1, x(x+1)-3=(2x-1)(x-1) 2, (x+2)(3x+1)-x^2+4=0 3, x(x-3)+2=x(1-x)+2x-2 14/07/2021 Bởi Savannah 1, x(x+1)-3=(2x-1)(x-1) 2, (x+2)(3x+1)-x^2+4=0 3, x(x-3)+2=x(1-x)+2x-2
Đáp án+Giải thích các bước giải: `1)` `x(x+1)-3=(2x-1)(x-1)` `→x^2+x-3=2x^2-2x-x+1` `→x^2+x-3=2x^2-3x+1` `→2x^2-3x+1-x^2-x+3=0` `→x^2-4x+4=0` `→x^2-2.x.2+2^2=0` `→(x-2)^2=0` `→x-2=0` `→x=2` Vậy `x=2` `2)` `(x+2)(3x+1)-x^2+4=0` `→3x^2+x+6x+2-x^2+4=0` `→2x^2+7x+6=0` `→2x^2+4x+3x+6=0` `→2x(x+2)+3(x+2)=0` `→(x+2)(2x+3)=0` \(→\left[ \begin{array}{l}x+2=0\\2x+3=0\end{array} \right.\) \(→\left[ \begin{array}{l}x=-2\\2x=-3\end{array} \right.\) \(→\left[ \begin{array}{l}x=-2\\x=-\dfrac{3}{2}\end{array} \right.\) Vậy `x∈\{-2;-3/2\}` `3)` `x(x-3)+2=x(1-x)+2x-2` `→x^2-3x+2=x-x^2+2x-2` `→x^2-3x+2=-x^2+3x-2` `→x^2-3x+2+x^2-3x+2=0` `→2x^2-6x+4=0` `→x^2-3x+2=0` `→x^2-x-2x+2=0` `→x(x-1)-2(x-1)=0` `→(x-1)(x-2)=0` \(→\left[ \begin{array}{l}x-1=0\\x-2=0\end{array} \right.\) \(→\left[ \begin{array}{l}x=1\\x=2\end{array} \right.\) Vậy `x∈\{1;2\}` Bình luận
Đáp án: Giải thích các bước giải: `1, x( x + 1 ) – 3 = ( 2x – 1 )( x – 1 )` `<=> x^2 + x – 3 = 2x^2 – 2x – x + 1` `<=> x^2 – 2x^2 + x + 2x + x – 3 – 1 = 0` `<=> ( x^2 – 2x^2 ) + ( x + 2x + x ) + ( – 3 – 1 ) = 0` `<=> -x^2 + 4x – 4 = 0` `<=> -x^2 + 2x + 2x – 4 = 0` `<=> ( -x^2 + 2x ) + ( 2x – 4 ) = 0` `<=> -x ( x – 2 ) + 2 ( x – 2 ) = 0` `<=> ( -x + 2 )( x – 2 ) = 0` `<=> – ( x – 2 )( x – 2 ) = 0` `<=> ( x – 2 )^2 = 0` `<=> x – 2 = 0` `<=> x = 2 ` `2, ( x + 2 )( 3x + 1 ) – x^2 + 4 = 0` `<=> 3x^2 + x + 6x + 2 – x^2 + 4 = 0` `<=> ( 3x^2 – x^2 ) + ( x + 6x ) + ( 2 + 4 ) = 0` `<=> 2x^2 + 7x + 6 = 0` `<=> 2x^2 + 2x + 3/2x + 6 = 0` `<=> ( 2x^2 + 4x ) + ( 3x + 6 ) = 0` `<=> 2x( x + 2 ) + 3 ( x + 2 ) = 0` `<=> ( x + 2 )( 2x + 3 ) = 0` `<=>` \(\left[ \begin{array}{l}x+2=0\\2x=3=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-2\\x=-3/2\end{array} \right.\) `3, x( x – 3 ) + 2 = x( 1 – x ) + 2x – 2` `<=> x^2 – 3x + 2 = x – x^2 + 2x – 2` `<=> x^2 – 3x + 2 – x + x^2 – 2x + 2 = 0` `<=> ( x^2 + x^2 ) – ( 3x + x + 2x ) + ( 2 + 2 ) = 0` `<=> 2x^2 – 6x + 4 = 0` `<=> 2 ( x^2 – 3x + 2 ) = 0` `<=> x^2 – x – 2x + 2 = 0` `<=> ( x^2 – x ) – ( 2x – 2 ) = 0` `<=> x ( x – 1 ) – 2 ( x – 1 ) = 0` `<=> ( x – 2 )( x – 1 ) = 0` `<=>` \(\left[ \begin{array}{l}x-2=0\\x-1=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=2\\x=1\end{array} \right.\) Bình luận
