Bài 1:

\(A=-x^2-2x+9\)

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

\(A=-\left(x^2+2x+1-10\right)\)

\(A=-\left(x+1\right)^2+10\)

Vì \(-\left(x+1\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x+1\right)^2+10\le10\)

\(\Rightarrow Amax=10\Leftrightarrow x=-1\)

\(B=-9x^2+6x+25\)

\(B=-\left(9x^2-6x-25\right)\)

\(B=-\left[\left(3x\right)^2-2.3x+1-26\right]\)

\(B=-\left(3x-1\right)^2+26\)

Vì \(-\left(3x-1\right)^2\le0\) với mọi x

\(\Rightarrow-\left(3x-1\right)^2+26\le26\)

\(\Rightarrow Bmax=26\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)

\(C=-x^2+x+1\)

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

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

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

Vì \(-\left(x-\dfrac{1}{2}\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)

\(\Rightarrow Cmax=\dfrac{5}{4}\Leftrightarrow x=\dfrac{1}{2}\)

\(D=-2x^2+3x+1\)

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

\(D=-2\left(x^2-2.x\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}-\dfrac{1}{2}\right)\)

\(D=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\)

Vì \(-2\left(x-\dfrac{3}{4}\right)^2\le0\) với mọi x

\(\Rightarrow-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\le\dfrac{17}{8}\)

\(\Rightarrow Dmax=\dfrac{17}{8}\Leftrightarrow x=\dfrac{3}{4}\)

\(E=-25x^2-10x+7\)

\(E=-\left(25x^2+10x-7\right)\)

\(E=-\left[\left(5x\right)^2+2.5x+1-8\right]\)

\(E=-\left(5x+1\right)^2+8\)

Vì \(-\left(5x+1\right)^2\le0\) với mọi x

\(\Rightarrow-\left(5x+1\right)^2+8\le8\)

\(\Rightarrow Emax=8\Leftrightarrow5x+1=0\Rightarrow x=-\dfrac{1}{5}\)

Bài 2:

\(A=9x^2+6x+4\)

\(A=\left(3x\right)^2+2.3x+1+3\)

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

Vì \(\left(3x+1\right)^2\ge0\) với mọi x

\(\Rightarrow\left(3x+1\right)^2+3\ge3\)

\(\Rightarrow Amin=3\Leftrightarrow x=-\dfrac{1}{3}\)

\(B=4x^2+4x+12\)

\(B=\left(2x\right)^2+2.2x+1+11\)

\(B=\left(2x+1\right)^2+11\)

Vì \(\left(2x+1\right)^2\ge0\) với mọi x

\(\Rightarrow\left(2x+1\right)^2+11\ge11\)

\(\Rightarrow Bmin=11\Leftrightarrow x=-\dfrac{1}{2}\)

\(C=x^2+x+3\)

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

\(C=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)

Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

\(\Rightarrow Cmin=\dfrac{11}{4}\Leftrightarrow x=-\dfrac{1}{2}\)

\(D=2x^2+3x+1\)

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

\(D=2\left(x^2+2.x.\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}+\dfrac{1}{2}\right)\)

\(D=2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\)

Vì \(2\left(x+\dfrac{3}{4}\right)^2\ge0\) với mọi x

\(\Rightarrow2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\ge-\dfrac{1}{8}\)

\(\Rightarrow Dmin=-\dfrac{1}{8}\Leftrightarrow x=-\dfrac{3}{4}\)

\(E=64x^2+16x+3\)

\(E=\left(8x\right)^2+2.8x+1+2\)

\(E=\left(8x+1\right)^2+2\)

Vì \(\left(8x+1\right)^2\ge0\) với mọi x

\(\Rightarrow\left(8x+1\right)^2+2\ge2\)

\(\Rightarrow Emin=2\Leftrightarrow x=-\dfrac{1}{8}\)

a) Ta có: \(x^4-16x^2=0\)

\(\Leftrightarrow x^2\left(x^2-16\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x-4=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{0;4;-4\right\}\)

b) Ta có: \(9x^2+6x+1=0\)

\(\Leftrightarrow\left(3x\right)^2+2\cdot3x\cdot1+1^2=0\)

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

\(\Leftrightarrow3x+1=0\)

