Ai giúp mình với ạh

a) \(\left(x-3\right)\)\(\left(x^2+3x+9\right)\)+\(x\left(x+2\right)\left(x-2\right)\) =1

\(\Leftrightarrow x^3\)\(-27\)+\(x\left(x^2-4\right)\) =1

\(\Leftrightarrow\)\(x^3\)\(-27\)\(+x^3\)\(-4x\) =1

\(\Leftrightarrow\)\(2x^3\)\(-4x-27\) = 1

Suy ra \(x\) =2,685673906

Bài 1:

a. A = x^2 - 5x - 1

\(=x^2-5x+\frac{25}{4}-\frac{29}{4}\)

\(=x^2-5x+\left(\frac{5}{2}\right)^2-\frac{29}{4}\)

\(=\left(x-\frac{5}{2}\right)^2-\frac{29}{4}\ge0-\frac{29}{4}=-\frac{29}{4}\)

Dấu = khi x=5/2

Vậy MinC=-29/4 khi x=5/2

2. Tìm x:
a. ( 2x - 3 )^2 - ( 4x + 1 )( 4x - 1 ) = ( 2x - 1 ).( 3 - 7x )

=>4x²-12x+9+1-16x²=-14x²+13x-3

=>-12x²-12x+10=-14x²+13x-3

=>2x²-25x+13=0

\(\Rightarrow2\left(x-\frac{25}{4}\right)^2-\frac{521}{8}=0\)

\(\Rightarrow\left(x-\frac{25}{4}\right)^2=\frac{521}{16}\)

\(\Rightarrow x-\frac{25}{4}=\pm\sqrt{\frac{521}{16}}\)

\(\Rightarrow x=\frac{25}{4}\pm\frac{\sqrt{521}}{4}\)

c. 4.( x - 3 ) - ( x + 2 ) = 0

=>4x-12-x-2=0

=>3x-14=0

=>3x=14

=>x=14/3

\(\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge\frac{2}{1+xy}\)

\(\frac{1+y^2+1+x^2}{\left(1+x^2\right)\left(1+y^2\right)}\ge\frac{2}{1+xy}\\ \frac{2+x^2+y^2}{\left(1+x^2\right)\left(1+y^2\right)}\ge\frac{2}{1+xy}\)

=>\(\left(2+x^2+y^2\right)\left(1+xy\right)\ge2\left(1+x^2\right)\left(1+y^2\right)\)

\(\left(2+x^2+y^2\right)+\left(2+x^2+y^2\right)xy\ge2\left(1+x^2\right)\left(1+y^2\right)\)

\(2+x^2+y^2+2xy+x^3y+y^3x\ge\left(2+2x^2\right)\left(1+y^2\right)\)

\(2+x^2+y^2+2xy+x^3y+y^3x\ge2+2x^2+\left(2+2x^2\right)y^2\)

\(2+x^2+y^2+2xy+x^3y+y^3x\ge2+2x^2+2y^2+2x^2y^2\)

\(2xy+x^3y+y^3x\ge x^2+y^2+2x^2y^2\)

\(2xy+x^3y+y^3x-x^2-y^2-2x^2y^2\ge0\)

\(x^3y-x^2+y^3x-y^2+2xy-2x^2y^2\ge0\)

\(x^2\left(xy-1\right)+y^2\left(xy-1\right)-2xy\left(xy-1\right)\)\(\ge0\)

\(\left(xy-1\right)\left(x^2-2xy+y^2\right)\ge0\)

\(\left(xy-1\right)\left(x-y\right)^2\ge0\)

\(Do\begin{cases}x,y\ge1=>xy\ge1=>xy-1\ge0\\\left(x-y\right)^2\ge0\end{cases}\)

\(=>\left(xy-1\right)\left(x-y\right)^2\ge0\left(dpcm\right)\)

ĐK: a > 0, a khác 1

\(M=\dfrac{a-1}{\sqrt{a}-1}=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}-1}=\sqrt{a}+1\)