Đáp án+Giải thích các bước giải:
`1)`
`x(x+1)-3=(2x-1)(x-1)`
`→x^2+x-3=2x^2-2x-x+1`
`→x^2+x-3=2x^2-3x+1`
`→2x^2-3x+1-x^2-x+3=0`
`→x^2-4x+4=0`
`→x^2-2.x.2+2^2=0`
`→(x-2)^2=0`
`→x-2=0`
`→x=2`
Vậy `x=2`
`2)`
`(x+2)(3x+1)-x^2+4=0`
`→3x^2+x+6x+2-x^2+4=0`
`→2x^2+7x+6=0`
`→2x^2+4x+3x+6=0`
`→2x(x+2)+3(x+2)=0`
`→(x+2)(2x+3)=0`
\(→\left[ \begin{array}{l}x+2=0\\2x+3=0\end{array} \right.\)
\(→\left[ \begin{array}{l}x=-2\\2x=-3\end{array} \right.\)
\(→\left[ \begin{array}{l}x=-2\\x=-\dfrac{3}{2}\end{array} \right.\)
Vậy `x∈\{-2;-3/2\}`
`3)`
`x(x-3)+2=x(1-x)+2x-2`
`→x^2-3x+2=x-x^2+2x-2`
`→x^2-3x+2=-x^2+3x-2`
`→x^2-3x+2+x^2-3x+2=0`
`→2x^2-6x+4=0`
`→x^2-3x+2=0`
`→x^2-x-2x+2=0`
`→x(x-1)-2(x-1)=0`
`→(x-1)(x-2)=0`
\(→\left[ \begin{array}{l}x-1=0\\x-2=0\end{array} \right.\)
\(→\left[ \begin{array}{l}x=1\\x=2\end{array} \right.\)
Vậy `x∈\{1;2\}`
Đáp án:
Giải thích các bước giải:
`1, x( x + 1 ) – 3 = ( 2x – 1 )( x – 1 )`
`<=> x^2 + x – 3 = 2x^2 – 2x – x + 1`
`<=> x^2 – 2x^2 + x + 2x + x – 3 – 1 = 0`
`<=> ( x^2 – 2x^2 ) + ( x + 2x + x ) + ( – 3 – 1 ) = 0`
`<=> -x^2 + 4x – 4 = 0`
`<=> -x^2 + 2x + 2x – 4 = 0`
`<=> ( -x^2 + 2x ) + ( 2x – 4 ) = 0`
`<=> -x ( x – 2 ) + 2 ( x – 2 ) = 0`
`<=> ( -x + 2 )( x – 2 ) = 0`
`<=> – ( x – 2 )( x – 2 ) = 0`
`<=> ( x – 2 )^2 = 0`
`<=> x – 2 = 0`
`<=> x = 2 `
`2, ( x + 2 )( 3x + 1 ) – x^2 + 4 = 0`
`<=> 3x^2 + x + 6x + 2 – x^2 + 4 = 0`
`<=> ( 3x^2 – x^2 ) + ( x + 6x ) + ( 2 + 4 ) = 0`
`<=> 2x^2 + 7x + 6 = 0`
`<=> 2x^2 + 2x + 3/2x + 6 = 0`
`<=> ( 2x^2 + 4x ) + ( 3x + 6 ) = 0`
`<=> 2x( x + 2 ) + 3 ( x + 2 ) = 0`
`<=> ( x + 2 )( 2x + 3 ) = 0`
`<=>` \(\left[ \begin{array}{l}x+2=0\\2x=3=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-2\\x=-3/2\end{array} \right.\)
`3, x( x – 3 ) + 2 = x( 1 – x ) + 2x – 2`
`<=> x^2 – 3x + 2 = x – x^2 + 2x – 2`
`<=> x^2 – 3x + 2 – x + x^2 – 2x + 2 = 0`
`<=> ( x^2 + x^2 ) – ( 3x + x + 2x ) + ( 2 + 2 ) = 0`
`<=> 2x^2 – 6x + 4 = 0`
`<=> 2 ( x^2 – 3x + 2 ) = 0`
`<=> x^2 – x – 2x + 2 = 0`
`<=> ( x^2 – x ) – ( 2x – 2 ) = 0`
`<=> x ( x – 1 ) – 2 ( x – 1 ) = 0`
`<=> ( x – 2 )( x – 1 ) = 0`
`<=>` \(\left[ \begin{array}{l}x-2=0\\x-1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2\\x=1\end{array} \right.\)