\(\Leftrightarrow3x=-1\)

hay \(x=-\frac{1}{3}\)

Vậy: \(x=-\frac{1}{3}\)

c) Ta có: \(x^2-6x=16\)

\(\Leftrightarrow x^2-6x-16=0\)

\(\Leftrightarrow x^2-8x+2x-16=0\)

\(\Leftrightarrow x\left(x-8\right)+2\left(x-8\right)=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{8;-2\right\}\)

d) Ta có: \(9x^2+6x=80\)

\(\Leftrightarrow9x^2+6x-80=0\)

\(\Leftrightarrow9x^2+6x+1-81=0\)

\(\Leftrightarrow\left(3x+1\right)^2-9^2=0\)

\(\Leftrightarrow\left(3x+1-9\right)\left(3x+1+9\right)=0\)

\(\Leftrightarrow\left(3x-8\right)\left(3x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-8=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=8\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=-\frac{10}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{8}{3};-\frac{10}{3}\right\}\)

e) Ta có: \(25\left(2x-1\right)^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(10x-5\right)^2-\left(3x+3\right)^2=0\)

\(\Leftrightarrow\left(10x-5-3x-3\right)\left(10x-5+3x+3\right)=0\)

\(\Leftrightarrow\left(7x-8\right)\left(13x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-8=0\\13x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=8\\13x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{7}\\x=\frac{2}{13}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{8}{7};\frac{2}{13}\right\}\)

a) \(\sqrt[]{x^2-4x+4}=x+3\)

\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)

\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)

Giải pt (1)

\(\Delta=9+32=41>0\)

Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)

Giải pt (2)

\(\Delta=9+48=57>0\)

Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)

Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)

a) x⁴ - 16x² = 0

<=> ( x² )² - ( 4x )² = 0

<=> ( x² - 4x )( x² + 4x ) = 0

<=> [ x( x - 4 ) ][ x( x + 4 ) ] = 0

<=> x( x - 4 )x( x + 4 ) = 0

<=> x²( x - 4 )( x + 4 ) = 0

<=> \(\hept{\begin{cases}x^2=0\\x-4=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)( thay bằng dấu hoặc hộ mình nhé )

b) 9x² + 6x + 1 = 0

<=> ( 3x )² + 2.3x.1 + 1² = 0

<=> ( 3x + 1 )² = 0

<=> 3x + 1 = 0

<=> 3x = -1

<=> x = -1/3

c) x² - 6x = 16

<=> x² - 6x - 16 = 0

<=> x² + 2x - 8x - 16 = 0

<=> x( x + 2 ) - 8( x + 2 ) = 0

<=> ( x + 2 )( x - 8 ) = 0

<=> \(\orbr{\begin{cases}x+2=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}\)

d) 9x² + 6x = 80

<=> 9x² + 6x - 80 = 0

<=> 9x² + 30x - 24x - 80 = 0

<=> 9x( x + 10/3 ) - 24( x + 10/3 ) = 0

<=> ( x + 10/3 )( 9x - 24 ) = 0

<=> \(\orbr{\begin{cases}x+\frac{10}{3}=0\\9x-24=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{10}{3}\\x=\frac{8}{3}\end{cases}}\)

e) Áp dụng công thức aⁿ.bⁿ = ( ab )ⁿ ta có :

25( 2x - 1 )² - 9( x + 1 )² = 0

<=> 5²( 2x - 1 )² - 3²( x + 1 )² = 0 

<=> [ 5( 2x - 1 ) ]² - [ 3( x + 1 ) ]² = 0

<=> ( 10x - 5 )² - ( 3x + 3 )² = 0

<=> [ ( 10x - 5 ) - ( 3x + 3 ) ][ ( 10x - 5 ) + ( 3x + 3 ) ] = 0

<=> ( 10x - 5 - 3x - 3 )( 10x - 5 + 3x + 3 ) = 0

<=> ( 7x - 8 )( 13x - 2 ) = 0

<=> \(\orbr{\begin{cases}7x-8=0\\13x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)

