a) \(y=2x^2\\\begin{array}{|c|c|c|}\hline x&-2&-1&0&1&2\\\hline y&8&2&0&2&8\\\hline\end{array}\)
\(→\) Hàm số \(y=2x^2\) đi qua điểm (-2;8);(-1;2);(0;0);(1;2);(2;8)
b) Đồ thị (P) đi qua điểm M có hoành độ là 1
\(→y=2.1^2\\↔y=2\)
\(→ M(1;2)\)
Vì đồ thị (d):y=4x+m đi qua điểm M
\(→4.1+m=2\\↔m+4=2\\↔m=-2\)
Vậy m=-2
Đáp án:
$m=-2$
Giải thích các bước giải:
$a,y=2x^2$
Ta có bảng:
$\begin{array}{|l|l|} \hline x&-2&-1&\,\,0\,\,&\,\,1\,\,&\,\,2\,\,\\\hline y&\,\,\,8&\,\,\,2&\,\,0&\,\,2&\,\,8\\\hline\end{array}$
$b,y=4x+m$
Phương trình hoành độ giao điểm:
$2x^2=4x+m$
Giao điểm $M$ có hoành độ là $1$
$⇒2.1^2=4.1+m$
$⇒m=-2$
Vậy $m=-2$.
Tìm x,biết a) 3( x + 1) + 4( x-2 ) = 7 b) 2( 3x -2 ) - 5 ( x+1) =10 c) 2x ( x+1 ) -2( x + 3 ) = 5 d) 3x ( x-2 )+ x ( 5 - 3x )= 10
giải giúp mình câu B ko cần câu A giải chi tiết giúp mình nhe
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