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\(A\left(-1\right)=m\cdot\left(-1\right)^2+2m\cdot\left(-1\right)-3=0\)
\(\Rightarrow m-2m-3=0\)
\(\Rightarrow-m-3=0\)
\(\Rightarrow-m=3\)
\(\Rightarrow m=-3\)
a) Đặt \(f_{\left(x\right)}=0\)
\(\Leftrightarrow x^3+3x^2-2x-2=0\)
\(\Leftrightarrow x^3-x^2+4x^2-4x+2x-2=0\)
\(\Leftrightarrow x^2\left(x-1\right)+4x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+4x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+4x+4-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+2\right)^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x+2=\sqrt{2}\\x+2=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{2}-2\\x=-\sqrt{2}-2\end{matrix}\right.\)
Vậy: \(S=\left\{1;\sqrt{2}-2;-\sqrt{2}-2\right\}\)
b) Đặt \(G_{\left(x\right)}=0\)
\(\Leftrightarrow3x+1=0\)
\(\Leftrightarrow3x=-1\)
hay \(x=\frac{-1}{3}\)
Vậy: \(S=\left\{-\frac{1}{3}\right\}\)
c) Đặt \(A_{\left(x\right)}=0\)
\(\Leftrightarrow2x^2-4=0\)
\(\Leftrightarrow2x^2=4\)
\(\Leftrightarrow x^2=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
Vậy: \(S=\left\{\sqrt{2};-\sqrt{2}\right\}\)
d) Đặt \(h_{\left(x\right)}=0\)
\(\Leftrightarrow2x^2+3x-5=0\)
\(\Leftrightarrow2x^2+5x-2x-5=0\)
\(\Leftrightarrow x\left(2x+5\right)-\left(2x+5\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-5\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-5}{2}\\x=1\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{-5}{2};1\right\}\)
e) Đặt P=0
\(\Leftrightarrow3x^2+4x^2+6x+3=0\)
\(\Leftrightarrow7x^2+6x+3=0\)
\(\Leftrightarrow7\left(x^2+\frac{6}{7}x+\frac{3}{7}\right)=0\)
mà 7>0
nên \(x^2+\frac{6}{7}x+\frac{3}{7}=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\frac{6}{14}+\frac{9}{49}+\frac{12}{49}=0\)
\(\Leftrightarrow\left(x+\frac{3}{7}\right)^2=-\frac{12}{49}\)(vô lý)
Vậy: S=∅
Bài 1:
Đặt \(h_{\left(x\right)}=0\)
\(\Leftrightarrow x^2-5x+5=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{5}{2}+\frac{25}{4}-\frac{5}{4}=0\)
\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=\frac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{5}{2}=\frac{\sqrt{5}}{2}\\x-\frac{5}{2}=-\frac{\sqrt{5}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{5}+5}{2}\\x=\frac{-\sqrt{5}+5}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{5+\sqrt{5}}{2};\frac{5-\sqrt{5}}{2}\right\}\)
Bài 2:
a) Đặt \(f_{\left(x\right)}=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Vậy: S={2}
b) Đặt \(g_{\left(x\right)}=0\)
\(\Leftrightarrow x^3-4x=0\)
\(\Leftrightarrow x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: S={0;2;-2}
c) Đặt \(h_{\left(x\right)}=0\)
\(\Leftrightarrow x^3+8=0\)
\(\Leftrightarrow x^3=-8\)
hay x=-2
Vậy: S={-2}
d) Đặt \(p_{\left(x\right)}=0\)
\(\Leftrightarrow x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x+1=0\)(vì \(x^2+1>0\forall x\))
hay x=-1
Vậy: S={-1}
Ta có: M(x) = 5x3 + 2x4 - x2 + 3x2 - x3 - x4 + 1 - 4x3
M(x) = (2x4 - x4) + (5x3 - x3 - 4x3) + (-x2 + 3x2) + 1
M(x) = x4 + 2x2 + 1
a) M(1) = 14 + 2.12 + 1 = 1 + 2 + 1 = 4
M(-1) = (-1)4 + 2.(-1)2 + 1 = 4
b) Ta có: x4 \(\ge\)0; 2x2 \(\ge\)0; 1 > 0
=> x4 + 2x2 + 1 > 0
=> M(x) > 0
=> M(x) ko có nghiệm
a. f(x) = g(x) - h(x)
= 4x2 + 3x + 1 - (3x2 - 2x - 3)
= 4x2 + 3x + 1 - 3x2 + 2x + 3
= (4x2 - 3x2) + (3x + 2x) + (1 + 3)
= x2 + 5x + 4
b. Xét đa thức f(x) = x2 + 5x + 4
f(-4) = (-4)2 + 5 . (-4) + 4 = 0
Vậy x = -4 là nghiệm của f(x)
c. Cho f(x) = 0
\(\Rightarrow\) x2 + 5x + 4 = 0
\(\Rightarrow\) x2 + x + 4x + 4 = 0
\(\Rightarrow\) x (x + 1) + 4 (x + 1) = 0
\(\Rightarrow\) (x + 1) (x + 4) = 0
\(\Rightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
Vậy f(x) có tập nghiệm là \(x\in\left\{-4;-1\right\}\).
f(x)=x^2+5x+4 (x+1)(x+4)=0 \(\hept{\begin{cases}x=-1\\x=-4\end{cases}}\) s={-1,-4}
1. S = { 3;4 }
2. S={ -2; 1}
3. S={\(\frac{1}{2}\) ; 2;-2}
4.S={\(\frac{4}{3}\) ;2}
S la tap ngo nhek , xin k nao
a) Để \(F_{\left(x\right)}=2mx-2\) có nghiệm là x=1 thì \(F_{\left(1\right)}=2\cdot m\cdot1-2=0\)
\(\Leftrightarrow2m-2=0\)
\(\Leftrightarrow2m=2\)
hay m=1
Vậy: Khi m=1 thì \(F_{\left(x\right)}=2mx-2\) có nghiệm là x=1
b) Để \(G_{\left(x\right)}=mx^2+2x+8\) có nghiệm là x=-1 thì \(G_{\left(-1\right)}=m\cdot\left(-1\right)^2+2\cdot\left(-1\right)+8=0\)
\(\Leftrightarrow m-2+8=0\)
\(\Leftrightarrow m+6=0\)
hay m=-6
Vậy: Khi m=-6 thì \(G_{\left(x\right)}=mx^2+2x+8\) có nghiệm là x=-1
c) Để \(H_{\left(x\right)}=x^4+3m^2x^3+2m^2+mx-1\) có nghiệm là x=1
thì \(H_{\left(1\right)}=1^4+3\cdot m^2\cdot1^3+2\cdot m^2+m\cdot1-1=0\)
\(\Leftrightarrow1+3m^2+2m^2+m-1=0\)
\(\Leftrightarrow5m^2+m=0\)
\(\Leftrightarrow m\left(5m+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=0\\5m+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=0\\5m=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=0\\m=\frac{-1}{5}\end{matrix}\right.\)
Vậy: Khi \(m\in\left\{0;\frac{-1}{5}\right\}\) thì \(H_{\left(x\right)}=x^4+3m^2x^3+2m^2+mx-1\) có nghiệm là x=1