\left(1+2+3+....+100\right)=205550\)
Ta tính tổng \(1+2+3+...+100\\ \) trước
Số các số hạng: \(\left[\left(100-1\right):1+1\right]=100\)
Tổng :\(\left[\left(100+1\right).100:2\right]=5050\)
Thay số vào ta có được:
\(\left(x.100\right)+5050=205550\\ \\ x.100=205550-5050\\ \\x.100=20500\\ \\x=20500:100\\ \\\Rightarrow x=2005\)
a) (x3 – 7x + 3 – x2) : (x – 3)
b) (2x4 – 3x2 – 3x2 – 2 + 6x) : (x2 – 2)
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a.
\(\left(x^3-7x+6\right):\left(x+3\right)\)
\(=\left(x^3+3x^2-3x^2-9x+2x+6\right):\left(x+3\right)\)
\(=\left[\left(x^3+3x^2\right)-\left(3x^2+9x\right)+\left(2x+6\right)\right]:\left(x+3\right)\)
\(=\left[x^2\left(x+3\right)-3x\left(x+3\right)+2\left(x+3\right)\right]:\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-3x+2\right):\left(x+3\right)\)
\(=x^2-3x+2\)
Đặt phép chia không được sao?