Mn giúp mk vs, mk cần gấp á

Thanks!

Giải phương trình: x^4-5x^3+6x^2-5x+1=0

x=2-căn bậc hai(3),

x=căn bậc hai(3)+2;

x = -(căn bậc hai(3)*i-1)/2;

x = (căn bậc hai(3)*i+1)/2;

Trả lời:

2ab⁴ - 8ab³ + 8ab² 

= 2ab² ( b² - 4b + 4 )

= 2ab² ( b - 2 )² 

\(\frac{1}{a+2b+3c}+\frac{1}{2a+3b+c}+\frac{1}{3a+b+2c}\)

\(\le\frac{1}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{b+c}+\frac{1}{a+b}+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{a+c}\right)\)

\(\le\frac{1}{36}\left(\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{c}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{a}+\frac{1}{c}\right)\)

\(=\frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(=\frac{1}{6}\left(\frac{a+b+c}{abc}\right)=\frac{1}{6}\)

Dấu \(=\)khi \(a=b=c=3\).

bài này áp dụng bđt phụ \(\frac{1}{x_1+x_2+x_3+x_4+x_5+x_6}\le\frac{1}{36}\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}+\frac{1}{x_6}\right)\)

là oke nhé =)) 

về cách chứng minh thì bạn dùng svacxo cho vp là xong 

A = ( 4x + 1 )² + 2 ≥ 2 > 0 ∀ x ( đpcm )

B = ( y - 5/2 )² + 7/4 ≥ 7/4 > 0 ∀ x ( đpcm )

C = 2( x - 1/2 )² + 3/2 ≥ 3/2 > 0 ∀ x ( đpcm )

D = ( 3x - 1 )² + ( 5y + 1 )² + 2 ≥ 2 > 0 ∀ x, y ( đpcm )

Trả lời:

a, \(A=16x^2+8x+3=\left(16x^2+8x+1\right)+2=\left(4x+1\right)^2+2\ge2>0\forall x\) 

Dấu "=" xảy ra khi x = - 1/4

Vậy bt A luôn dương với mọi x.

b, \(B=y^2-5y+8=x^2-2.y.\frac{5}{2}+\frac{25}{4}+\frac{7}{4}=\left(x-\frac{5}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}>0\forall y\)

Dấu "=" xảy ra khi x = 5/2

Vậy bt B luôn dương với mọi y.

c, 

\(C=2x^2-2x+2=2\left(x^2-x+1\right)=2\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)\)

\(=2\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]=2\left(x-\frac{1}{2}\right)^2+\frac{3}{2}\ge\frac{3}{2}>0\forall x\)

Dấu "=" xảy ra khi x = 1/2

Vậy bt C luôn dương với mọi x.

d, \(D=9x^2-6x+25y^2+10y+4\)

\(=9x^2-6x+25y^2+10y+1+1+2\)

\(=\left(9x^2-6x+1\right)+\left(25y^2+10y+1\right)+2\)

\(=\left(3x-1\right)^2+\left(5y+1\right)^2+2\ge2>0\forall x;y\)

Dấu "=" xảy ra khi x = 1/3; y = - 1/5

Vậy bt D luôn dương với mọi x;y

Trả lời:

\(A=\left(1-\frac{x^2-x}{x-1}\right)\left(1+\frac{x^2+x}{x+1}\right)+x^2\) \(\left(ĐKXĐ:x\ne\pm1\right)\)

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

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

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

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

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

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

\(B=\left(2-\frac{x^2-x}{x-1}\right)\left(2+\frac{x^2+x}{x+1}\right)\) \(\left(ĐKXĐ:x\ne\pm1\right)\)

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

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

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

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

\(=\frac{-\left[x\left(x-1\right)-2\left(x-1\right)\right]}{x-1}.\frac{x\left(x+1\right)+2\left(x+1\right)}{x+1}\)

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

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

\(a.\left(x^2+2x+x+2\right)\left(x^2+5x+6x+30\right)-5\)

\(=\left(x+1\right)\left(x+2\right)\left(x+5\right)\left(x+6\right)-5=\left(x^2+7x+6\right)\left(x^2+7x+10\right)\)