\(N=\dfrac{a-1}{\sqrt{a}+1}=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}+1}=\sqrt{a}-1\)

\(P=\dfrac{a\sqrt{a}-1}{\sqrt{a}-1}=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}-1}=a+\sqrt{a}+1\)

\(Q=\dfrac{a\sqrt{a}+1}{\sqrt{a}+1}=\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}+1}=a-\sqrt{a}+1\)

a)Đặt \(T=\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}=1\) (*)

Từ \(abc=1\Rightarrow c=\frac{1}{ab}\).Thay vào (*) ta có:

\(T=\frac{1}{1+a+ab}+\frac{1}{1+b+\frac{1}{a}}+\frac{1}{1+\frac{1}{ab}+\frac{1}{b}}\)

\(=\frac{1}{1+a+ab}+\frac{1}{\frac{a+ab+1}{a}}+\frac{1}{\frac{ab+1+a}{ab}}\)

\(=\frac{1}{a+ab+1}+\frac{a}{a+ab+1}+\frac{ab}{a+ab+1}\)

\(=\frac{a+ab+1}{a+ab+1}=1=VP\) (Đpcm)

b)Áp dụng Bđt Cô-si ta có:

\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\sqrt{\frac{a^2}{b^2}\cdot\frac{b^2}{c^2}}=\frac{2a}{c}\)

\(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge2\sqrt{\frac{b^2}{c^2}\cdot\frac{c^2}{a^2}}=\frac{2b}{a}\)

\(\frac{a^2}{b^2}+\frac{c^2}{a^2}\ge2\sqrt{\frac{a^2}{b^2}\cdot\frac{c^2}{a^2}}=\frac{2c}{b}\)

Cộng theo vế ta có:

\(\frac{2a^2}{b^2}+\frac{2b^2}{c^2}+\frac{2c^2}{a^2}\ge\frac{2a}{c}+\frac{2b}{a}+\frac{2c}{b}\)

\(\Leftrightarrow2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge2\left(\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\right)\)

\(\Leftrightarrow\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\) (Đpcm)

Dấu = khi a=b=c

Bài 1:

\(6x^2-2\left(x-y\right)^2-6y^2\)

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

\(=2\left(x-y\right)\left(3x+3-x+y\right)\)

\(=2\left(x-y\right)\left(2x+3+y\right)\)

Bài 2:

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

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

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

b) Thay \(x=\frac{9}{4}\)vào (1) ta được: 

\(\left(2.\frac{9}{4}-2\right)^2\)

\(=\frac{25}{4}\)

Vậy giá trị của P \(=\frac{25}{4}\)khi \(x=\frac{9}{4}\)

Bài 3:

Ta có: \(M=x^2+4x+5\)

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

Vì \(\left(x+2\right)^2\ge0;\forall x\)

\(\Rightarrow\left(x+2\right)^2+1\ge0+1;\forall x\)

Hay \(M\ge1;\forall x\)

Dấu"="xảy ra \(\Leftrightarrow\left(x+2\right)^2=0\)

                       \(\Leftrightarrow x=-2\)

Vậy \(M_{min}=1\Leftrightarrow x=-2\)

Bài 1 : trên là sai nha mình làm lại

\(6x^2-2\left(x-y\right)^2-6y^2\)

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

\(=2\left(x-y\right)\left(3x+3y-x+y\right)\)

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

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

làm nốt

d) (2x-1)(3x+2)(3-x)

=(6x²+x-2)(3-x)

=-6x³+17x²+5x-6

e) (x+3)(x²+3x-5)

=x³+6x²+4x-15

f) (xy-2)(x³-2x-6)

=x⁴y-2x³-2x²y-6xy+4x+12

g) (5x³-x2+2x-3)(4x²-x+2)

=20x⁵-9x⁴+19x³-16x²+7x-6

Bài 1:

a) (x-2)(x2+3x+4)

=x(5x+4)-2(5x+4)

= 5x²+4x-10x-8

=5x²-6x-8