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

\(1^2+2^2+\cdots+98^2\)

\(=\frac{98\left(98+1\right)\left(2\cdot98+1\right)}{6}\)

\(=\frac{98\cdot99\cdot197}{6}=49\cdot33\cdot197=318549\)

a: Sửa đề: Chiều dài là 57m

Chiều rộng là \(57\times\frac23=38\left(m\right)\)

Diện tích mảnh đất là: \(57\times38=2166\left(m^2\right)\)

Độ dài bán kính bể là 6:2=3(m)

Diện tích bể hình tròn là: \(3\times3\times3,14=28,26\left(m^2\right)\)

Diện tích phần đất còn lại là:

2166-28,26=2137,74\(\left(m^2\right)\)

b: Khối lượng khoai người ta thu hoạch được là:

2137,74:100x35=748,209(kg)

BÀi 4:

1: \(x\left(1-x\right)+\left(x-1\right)^2\)

\(=x-x^2+x^2-2x+1\)

=-x+1

2: \(\left(x-3\right)^2-x^2+10x-7\)

\(=x^2-6x+9-x^2+10x-7\)

=4x+2

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

\(=x^2+4x+4-\left(x^2-2x-3\right)\)

\(=x^2+4x+4-x^2+2x+3=6x+7\)

4: (x+4)(x-2)-\(\left(x-3\right)^2\)

\(=x^2-2x+4x-8-\left(x^2-6x+9\right)\)

\(=x^2+2x-8-x^2+6x-9=8x-17\)

5: \(\left(x-2\right)^2+\left(x-1\right)\left(x+5\right)\)

\(=x^2-4x+4+x^2+4x-5=2x^2-1\)

6: \(\left(x+3\right)\left(x-3\right)-x\left(x+23\right)\)

\(=x^2-9-x^2-23x\)

=-23x-9

Bài 3:

1: \(\left(2x+1\right)^2+\left(2x-1\right)^2\)

\(=4x^2-4x+1+4x^2+4x+1=8x^2+2\)

2: \(-\left(x+1\right)^2-\left(x-1\right)^2\)

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

\(=-x^2-2x-1-x^2+2x-1=-2x^2-2\)

3: \(\left(x+2y\right)^2-\left(x-2y\right)^2\)

\(=\left(x+2y-x+2y\right)\left(x+2y+x-2y\right)\)

\(=4y\cdot2x=8xy\)

4: \(\left(3x+y\right)^2+\left(x-y\right)^2\)

\(=9x^2+6xy+y^2+x^2-2xy+y^2=10x^2+4xy+2y^2\)

5: \(-\left(x+5\right)^2-\left(x-3\right)^2\)

\(=-\left(x^2+10x+25\right)-\left(x^2-6x+9\right)\)

\(=-x^2-10x-25-x^2+6x-9=-2x^2-4x-34\)

6: \(\left(3x-2\right)^2-\left(3x-1\right)^2\)

\(=\left(3x-2-3x+1\right)\left(3x-2+3x-1\right)=-1\left(6x-3\right)=-6x+3\)

7: \(\left(x-4y\right)^2+\left(x+4y\right)^2\)

\(=x^2-8xy+16y^2+x^2+8xy+16y^2=2x^2+32y^2\)

8: \(-\left(-2x+3\right)^2-\left(5x-3\right)^2\)

\(=-\left(4x^2-12x+9\right)-\left(25x^2-30x+9\right)\)

\(=-4x^2+12x-9-25x^2+30x-9=-29x^2+42x-18\)

9: \(\left(-2x+3\right)^2-\left(5x-3\right)^2\)

\(=4x^2-12x+9-\left(25x^2-30x+9\right)\)

\(=4x_{}^2-12x+9-25x^2+30x-9=-21x^2+18x\)

10: \(\left(2x+1\right)^2+\left(-3x-1\right)^2\)

\(=4x^2+4x+1+9x^2+6x+1=13x^2+10x+2\)

11: \(-\left(x-y\right)^2-\left(2x+y\right)^2\)

\(=-\left(x^2-2xy+y^2\right)-\left(4x^2+4xy+y^2\right)\)

\(=-x^2+2xy-y^2-4x^2-4xy-4y^2=-5x^2-2xy-5y^2\)

12: \(-\left(x+1\right)^2+\left(x-1\right)^2\)

\(=-x^2-2x-1+x^2-2x+1=-4x\)

13: \(\left(2x+7\right)^2+\left(-2x-3\right)^2\)

\(=4x^2+28x+49+4x_{}^2-12x+9=8x^2+16x+58\)

14: \(-\left(2x-y\right)^2-\left(x+3y\right)^2\)

\(=-\left(4x^2-4xy+y^2\right)-\left(x^2+6xy+9y^2\right)\)

\(=-4x^2+4xy-y^2-x^2-6xy-9y^2=-5x^2-2xy-10y^2\)

15: \(-\left(2x+7\right)^2+\left(-2x-3\right)^2\)

\(=-4x^2-28x-49+4x^2+12x+9=-16x-40\)

a: Xét ΔMNP có \(\hat{MNP}+\hat{MPN}+\hat{PMN}=180^0\)

=>\(\hat{PMN}=180^0-65^0-35^0=80^0\)

MQ là phân giác ngoài tại đỉnh M của ΔMNP

=>\(\hat{QMN}=\frac{180^0-\hat{NMP}}{2}=\frac{180^0-80^0}{2}=50^0\)

Ta có: \(\hat{MNQ}+\hat{MNP}=180^0\) (hai góc kề bù)

=>\(\hat{MNQ}=180^0-65^0=115^0\)

Xét ΔMNQ có \(\hat{NMQ}+\hat{NQM}+\hat{MNQ}=180^0\)

=>\(\hat{MQN}=180^0-115^0-50^0=180^0-165^0=15^0\)

=>\(\hat{MQP}=15^0\)

b: MQ là phân giác ngoài tại đỉnh M của ΔMNP

=>\(\hat{QMN}=\frac{180^0-\hat{NMP}}{2}=\frac{180^0-70^0}{2}=\frac{110^0}{2}=55^0\)

Xét ΔMNQ có \(\hat{MNP}\) là góc ngoài tại đỉnh N

nên \(\hat{MNP}=\hat{NMQ}+\hat{NQM}=55^0+25^0=80^0\)

Xét ΔMNP có \(\hat{MNP}+\hat{MPN}+\hat{NMP}=180^0\)

=>\(\hat{MPN}=180^0-80^0-70^0=30^0\)