Câu hỏi của Zero Two

Question

Tìm x , biết :a) x4 - 16x2 = 0b) 9x2 + 6x + 1 = 0c) x2 - 6x = 16 d) 9x2 + 6x = 80e) 25(2x - 1)2 - 9(x + 1)2 = 0

l҉o҉n҉g҉ d҉z҉ · Accepted Answer

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 = -1/3c) 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<=> $\orbr{\begin{cases}7x-8=0\13x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{7}\x=\frac{2}{13}\end{cases}}$Câu trả lời: 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 = -1/3c) 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<=> $\orbr{\begin{cases}7x-8=0\13x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{7}\x=\frac{2}{13}\end{cases}}$

Bellion · Answer

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=±4b) 9x2 + 6x + 1 = 0<=> ( 3x )2 + 2.3x.1 + 12 = 0<=> ( 3x + 1 )2 = 0<=> 3x + 1 = 0<=> 3x = -1<=> x = -1/3c) 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 $\Leftrightarrow\orbr{\begin{cases}7x-8=0\13x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{7}\x=\frac{2}{13}\end{cases}}$Câu trả lời: 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=±4b) 9x2 + 6x + 1 = 0<=> ( 3x )2 + 2.3x.1 + 12 = 0<=> ( 3x + 1 )2 = 0<=> 3x + 1 = 0<=> 3x = -1<=> x = -1/3c) 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 $\Leftrightarrow\orbr{\begin{cases}7x-8=0\13x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{7}\x=\frac{2}{13}\end{cases}}$

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