1: Thay x=2 vào (1), ta được:

\(2^2-2\left(m+1\right)\cdot2+2m-5=0\)

=>4-4(m+1)+2m-5=0

=>4-4m-4+2m-5=0

=>-2m-5=0

=>m=-5/2

Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=2\left(m+1\right)\)

=>\(x_2+2=2\left(-\frac52+1\right)=2\cdot\frac{-3}{2}=-3\)

=>\(x_2=-5\)

2: \(\Delta=\left\lbrack-2\left(m+1\right)\right\rbrack^2-4\cdot1\cdot\left(2m-5\right)\)

\(=4\left(m^2+2m+1\right)-4\left(2m-5\right)=4\left(m^2+2m+1-2m+5\right)=4\left(m^2+6\right)\ge24>0\forall m\)

=>(P) luôn cắt (d) tại hai điểm phân biệt

Theo Vi-et, ta có:\(x_1+x_2=-\frac{b}{a}=2\left(m+1\right)=2m+2;x_1x_2=\frac{c}{a}=2m-5\)

\(x_1^2+\left(2m+2\right)\cdot x_2-7=0\)

=>\(x_1^2+x_2\left(x_1+x_2\right)-7=0\)

=>\(x_1^2+x_2^2+x_1x_2-7=0\)

=>\(\left(x_1+x_2\right)^2-x_1x_2=7\)

=>\(\left(2m+2\right)^2-\left(2m-5\right)=7\)

=>\(4m^2+8m+4-2m+5-7=0\)

=>\(4m^2+6m+2=0\)

=>\(2m^2+3m+1=0\)

=>(m+1)(2m+1)=0

=>m=-1 hoặc m=-1/2

a: Diện tích tam giác đều cạnh 3cm là:

\(S=3^2\cdot\frac{\sqrt3}{4}=\frac{9\sqrt3}{4}\left(\operatorname{cm}^2\right)\)

Nửa chu vi là: \(p=\frac{3+3+3}{2}=\frac92\left(\operatorname{cm}\right)\)

Ta có: \(S=p\cdot r\)

=>\(r=\frac{9\sqrt3}{4}:\frac92=\frac{9\sqrt3}{4}\cdot\frac29=\frac{2\sqrt3}{4}=\frac{\sqrt3}{2}\left(\operatorname{cm}\right)\)

Diện tích hình tròn nội tiếp tam giác đều cạnh 3cm là:

\(S_{tròn}=\left(\frac{\sqrt3}{2}\right)^2\cdot\pi=\frac34\pi\left(\operatorname{cm}^2\right)\)

b: Với ngũ giác đều thì số đo của góc ở tâm chắn mỗi cạnh là: \(\frac{360^0}{5}=72^0\)

=>\(R=\frac{a}{2\cdot\sin\left(\frac{180^0}{5}\right)}=\frac{4}{2\cdot sin36^0}=\frac{2}{\sin36^0}\left(dm\right)\)

Diện tích hình tròn là:

\(S_2=\pi\cdot R^2=\pi\cdot\frac{4}{\sin^236^0}\left(dm^2\right)\)

\(\frac{\sin a}{\sin a-cosa}-\frac{cosa}{cosa-\sin a}\)

\(=\frac{\sin a}{\sin a-cosa}+\frac{cosa}{\sin a-cosa}=\frac{\sin a+cosa}{\sin a-cosa}\)

\(\frac{1+\cot^{}a}{1-\cot^{}a}\)

\(=\left(1+\frac{cosa}{\sin^{}a}\right):\left(1-\frac{cos^{}a}{\sin^{}a}\right)\)

\(=\frac{\sin a+cosa}{\sin a}:\frac{\sin a-cosa}{\sin a}=\frac{\sin a+cosa}{\sin a-cosa}\)

Do đó: \(\frac{\sin a}{\sin a-cosa}-\frac{cosa}{cosa-\sin a}=\frac{1+\cot a}{1-\cot a}\)

Câu 5:

OB=OD+DB=1,2+22=23,2(m)

Xét ΔODC vuông tại D và ΔOBA vuông tại B có

\(\hat{DOC}\) chung

Do đó: ΔODC~ΔOBA

=>\(\frac{DC}{BA}=\frac{OD}{OB}\)

=>\(\frac{1.2}{23.2}=\frac{1.5}{AB}\)

=>\(AB=23,2\cdot\frac{1.5}{1,2}=29\left(m\right)\)

Ta có: \(A=\frac{4-x}{x-2}\)

\(=\frac{-x+2+2}{x-2}=-1+\frac{2}{x-2}\)

Để A có giá trị nhỏ nhất thì \(\frac{2}{x-2}\) nhỏ nhất

=>x-2=-1

=>x=1

Gọi mẫu là x

Theo đề, ta có: \(\frac{-11}{13}<\frac{9}{x}<\frac{-11}{15}\)

=>\(\frac{-99}{117}<\frac{99}{11x}<\frac{-99}{135}\)

=>\(\frac{-99}{117}<\frac{-99}{-11x}<\frac{-99}{135}\)

=>\(\frac{99}{117}>\frac{99}{-11x}>\frac{99}{135}\)

