với x=11

Bài làm:

Ta có: Tại x = 11 thì giá trị của B là

\(B=x\left(x^2-3x+3\right)=11\left(11^2-3.11+3\right)\)

\(=11.91=1001\)

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

=>       \(A=x^2-9+2\left(4x^2+4x+1\right)\)

=>       \(A=x^2-9+8x^2+8x+2\)

=>       \(A=9x^2+8x-7\)

=>       \(A=\left(3x+\frac{4}{3}\right)^2-\frac{79}{9}\)

Có:      \(\left(3x+\frac{4}{3}\right)^2\ge0\forall x\Rightarrow\left(3x+\frac{4}{3}\right)^2-\frac{79}{9}\ge-\frac{79}{9}\)

=>      \(A\ge-\frac{79}{9}\)

DẤU "=" XẢY RA <=>     \(\left(3x+\frac{4}{3}\right)^2=0\)

<=>     \(x=-\frac{4}{9}\)

Vậy A min =     \(-\frac{79}{9}\)      <=>       \(x=-\frac{4}{9}\)

( x - 3 )( x + 3 ) + 2( 2x + 1 )²

= x² - 9 + 2( 4x² + 4x + 1 )

= x² - 9 + 8x² + 8x + 2

= 9x² + 8x - 7

= 9x² + 8x + 16/9 - 79/9

= ( 3x + 4/3 )² - 79/9

\(\left(3x+\frac{4}{3}\right)^2\ge0\forall x\Rightarrow\left(3x+\frac{4}{3}\right)^2-\frac{79}{9}\ge-\frac{79}{9}\)

Dấu " = " xảy ra <=> 3x + 4/3 = 0 => x = -4/9

=> GTNN của biểu thức = -79/9 <=> x =  -4/9

\(\Leftrightarrow6x^2-14x+4-6x^2-12x+18-7x+3=0\)

\(\Leftrightarrow-33x=-25\Rightarrow x=\frac{25}{33}\)

2( 3x - 1 )( x - 2 ) - 6( x - 1 )( x + 3 ) = 7x - 3 

<=> 2( 3x² - 7x + 2 ) - 6( x² + 2x - 3 ) = 7x - 3

<=> 6x² - 14x + 4 - 6x² - 12x + 18 = 7x - 3

<=> -26x + 22 = 7x - 3

<=> -26x - 7x = -3 - 22

<=> -33x = -25

<=> x = 25/33

<=> -36x = 

\(B=-2x^2-3x+4=-2\left(x^2+\frac{3}{2}x+\frac{9}{16}\right)+\frac{41}{8}\)

\(\Rightarrow B=-2\left(x+\frac{3}{4}\right)^2+\frac{41}{8}\le\frac{41}{8}\)

\("="\Leftrightarrow x=-\frac{3}{4}\)

B = -2x² - 3x + 5

B = -2( x² + 3/2x + 9/16 ) + 49/8

B = -2( x + 3/4 )² + 49/8

\(-2\left(x+\frac{3}{4}\right)^2\le0\forall x\Rightarrow-2\left(x+\frac{3}{4}\right)^2+\frac{49}{8}\le\frac{49}{8}\)

Dấu " = " xảy ra <=> x + 3/4 = 0 => x = -3/4

=> MaxB = 49/8 <=> x = -3/4

6x(3x + 5) - 2x(3x - 2) + (17 - x)(x - 1) + x(x - 18) = 0

=> (18x² - 6x² - x² + x²) + (30x + 4x - 16x - 18x) - 17 = 0

=> 12x² - 17 = 0

=> 12x² = 17

=> x² = 17/12

=> \(\orbr{\begin{cases}x=\sqrt{\frac{17}{12}}\\x=-\sqrt{\frac{17}{12}}\end{cases}}\)

\(6x\left(3x+5\right)-2x\left(3x-2\right)+\left(17-x\right)\left(x-1\right)+x\left(x-18\right)=0\)

\(\Leftrightarrow9x^2+30x-6x^2+4x+17x-17-x^2+x+x^2-18x=0\)

\(\Leftrightarrow3x^2+34x-17=0\) ( vô nghiệm ) 

\(27x^3-9x^2+x-\frac{1}{27}=\left(3x\right)^3-3.3^2.\frac{1}{3}x^2+3.3.\left(\frac{1}{3}\right)^2x-\left(\frac{1}{3}\right)^2\)

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

thank bn

Bài làm:

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

\(=\left(x-3\right)\left(x^2-6x+9-x^2-3x-9\right)-2\)

\(=-9x\left(x-3\right)-2\)

\(=27x-9x^2-2\)

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

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

\(=\left(x-1\right).0=0\)

=> đpcm

3) \(\frac{68^3-52^3}{16}-68.52\)

\(=\frac{\left(68-52\right)\left(68^2+68.52+52^2\right)}{16}-68.52\)

\(=\frac{16\left(4624+68.52+2704\right)}{16}-68.52\)

\(=7328+68.52-68.52=7328\)

Bài làm:

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

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

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

\(A=\left(2x\right)^2\)

Với x = -1/3 ta được:

\(A=\left(2.\frac{-1}{3}\right)^2=\frac{4}{9}\)

A=(x+y)²+(x-y)²-2(x+y)(y-x)

A=(x+y)²+2(x+y)(x-y)+(x-y)²

A=(x+y+x-y)²

A=(2x)²

với x+-1/3 ta được:

A=(2.-1/3)²=4/9

\(x^4+2x^2-8x+5=0\)

\(\Leftrightarrow\left(x^3+x^2+3x-5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2+2x+5\right)\left(x-1\right)^2=0\)

\(\Leftrightarrow x=1\)

Bài làm:

\(x^4+2x^2-8x+5=0\)

\(\Leftrightarrow\left(x^4-x^3\right)+\left(x^3-x^2\right)+\left(3x^2-3x\right)-\left(5x-5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+3x-5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x^3-x^2\right)+\left(2x^2-2x\right)+\left(5x-5\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x^2+2x+5\right)=0\)

Mà \(x^2+2x+5=\left(x+1\right)^2+4>0\left(\forall x\right)\)

\(\Rightarrow\left(x-1\right)^2=0\Rightarrow x=1\)

\(x^3+4x^2-9=0\)

\(\Leftrightarrow\left(x^2+x-3\ne0\right)\left(x+3\right)=0\)

\(\Leftrightarrow x=-3\)

\(x^3+4x^2-9=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+x-3\right)=0\)

<=> x + 3 = 0 hoặc x^2 + x - 3 = 0

<=> x = - 3 hoặc x ( x + 1 ) = - 3

<=> x = 3 hoặc \(x=-\frac{1-\sqrt{13}}{2};x=-\frac{1+\sqrt{13}}{2}\)