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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) 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
=> \(\orbr{\begin{cases}7x=8\\13x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)
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
a: \(A=\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{2}{x-2}\)
b: \(B=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\dfrac{x+3}{x-3}\)
c: \(C=\dfrac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\dfrac{3x+4}{x}\)
d: \(D=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}=\dfrac{x+2}{2}\)
e: \(E=\dfrac{-x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{-x}{x+2}\)
f: \(F=\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}\)
\(a,=\left(x-5\right)\left(x+5\right)\\ b,=\left(x-3\right)^2\\ c,=\left(3x-2\right)\left(3x+2\right)\\ d,=\left(x+1\right)^2\\ e,=\left(x-10\right)\left(x+10\right)\)