a) 3x(x – 1) + x-1 = 0 b) 2( x+3) + x^2 + 3x = 0 c) (2x – 1)^2 – (3x + 1)^2 = 0 d) x^2 -2x + (x-2) = 0 24/09/2021 Bởi Cora a) 3x(x – 1) + x-1 = 0 b) 2( x+3) + x^2 + 3x = 0 c) (2x – 1)^2 – (3x + 1)^2 = 0 d) x^2 -2x + (x-2) = 0
Đáp án: a) x=1, x=1/3 b) x=-3, x=-2 c) x=0, x=-2 d) x=2, x=-1. Giải thích các bước giải: \(\eqalign{ & a)\,\,3x\left( {x – 1} \right) + \left( {x – 1} \right) = 0 \cr & \Leftrightarrow \left( {x – 1} \right)\left( {3x + 1} \right) = 0 \cr & \Leftrightarrow \left[ \matrix{ x – 1 = 0 \hfill \cr 3x + 1 = 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = 1 \hfill \cr x = {1 \over 3} \hfill \cr} \right. \cr & b)\,\,2\left( {x + 3} \right) + {x^2} + 3x = 0 \cr & \Leftrightarrow 2\left( {x + 3} \right) + x\left( {x + 3} \right) = 0 \cr & \Leftrightarrow \left( {x + 3} \right)\left( {x + 2} \right) = 0 \cr & \Leftrightarrow \left[ \matrix{ x + 3 = 0 \hfill \cr x + 2 = 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = – 3 \hfill \cr x = – 2 \hfill \cr} \right. \cr & c)\,\,{\left( {2x – 1} \right)^2} – {\left( {3x + 1} \right)^2} = 0 \cr & \Leftrightarrow \left( {2x – 1 + 3x + 1} \right)\left( {2x – 1 – 3x – 1} \right) = 0 \cr & \Leftrightarrow 5x\left( { – x – 2} \right) = 0 \cr & \Leftrightarrow \left[ \matrix{ 5x = 0 \hfill \cr – x – 2 = 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = 0 \hfill \cr x = – 2 \hfill \cr} \right. \cr & d)\,\,{x^2} – 2x + \left( {x – 2} \right) = 0 \cr & \Leftrightarrow x\left( {x – 2} \right) + \left( {x – 2} \right) = 0 \cr & \Leftrightarrow \left( {x – 2} \right)\left( {x + 1} \right) = 0 \cr & \Leftrightarrow \left[ \matrix{ x – 2 = 0 \hfill \cr x + 1 = 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = 2 \hfill \cr x = – 1 \hfill \cr} \right. \cr} \) Bình luận
Đáp án:
a) x=1, x=1/3
b) x=-3, x=-2
c) x=0, x=-2
d) x=2, x=-1.
Giải thích các bước giải:
\(\eqalign{
& a)\,\,3x\left( {x – 1} \right) + \left( {x – 1} \right) = 0 \cr
& \Leftrightarrow \left( {x – 1} \right)\left( {3x + 1} \right) = 0 \cr
& \Leftrightarrow \left[ \matrix{
x – 1 = 0 \hfill \cr
3x + 1 = 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = 1 \hfill \cr
x = {1 \over 3} \hfill \cr} \right. \cr
& b)\,\,2\left( {x + 3} \right) + {x^2} + 3x = 0 \cr
& \Leftrightarrow 2\left( {x + 3} \right) + x\left( {x + 3} \right) = 0 \cr
& \Leftrightarrow \left( {x + 3} \right)\left( {x + 2} \right) = 0 \cr
& \Leftrightarrow \left[ \matrix{
x + 3 = 0 \hfill \cr
x + 2 = 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = – 3 \hfill \cr
x = – 2 \hfill \cr} \right. \cr
& c)\,\,{\left( {2x – 1} \right)^2} – {\left( {3x + 1} \right)^2} = 0 \cr
& \Leftrightarrow \left( {2x – 1 + 3x + 1} \right)\left( {2x – 1 – 3x – 1} \right) = 0 \cr
& \Leftrightarrow 5x\left( { – x – 2} \right) = 0 \cr
& \Leftrightarrow \left[ \matrix{
5x = 0 \hfill \cr
– x – 2 = 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = 0 \hfill \cr
x = – 2 \hfill \cr} \right. \cr
& d)\,\,{x^2} – 2x + \left( {x – 2} \right) = 0 \cr
& \Leftrightarrow x\left( {x – 2} \right) + \left( {x – 2} \right) = 0 \cr
& \Leftrightarrow \left( {x – 2} \right)\left( {x + 1} \right) = 0 \cr
& \Leftrightarrow \left[ \matrix{
x – 2 = 0 \hfill \cr
x + 1 = 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = 2 \hfill \cr
x = – 1 \hfill \cr} \right. \cr} \)
Đáp án: giải dạng phức tạp
Giải thích các bước giải: