\(\dfrac{-3}{x^2-5x+1}\)
K
Khách

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14 tháng 7 2017

1) \(\left(x-3\right)\left(x-5\right)+44\)

\(=x^2-3x-5x+15+44\)

\(=x^2-8x+59\)

\(=x^2-2.x.4+4^2+43\)

\(=\left(x-4\right)^2+43\ge43>0\)

\(\rightarrowĐPCM.\)

2) \(x^2+y^2-8x+4y+31\)

\(=\left(x^2-8x\right)+\left(y^2+4y\right)+31\)

\(=\left(x^2-2.x.4+4^2\right)-16+\left(y^2+2.y.2+2^2\right)-4+31\)

\(=\left(x-4\right)^2+\left(y+2\right)^2+11\ge11>0\)

\(\rightarrowĐPCM.\)

3)\(16x^2+6x+25\)

\(=16\left(x^2+\dfrac{3}{8}x+\dfrac{25}{16}\right)\)

\(=16\left(x^2+2.x.\dfrac{3}{16}+\dfrac{9}{256}-\dfrac{9}{256}+\dfrac{25}{16}\right)\)

\(=16\left[\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{256}\right]\)

\(=16\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{16}>0\)

-> ĐPCM.

4) Tương tự câu 3)

5) \(x^2+\dfrac{2}{3}x+\dfrac{1}{2}\)

\(=x^2+2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{2}\)

\(=\left(x+\dfrac{1}{3}\right)^2+\dfrac{7}{18}>0\)

-> ĐPCM.

6) Tương tự câu 5)

7) 8) 9) Tương tự câu 3).

15 tháng 7 2017

Giải rõ giúp mình với

10 tháng 8 2017

a) \(\dfrac{2x+3}{x-5}=\dfrac{2\left(x-5\right)+13}{x-5}=2+\dfrac{13}{x-5}\)

Để \(2+\dfrac{13}{x-5}\in Z\)

thì \(\dfrac{13}{x-5}\in Z\Rightarrow13⋮x-5\)

\(\Rightarrow x-5\inƯ\left(13\right)\)

\(\Rightarrow x-5\in\left\{\pm1;\pm13\right\}\)

Xét các trường hợp...

b) \(\dfrac{x^3-x^2+2}{x-1}=\dfrac{x^2\left(x-1\right)+2}{x-1}=x^2+\dfrac{2}{x-1}\)

Tương tự câu a)

c) \(\dfrac{x^3-2x^2+4}{x-2}=\dfrac{x^2\left(x-2\right)+4}{x-2}=x^2+\dfrac{4}{x-2}\)

...

d) \(\dfrac{2x^3+x^2+2x+2}{2x+1}=\dfrac{x^2\left(2x+1\right)+2x+2}{2x+1}=x^2+\dfrac{2x+2}{2x+1}\)

Khi đó lí luận cho \(2x+2⋮2x+1\)

\(\Rightarrow\left(2x+1\right)+1⋮2x+1\)

\(\Rightarrow1⋮2x+1\)

\(\Rightarrow2x+1\inƯ\left(1\right)\)

...

e) \(\dfrac{3x^3-7x^2+11x-1}{3x-1}=\dfrac{x^2\left(3x-1\right)-2x\left(3x-1\right)+3\left(3x-1\right)+2}{3x-1}\)

\(=\dfrac{\left(x^2-2x+3\right)\left(3x-1\right)+2}{3x-1}=\left(x^2-2x+3\right)+\dfrac{2}{3x-1}\)

...

f) \(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}=\dfrac{\left(x^2\right)^2-4^2}{\left(x-2\right)^2\left(x^2+4\right)}\)

\(=\dfrac{\left(x^2-4\right)\left(x^2+4\right)}{\left(x-2\right)^2\left(x^2+4\right)}=\dfrac{x^2-4}{\left(x-2\right)^2}=\dfrac{x+2}{x-2}=\dfrac{\left(x-2\right)+4}{x-2}=1+\dfrac{4}{x-2}\)

....

10 tháng 8 2017

thank you

15 tháng 7 2016

a) \(A=-x^2+4x+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\ge7\)

Dấu "=" xảy ra khi và chỉ khi x = 2

Vậy Max A = 7 <=> x = 2

b) \(B=-x^2+x=-\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Dấu "=" xảy ra khi và chỉ khi x = \(\frac{1}{2}\)

Vậy Max B = \(\frac{1}{4}\Leftrightarrow x=\frac{1}{2}\)

c) \(C=-2x^2+2x-5=-2\left(x^2-x\right)-5=-2\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{2}-5\)

\(=-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\le-\frac{9}{2}\)

Dấu "=" xảy ra khi và chỉ khi x = \(\frac{1}{2}\)

Vậy Max C = \(-\frac{9}{2}\Leftrightarrow x=\frac{1}{2}\)

24 tháng 6 2017

\(a,A=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\) Vậy \(Max_A=7\) khi \(x-2=0\Rightarrow x=2\)