=>117<-11x<135

=>\(-\frac{117}{11}>x>-\frac{135}{11}\)

mà x là số nguyên

nên x∈{-11;-12}

vậy: Các phân số cần tìm là 9/-11;9/-12

Sửa đề: \(A=\left(2x-1\right)\left(2x^2-3x-1\right)\left(x-1\right)+2005\)

\(=\left(2x^2-3x+2\right)\left(2x^2-3x-1\right)+2005\)

\(=\left(2x^2-3x\right)^2+\left(2x^2-3x\right)+2+2005\)

\(=\left(2x^2-3x\right)^2+\left(2x^2-3x\right)+\frac14+2006,75=\left(2x^2-3x+\frac12\right)^2+2006,75\ge2006,75\forall x\)

Dấu '=' xảy ra khi \(2x^2-3x+\frac12=0\)

=>\(x^2-\frac32x+\frac14=0\)

=>\(x^2-2\cdot x\cdot\frac34+\frac{9}{16}-\frac{7}{16}=0\)

=>\(\left(x-\frac34\right)^2=\frac{7}{16}\)

=>\(\left[\begin{array}{l}x-\frac34=\frac{\sqrt7}{4}\\ x-\frac34=-\frac{\sqrt7}{4}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt7+3}{4}\\ x=\frac{-\sqrt7+3}{4}\end{array}\right.\)

Đặt \(A=\frac{1}{2^2}+\frac{1}{4^2}+\cdots+\frac{1}{4010^2}\)

\(=\frac{1}{2^2}\left(\frac{1}{1^2}+\frac{1}{2^2}+\cdots+\frac{1}{2005^2}\right)\)

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{2005^2}<\frac{1}{2004\cdot2005}=\frac{1}{2004}-\frac{1}{2005}\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2005^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{2004}-\frac{1}{2005}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2005^2}<1-\frac{1}{2005}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2005^2}<1\)

=>\(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2005^2}<2\)

=>\(\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2005^2}\right)<2\cdot\frac14\)

=>A<1/2(ĐPCM)

a: \(\frac{23}{55}=1-\frac{32}{55};\frac{1978}{2010}=1-\frac{32}{2010}\)

Ta có: 55<2010

=>\(\frac{32}{55}>\frac{32}{2010}\)

=>\(-\frac{32}{55}<-\frac{32}{2010}\)

=>\(-\frac{32}{55}+1<-\frac{32}{2010}+1\)

=>\(\frac{23}{55}<\frac{1978}{2010}\)

b: \(\frac{2003\cdot2004-1}{2003\cdot2004}=1-\frac{1}{2003\cdot2004}\)

\(\frac{2004\cdot2005-1}{2004\cdot2005}=1-\frac{1}{2004\cdot2005}\)

Ta có: \(2003\cdot2004<2004\cdot2005\)

=>\(\frac{1}{2003\cdot2004}>\frac{1}{2004\cdot2005}\)

=>\(\frac{-1}{2003\cdot2004}<\frac{-1}{2004\cdot2005}\)

=>\(\frac{-1}{2003\cdot2004}+1<\frac{-1}{2004\cdot2005}+1\)

=>\(\frac{2003\cdot2004-1}{2003\cdot2004}<\frac{2004\cdot2005-1}{2004\cdot2005}\)

c: \(\frac{a}{b}-\frac{a+m}{b+m}=\frac{a\left(b+m\right)-b\left(a+m\right)}{b\left(b+m\right)}=\frac{ab+am-ab-bm}{b\left(b+m\right)}\)

\(=\frac{m\left(a-b\right)}{b\left(b+m\right)}>0\)

=>\(\frac{a}{b}>\frac{a+m}{b+m}\)

d: a<b

=>a-b<0

\(\frac{a}{b}-\frac{a+m}{b+m}=\frac{a\left(b+m\right)-b\left(a+m\right)}{b\left(b+m\right)}=\frac{ab+am-ab-bm}{b\left(b+m\right)}\)

\(=\frac{m\left(a-b\right)}{b\left(b+m\right)}<0\)

=>\(\frac{a}{b}<\frac{a+m}{b+m}\)

Bài 1:

a: A(x)=M(x)-N(x)

\(=-x^3+2,5x^2-0,5x-1\frac12+x^3-2,5x^2-2x+6=-2,5x+4,5\)

Đặt A(x)=0

=>-2,5x+4,5=0

=>-2,5x=-4,5

=>x=1,8

b: B(x)=M(x)+N(x)

\(=-x^3+2,5x^2-0,5x+1,5-x^3+2,5x^2+2x-6=-2x^3+5x^2+1,5x-4,5\)

Bậc là 3

Bài 3:

a: Đặt f(x)=0

=>\(x^2-4x+3=0\)

=>(x-1)(x-3)=0

=>x=1 hoặc x=3

b: Đặt f(x)=0

=>\(x^2-7x+12=0\)

=>(x-3)(x-4)=0

=>x=3 hoặc x=4

c: Đặt f(x)=0

=>\(x^2+2x+1=0\)

=>\(\left(x+1\right)^2=0\)

=>x+1=0

=>x=-1

d: \(x^4\ge0\forall x\)

=>\(x^4+2\ge2>0\forall x\)

=>f(x) vô nghiệm