\(x^3+9x+26=0\)

\(\Leftrightarrow x^3+2x^2-2x^2-4x+13x+26=0\)

\(\Leftrightarrow x^2\left(x+2\right)-2x\left(x+2\right)+13\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+13\right)=0\) (1)

Ta có: \(x^2-2x+13=\left(x-1\right)^2+12\) >0 với mọi x

Khi đó: \(\left(1\right)\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

Vậy pt đã cho có nghiệm x=-2

\(y=x^3-3x^2-9x+35\)

\(y'=3x^2-6x-9\)

\(y'=0\Leftrightarrow3x^2-6x-9=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

\(y\left(-4\right)=-41;y\left(-1\right)=40;y\left(3\right)=8;y\left(4\right)=52\)

\(\Rightarrow y_{max}=y\left(4\right)=52;y_{min}=y\left(-4\right)=-41\) trên đoạn \(\left[-4;4\right]\)

\(A=\dfrac{3x^2+9x+17}{3x^2+9x+7}=1+\dfrac{10}{3x^2+9x+7}=1+\dfrac{10}{3\left(x^2+2.x.\dfrac{9}{2}+\dfrac{81}{4}\right)-\dfrac{215}{4}}\\ =1+\dfrac{10}{3\left(x+\dfrac{9}{2}\right)^2-\dfrac{215}{4}}\le\dfrac{35}{43}\)

Câu khác giải TT

A=3(x^2+2/3x-1)

=3(x^2+2*x*1/3+1/9-10/9)

=3(x+1/3)^2-10/3>=-10/3

Dấu = xảy ra khi x=-1/3

\(B=1+\dfrac{15}{x^2+x+5}=1+\dfrac{15}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}}< =1+15:\dfrac{19}{4}=1+\dfrac{60}{19}=\dfrac{79}{19}\)

Dấu = xảy ra khi x=-1/2

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

\(maxN=-\dfrac{3}{4}\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(B=3x^2+4x-13=3\left(x^2+\dfrac{4}{3}x+\dfrac{4}{9}\right)-\dfrac{35}{3}=3\left(x+\dfrac{2}{3}\right)^2-\dfrac{35}{3}\ge-\dfrac{35}{3}\)

\(minB=-\dfrac{35}{3}\Leftrightarrow x=-\dfrac{2}{3}\)

a: Ta có: \(N=-x^2-x-1\)

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

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

\(=-\left(x+\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)

b: ta có: \(B=3x^2+4x-13\)

\(=3\left(x^2+\dfrac{4}{3}x-\dfrac{13}{3}\right)\)

\(=3\left(x^2+2\cdot x\cdot\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{43}{9}\right)\)

\(=3\left(x+\dfrac{2}{3}\right)^2-\dfrac{43}{3}\ge-\dfrac{43}{3}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{2}{3}\)

2. a. \(A=2x^2-8x-10=2\left(x^2-4x+4\right)-18\)

\(=2\left(x-2\right)^2-18\)

Vì \(\left(x-2\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-2\right)^2-18\ge-18\)

Dấu "=" xảy ra \(\Leftrightarrow2\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy minA = - 18 <=> x = 2

b. \(B=9x-3x^2=-3\left(x^2-3x+\frac{9}{4}\right)+\frac{27}{4}\)

\(=-3\left(x-\frac{3}{2}\right)^2+\frac{27}{4}\)

Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)\(\Rightarrow-3\left(x-\frac{3}{2}\right)^2+\frac{27}{4}\le\frac{27}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow-3\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)

Vậy maxB = 27/4 <=> x = 3/2

Sửa đề:x³-3x²-4x+12

a,x³-3x²-4x+12

=(x³-3x²)-(4x+12)

=x²(x-3)-4(x-3)

=(x²-4)(x-3)

b,x⁴- 5x² +4

x⁴-4x²-x²+4

(x⁴-x²)-(4x²+4)

x²(x²-1)-4(x²-1)

(x²-4)(x²-1)

Bài 1.

a) x³ - 3x² - 4x + 12 ( mạn phép sửa 13 thành 12, chứ để 13 là không phân tích được :> )

= x²( x - 3 ) - 4( x - 3 )

= ( x - 3 )( x² - 4 )

= ( x - 3 )( x - 2 )( x + 2 )

b) x⁴ - 5x² + 4

Đặt t = x²

Đa thức <=> t² - 5t + 4

= t² - t - 4t + 4

= t( t - 1 ) - 4( t - 1 )

= ( t - 1 )( t - 4 )

= ( x² - 1 )( x² - 4 )

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

c) ( x + y + z )³ - x³ - y³ - z³

= ( x + y + z )³ - ( x³ + y³ + z³ )

= ( x + y + z )³ - [ ( x + y + z )³ - 3( x + y )( y + z )( z + x ) ] ( chỗ này bạn xem HĐT tổng ba lập phương nhé )

= ( x + y + z )³ - ( x + y + z )³ + 3( x + y )( y + z )( z + x )

= 3( x + y )( y + z )( z + x )

d) 45 + x³ - 5x² - 9x 

= ( x³ - 5x² ) - ( 9x - 45 )

= x²( x - 5 ) - 9( x - 5 )

= ( x - 5 )( x² - 9 )

= ( x - 5 )( x - 3 )( x + 3 )

e) x⁴ - 2x³ + 3x² - 2x - 3 ( sửa -3x³ -> 3x² )

= x⁴ - x³ - x³ + 3x² - x² + x² - 3x + x - 3

= ( x⁴ - x³ + 3x² ) - ( x³ - x² + 3x ) - ( x² - x + 3 )

= x²( x² - x + 3 ) - x( x² - x + 3 ) - 1( x² - x + 3 )

= ( x² - x - 1 )( x² - x + 3 )

Bài 2.

A = 2x² - 8x - 10

= 2( x² - 4x + 4 ) - 18

= 2( x - 2 )² - 18 

2( x - 2 )² ≥ 0 ∀ x => 2( x - 2 )² - 18 ≥ -18

Đẳng thức xảy ra <=> x - 2 = 0 => x = 2

=> MinA = -18 <=> x = 2

B = 9x - 3x²

= -3( x² - 3x + 9/4 ) + 27/4

= -3( x - 3/2 )² + 27/4

-3( x - 3/2 )² ≤ 0 ∀ x => -3( x - 3/2 )² + 27/4 ≤ 27/4

Đẳng thức xảy ra <=> x - 3/2 = 0 => x = 3/2

=> MaxB = 27/4 <=> x = 3/2

\(M=-x^2+12x+8=-\left(x-6\right)^2+44\le44\)

\(M_{max}=44\) khi \(x=6\)

\(N=a^2+9b^2+5a-6b=\left(a+\dfrac{5}{2}\right)^2+\left(3b-1\right)^2-\dfrac{41}{4}\ge-\dfrac{41}{4}\)

\(N_{min}=-\dfrac{41}{4}\) khi \(\left(a;b\right)=\left(-\dfrac{5}{2};\dfrac{1}{3}\right)\)

\(Q=3\left(a-5\right)^2-82\ge-82\)

\(Q_{min}=-82\) khi \(a=5\)