\(b,x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)Vậy \(Max_B=\dfrac{1}{4}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)

\(c,2x-2x^2+5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-\left(x-\dfrac{1}{2}\right)-\dfrac{9}{2}\le\dfrac{-9}{2}\)Vậy \(Max_C=\dfrac{-9}{2}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)

28 tháng 3 2018

a) \(ĐKXĐ:x\ne\pm3;x\ne-6\)

Với \(x\ne\pm3;x\ne-6\), ta có:

\(P=\left(\dfrac{x}{x-3}-\dfrac{2}{x+3}+\dfrac{x^2}{9-x^2}\right):\dfrac{x+6}{3x+9}\\ =\left(\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2}{\left(x+3\right)\left(x-3\right)}\right)\cdot\dfrac{3\left(x+3\right)}{x+6}\\ =\dfrac{x^2+3x-2x+6-x^2}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{3\left(x+3\right)}{x+6}\\ =\dfrac{x+6}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{3\left(x+3\right)}{x+6}\\ =\dfrac{3}{x-3}\)

Vậy \(P=\dfrac{3}{x-3}\) với \(x\ne\pm3;x\ne-6\)

b) Ta có: \(2x-\left|4-x\right|=5\)

+) Nếu \(x\le4\Leftrightarrow2x-\left(4-x\right)=5\)

\(\Leftrightarrow2x-4+x=5\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\left(Tm\right)\)

+) Nếu \(x>4\Leftrightarrow2x-\left(x-4\right)=5\)

\(\Leftrightarrow2x-x+4=5\\ \Leftrightarrow x=1\left(Ktm\right)\)

Với \(x\ne\pm3;x\ne-6\)

Khi \(x=3\left(Ktm\right)\rightarrow\text{loại}\)

Vậy khi \(2x-\left|4-x\right|=5\) không có giá trị.

28 tháng 3 2018

c) Với \(x\ne\pm3;x\ne-6\)

Để P nhận giá trị nguyên

thì \(\Rightarrow\dfrac{3}{x-3}\in Z\)

\(\Rightarrow3⋮x-3\\ \Rightarrow x-3\inƯ_{\left(3\right)}\)

\(Ư_{\left(3\right)}=\left\{\pm1;\pm3\right\}\)

Lập bảng giá trị:

\(x-3\) \(-3\) \(-1\) \(1\) \(3\)
\(x\) \(0\left(TM\right)\) \(2\left(TM\right)\) \(4\left(TM\right)\) \(6\left(KTM\right)\)

Vậy để P nhận giá trị nguyên

thì \(x\in\left\{0;2;4\right\}\)

d) Với \(x\ne\pm3;x\ne-6\)

Ta có : \(P^2-P+1=\dfrac{9}{\left(x-3\right)^2}-\dfrac{3}{x-3}+1\)

Đặt \(\dfrac{3}{x-3}=y\)

\(\Rightarrow P^2-P+1=y^2-y+1\\ =y^2-y+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(y^2-y+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Do \(\left(y-\dfrac{1}{2}\right)^2\ge0\forall y\)

\(\Rightarrow P^2-P+1=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall y\)

Dấu "=" xảy ra khi:

\(\left(y-\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow y-\dfrac{1}{2}=0\\ \Leftrightarrow y=\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{x-3}=\dfrac{1}{2}\\ \Leftrightarrow x-3=6\\ \Leftrightarrow x=9\left(TM\right)\)

Vậy \(GTNN\) của biểu thức là \(\dfrac{3}{4}\) khi \(x=9\)

13 tháng 8 2018

giải bất phương trình

a: =>-4x>16

=>x<-4

c: =>20x-25<=21-3x

=>23x<=46

=>x<=2

d: =>20(2x-5)-30(3x-1)<12(3-x)-15(2x-1)

=>40x-100-90x+30<36-12x-30x+15

=>-50x-70<-42x+51

=>-8x<121

=>x>-121/8

Bài 1: 

a: \(A=\dfrac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)

\(=\dfrac{\left(x+1\right)\left(x^3+1\right)}{\left(x^2-x+1\right)\left(x^2+1\right)}=\dfrac{\left(x+1\right)^2}{x^2+1}\)

Để A=0 thì x+1=0

hay x=-1

b: \(B=\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}=\dfrac{x^2-4}{x^2-9}\)

Để B=0 thi (x-2)(x+2)=0

=>x=2 hoặc x=-2

30 tháng 7 2018

\(A=\dfrac{1}{-x^2+2x-2}\)

A min \(\Leftrightarrow\dfrac{1}{A}\)max

ta có \(\dfrac{1}{A}=-x^2+2x-2=-\left(x^2-2x+2\right)=-\left(x-1\right)^2-1\le-1\)

\(\dfrac{1}{A}\)max= -1 tại x=1

=> A min = -1 tại x=1

\(B=\dfrac{2}{-4x^2+8x-5}\) ( phải là -4x2 nha bn)

