Cho g ( x ) = 2 x 2 - 4 m x + m 2 - 2 m - 1 x - m 2 . Tìm m để g ( x ) ≤ 0 với ∀ x ∈ 1 ; 3 .
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![](https://rs.olm.vn/images/avt/0.png?1311)
1: f(-1)=0
=>1+m-1+3m-2=0 và
=>4m-2=0
=>m=1/2
2: g(2)=0
=>2^2-4(m+1)-5m+1=0
=>4-5m+1-4m-4=0
=>-9m+1=0
=>m=1/9
4: f(1)=g(2)
=>1-(m-1)+3m-2=4-4(m+1)-5m+1
=>1-m+1+3m-2=4-4m-4-5m+1
=>2m-2=-9m+1
=>11m=3
=>m=3/11
3:
H(-1)=0
=>-2-m-7m+3=0
=>-8m=-1
=>m=1/8
5: g(1)=h(-2)
=>1-2(m+1)-5m+1=-8-2m-7m+3
=>-5m+2-2m-2=-9m-5
=>-7m=-9m-5
=>2m=-5
=>m=-5/2
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
\(f(x)=x^2+3mx+m^2\Rightarrow f(1)=1+3m+m^2\)
\(g(x)=x^2+(2m-1)x+m^2\Rightarrow g(1)=1+(2m-1)+m^2=m^2+2m\)
Để \(f(1)=g(1)\Leftrightarrow 1+3m+m^2=m^2+2m\)
\(\Leftrightarrow 1+m=0\Leftrightarrow m=-1\)
Vậy \(m=-1\)
\(\left\{{}\begin{matrix}f\left(x\right)=x^2+3mx+m^2\\g\left(x\right)=x^2+\left(2m-1\right)x+m^2\end{matrix}\right.\)
\(h\left(x\right)=f\left(x\right)-g\left(x\right)=\left[3m-\left(2m-1\right)\right]x=\left(m+1\right)x\)
\(f\left(1\right)=g\left(1\right)\Rightarrow f\left(1\right)-g\left(1\right)=0\Rightarrow h\left(1\right)=0\)
\(\Rightarrow\left(m+1\right).1=0\Rightarrow m=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 4:
Để f(x) chia hết cho g(x) thì \(x^2+5x+a⋮x+1\)
\(\Leftrightarrow x^2+x+4x+4+a-4⋮x+1\)
=>a-4=0
hay a=4
Câu 5:
Đêt f(x) chia hết cho g(x) thì \(2x^2+3x+a⋮x+2\)
\(\Leftrightarrow2x^2+4x-x-2+a+2⋮x+2\)
=>a+2=0
hay a=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(f\left(1\right)=1-m+1+3m-2=2m\)
\(g\left(2\right)=4-4\left(m+1\right)-5m+1=4-4m-4-5m+1=-9m+1\)
mà \(f\left(1\right)=g\left(2\right)\)hay \(2m=-9m+1\Leftrightarrow11m=1\Leftrightarrow m=\frac{1}{11}\)
Trả lời:
f(1)=g(2)
<=> 12-(m-1).1 +3m -2= 22-2(m+1).2-5m+1
<=>1-m+1+3m=4-4m-4-5m+1
<=> 2m+2=-9m+1
<=> 11m=1
=> m=1/11
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Để \(x^2-\left(m+1\right)x+4⋮x-1\Leftrightarrow4-m=0\Leftrightarrow m=4\)
Bài 2:
Sửa đề bài chút: g(x) = x^2 -3x +2
Để f(x) chia hết g(x) <=> a - 4 = 0 <=> a = 4
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có f(1) = 12 -(m - 1).1 + 3m - 2 = 2m
g(2) = 22 - 2(m + 1).2 - 5m + 1 = -9m + 1
Vì f(1) = g(2) ⇒ 2m = -9m + 1 ⇒ 11m = 1 ⇒ m = 1/11. Chọn D
![](https://rs.olm.vn/images/avt/0.png?1311)
h(x) + g(x) = f(x)
=> h(x)= f(x) - g(x) = \(3x^4+2x^2-2x^4+x^2-5x-\left(x^4-x^2-2x+6+3x^2\right)=x^2-3x-6\)\(h\left(-\dfrac{1}{3}\right)=\left(-\dfrac{1}{3}\right)^2-3\left(-\dfrac{1}{3}\right)-6=\dfrac{-44}{9}\)
\(h\left(\dfrac{3}{2}\right)=\left(\dfrac{3}{2}\right)^2-3\cdot\dfrac{3}{2}-6=-\dfrac{33}{4}\)
\(x^2-3x-6=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{33}}{6}\\x=\dfrac{3-\sqrt{33}}{6}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2:
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}a+b=2\\3^7\cdot a+b=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{1093}\\b\simeq2\end{matrix}\right.\)
Câu 3:
a: \(f\left(x\right)=4x^2+3x+1-3x^2+2x+3=x^2+5x+4\)
b: f(-4)=16-20+4=0
=>x=-4 là nghiệm
c: Đặt f(x)=0
=>(x+4)(x+1)=0
=>x=-4 hoặc x=-1
Chọn A