             Bài làm :

a) x⁴ - 16x² = 0

<=> ( x² )² - ( 4x )² = 0

<=> ( x² - 4x )( x² + 4x ) = 0

<=> [ x( x - 4 ) ][ x( x + 4 ) ] = 0

<=> x( x - 4 )x( x + 4 ) = 0

<=> x²( x - 4 )( x + 4 ) = 0

 Vậy x=0 hoặc x=±4

b) 9x² + 6x + 1 = 0

<=> ( 3x )² + 2.3x.1 + 1² = 0

<=> ( 3x + 1 )² = 0

<=> 3x + 1 = 0

<=> 3x = -1

<=> x = -1/3

c) x² - 6x = 16

<=> x² - 6x - 16 = 0

<=> x² + 2x - 8x - 16 = 0

<=> x( x + 2 ) - 8( x + 2 ) = 0

<=> ( x + 2 )( x - 8 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}\)

d) 9x² + 6x = 80

<=> 9x² + 6x - 80 = 0

<=> 9x² + 30x - 24x - 80 = 0

<=> 9x( x + 10/3 ) - 24( x + 10/3 ) = 0

<=> ( x + 10/3 )( 9x - 24 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}x+\frac{10}{3}=0\\9x-24=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{10}{3}\\x=\frac{8}{3}\end{cases}}\)

e) 25( 2x - 1 )² - 9( x + 1 )² = 0

<=> 5²( 2x - 1 )² - 3²( x + 1 )² = 0 

<=> [ 5( 2x - 1 ) ]² - [ 3( x + 1 ) ]² = 0

<=> ( 10x - 5 )² - ( 3x + 3 )² = 0

<=> [ ( 10x - 5 ) - ( 3x + 3 ) ][ ( 10x - 5 ) + ( 3x + 3 ) ] = 0

<=> ( 10x - 5 - 3x - 3 )( 10x - 5 + 3x + 3 ) = 0

<=> ( 7x - 8 )( 13x - 2 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}7x-8=0\\13x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)

a) Ta có : x⁴ - 16x² = 0

=> x⁴ - 8x² - 8x² + 64 = 64

=> x²(x² - 8) - 8(x² - 8) = 64

=> (x² - 8)² = 64

=> \(\orbr{\begin{cases}x^2-8=8\\x^2-8=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=16\\x^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\pm4\\x=0\end{cases}}\Rightarrow x\in\left\{4;-4;0\right\}\)

b) Ta có 9x² + 6x + 1 = 0

=> 9x² + 3x + 3x + 1 = 0

=> 3x(3x + 1) + (3x + 1) = 0

=> (3x + 1)² = 0

=> 3x + 1 = 0

=> x = -1/3

c) Ta có x² - 6x = 16

=> x² - 6x + 9 = 25

=> (x - 3)² = 25

=> \(\orbr{\begin{cases}x-3=5\\x-3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-2\end{cases}}\Rightarrow x\in\left\{8;-2\right\}\)

d) 9x² + 6x = 80

=> 9x² + 6x + 1 = 81

=> (3x + 1)² = 81

=> \(\orbr{\begin{cases}3x+1=9\\3x+1=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=-\frac{10}{3}\end{cases}\Rightarrow x\in}\left\{\frac{8}{3};\frac{-10}{3}\right\}\)

e) 25(2x - 1)² - 9(x + 1)² = 0

=> [5(2x - 1)]² - [3(x + 1)]² = 0

=> (10x - 5)² - (3x + 3)² = 0

=> (10x - 5 - 3x - 3)(10x - 5 + 3x + 3) = 0

=> (7x - 8)(13x - 2) = 0

=> \(\orbr{\begin{cases}7x=8\\13x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)

ko bt lm:)

b, x = -5/3 hoặc x = 4/3.

c, x = 0 hoặc x = 3, -3.

d, x = 0 hoặc x = 2, -2.

e, x = 1 hoặc x = \(\dfrac{-1}{2}\).

a: \(\Leftrightarrow x^2-40x+400-x^2-4x-3=-7\)

=>-44x+397=-7

=>-44x=-404

hay x=101

b: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=0\\4-3x=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{5}{3};\dfrac{4}{3}\right\}\)

c: \(\Leftrightarrow x\left(x^2-9\right)=0\)

=>x(x-3)(x+3)=0

hay \(x\in\left\{0;3;-3\right\}\)

d: \(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)

hay \(x\in\left\{0;2;-2\right\}\)

e: =>(2x+1)(1-x)=0

=>x=-1/2 hoặc x=1