B min \(\Leftrightarrow\dfrac{1}{B}\) max

ta có \(\dfrac{1}{B}=\dfrac{-4x^2+8x-5}{2}=\dfrac{-\left(4x^2-8x+5\right)}{2}=\dfrac{-\left(2x-4\right)^2+11}{2}=\dfrac{\left(-2x-4\right)^2}{2}+\dfrac{11}{2}\le\dfrac{11}{2}\)

\(\dfrac{1}{B}\)max=\(\dfrac{11}{2}\) tại x=2

\(\Rightarrow B\) min = \(\dfrac{1}{\dfrac{11}{2}}=\dfrac{2}{11}\) tại x=2

\(A=\dfrac{3}{2x^2+2x+3}=\dfrac{3}{2\left(x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{5}{2}}=\dfrac{3}{2\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{2}}\)

A max \(\Leftrightarrow\dfrac{1}{A}\) min

\(\Leftrightarrow\dfrac{2\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{2}}{3}=\dfrac{2\left(x+\dfrac{1}{2}\right)^2}{3}+\dfrac{\dfrac{5}{2}}{3}=\dfrac{2\left(x+\dfrac{1}{2}\right)^2}{3}+\dfrac{5}{6}\ge\dfrac{5}{6}\)

\(\dfrac{1}{A}\) min = \(\dfrac{5}{6}\)tại x= \(-\dfrac{1}{2}\)

\(\Rightarrow A\)max = \(\dfrac{6}{5}\) tại x= \(-\dfrac{1}{2}\)

B\(=\dfrac{5}{3x^2+4x+15}=\dfrac{5}{3.\left(x^2+\dfrac{4}{3}x+5\right)}=\dfrac{5}{3\left(x^2+2.x.\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{41}{9}\right)}=\dfrac{5}{3\left(x+\dfrac{2}{3}\right)^2+\dfrac{41}{3}}\)

B max \(\Leftrightarrow\dfrac{1}{B}\) min

\(\Leftrightarrow\dfrac{3\left(x+\dfrac{2}{3}\right)^2+\dfrac{41}{3}}{5}=\dfrac{3\left(x+\dfrac{2}{3}\right)^2}{5}+\dfrac{41}{15}\ge\dfrac{41}{15}\)

\(\dfrac{1}{B}\) min = \(\dfrac{41}{15}\) tại x=\(-\dfrac{2}{3}\)

=> \(B\) max = \(\dfrac{15}{41}\) tại x=\(-\dfrac{2}{3}\)

Đây chỉ là gợi ý !! bn pải tự lí luận nha

tik thanghoa

23 tháng 2 2019

a) Đk : \(x\ne0;\ne1\)

\(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=\dfrac{2\left(x^2+x-1\right)}{x\left(x+1\right)}\)

\(\Rightarrow\dfrac{x^2+3x}{x\left(x+1\right)}+\dfrac{x^2-x-2}{x\left(x+1\right)}-\dfrac{2x^2+2x-2}{x\left(x+1\right)}=0\)

\(\Rightarrow\dfrac{x^2+3x+x^2-x-2-2x^2-2x+2}{x\left(x-1\right)}=0\)

\(\Rightarrow\dfrac{0}{x-1}=0\)

=> Phương trình có vô số nghiệm x

b) Đk : \(x\ne2;x\ne3\)

\(\dfrac{2}{x-2}-\dfrac{x}{x+3}=\dfrac{5x}{\left(x-2\right)\left(x+3\right)}-1\)

\(\Rightarrow\dfrac{2x+6}{\left(x-2\right)\left(x+3\right)}-\dfrac{x^2-2x}{\left(x-2\right)\left(x+3\right)}-\dfrac{5x}{\left(x-2\right)\left(x+3\right)}+\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}\)

=0

\(\Rightarrow\dfrac{2x+6-x^2+2x-5x+x^2+x+6}{\left(x-2\right)\left(x+3\right)}=0\)

\(\Rightarrow\dfrac{12}{\left(x-2\right)\left(x+3\right)}=0\)

=> Phương trình vô nghiệm

c)

\(\Leftrightarrow\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+x+1}{x^4+x^2+1}-\dfrac{1-2x}{x^4+x^2+1}=0\)

\(\Rightarrow\dfrac{x^2-x+1-x^2-x-1-1+2x}{x^4+x^2+1}=0\)

\(\Rightarrow\dfrac{-1}{x^4+x^2+1}=0\)

=> PTVN

d) Thôi tự làm đi, câu này dễ :Vvv

e)

\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)\)=40

\(\Rightarrow\left[\left(x+1\right)\left(x+5\right)\right]\cdot\left[\left(x+2\right)\left(x+4\right)\right]=40\)

\(\Rightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)

Đặt

\(x^2+6x+7=t\)

Phương trình tương đương

\(\left(t-1\right)\left(t+1\right)=40\)

\(t^2=41\)

\(\)\(t=\pm\sqrt{41}\)

Thay vào tìm x.

24 tháng 2 2019

Thanks ;)