Đặt \(x^2+7x+8=a\Rightarrow\text{Biểu thức }=\left(a-2\right)\left(a+2\right)-5=a^2-9=\left(a-3\right)\left(a+3\right)\)

nên : \(BT=\left(x^2+7x+5\right)\left(x^2+7x+11\right)\)

b.\(BT=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\)

Đặt \(x^2+5ax+5a^2=y\Rightarrow BT=\left(y-a^2\right)\left(y+a^2\right)+a^4=y^2=\left(x^2+5ax+5a^2\right)^2\)

a) \(A=4x^2+3y^2-8x+11=\left(4x^2-8x+4\right)+3y^2+7\)

\(A=4\left(x-1\right)^2+3y^2+7\ge7\Rightarrow minA=7\)

b) \(B=\left(x-2y-4\right)^2+4xy+34=x^2+4y^2-8x+16y+34+16\)

\(B=\left(x^2-8x+16\right)+\left(4y^2+16y+16\right)+18\)

\(B=\left(x-4\right)^2+4\left(y+2\right)^2+18\ge18\Rightarrow minB=18\)

c) \(C=9x^2+y^2-6x+4y-24=\left(9x^2-6x+1\right)+\left(y^2+4y+4\right)-29\)

\(\left(3x-1\right)^2+\left(y+2\right)^2-29\ge-29\Rightarrow minC=-29\)

d) \(D=\left(2x-5y\right)^2+20xy-8x+10y-121\)

\(D=4x^2-20xy+25y^2+20xy-8x+10y-121\)

\(D=\left(4x^2-8x+4\right)+\left(25y^2+10y+1\right)-126\)

\(D=4\left(x-1\right)^2+\left(5y+1\right)^2-126\ge-126\Rightarrow minD=-126\)

Trả lời:

a, \(2x-1\ge1\)

\(\Leftrightarrow2x\ge2\)

\(\Leftrightarrow x\ge1\)

Vậy \(x\ge1\) là nghiệm của pt.

0 x 1

b, \(3x-2\ge1\)

\(\Leftrightarrow3x\ge3\)

\(\Leftrightarrow x\ge1\)

Vậy \(x\ge1\) là nghiệm của pt.

( biểu diễn giống ý trên )

c, \(2-2x< 3\)

\(\Leftrightarrow-2x< 1\)

\(\Leftrightarrow x>-\frac{1}{2}\)

Vậy \(x>-\frac{1}{2}\) là nghiệm của pt.

0 x -1/2

d, \(4-3x< 5\)

\(\Leftrightarrow-3x< 1\)

\(\Leftrightarrow x>-\frac{1}{3}\)

Vậy \(x>-\frac{1}{3}\) là nghiệm của pt.

0 x -1/3

\(a,VT=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)

\(VT+3=\frac{a}{b+c}+1+\frac{b}{a+c}+1+\frac{c}{a+b}\)

\(VT+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)

\(VT+3=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

\(2\left(VT+3\right)=\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

áp dụng bđt Cô si :

\(\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)

\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge3\sqrt[3]{\frac{1}{a+b}\cdot\frac{1}{b+c}\cdot\frac{1}{c+a}}\)

\(\Rightarrow2\left(VT+3\right)\ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\cdot3\sqrt[3]{\frac{1}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

\(\Rightarrow2\left(VT+3\right)\ge9\) \(\Leftrightarrow VT\ge\frac{3}{2}\)

chưa có dấu = xảy ra kìa, khi a=b=c > 0

\(b,\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}=\frac{a^2}{a\left(b+2c\right)}+\frac{b^2}{b\left(c+2a\right)}+\frac{c^2}{c\left(a+2b\right)}\)

\(\Rightarrow vt\ge\frac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\)

có : \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca\ge3\left(ab+bc+ca\right)\)

\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

nên \(vt\ge\frac{3\left(ab+bc+ca\right)}{3\left(ab+cb+ca\right)}=1\)

dấu = xảy ra khi a=b=c