Toán Lớp 8: k, (x+3)(2x-1)=(x+3)(x-2) m, x ²-4x+4=(2-x)(3x+1) Chi tiết đầy đủ hứa vote 5*

Question

Toán Lớp 8:  k, (x+3)(2x-1)=(x+3)(x-2) m, x ²-4x+4=(2-x)(3x+1) Chi tiết đầy đủ hứa vote 5*

thaolanphuobg · Answer

Giải đáp + Lời giải và giải thích chi tiết:

k) $(x + 3)(2x – 1) = (x + 3)(x – 2)$
$\Leftrightarrow (x + 3)(2x – 1) – (x + 3)(x – 2) = 0$

$\Leftrightarrow (x + 3)(2x – 1 – x + 2) = 0$

$\Leftrightarrow (x + 3)(x + 1) = 0$

$\Leftrightarrow$ $\left[ \begin{array}{l}x+3=0\\x+1=0\end{array} \right.$

$\Leftrightarrow$ $\left[ \begin{array}{l}x=-3\\x=-1\end{array} \right.$

Vậy S = {-3; -1}

m) $x^2 – 4x + 4 = (2 – x)(3x + 1)$

$(x – 2)(x – 2) = (2 – x)(3x + 1)$

$(x – 2)(x – 2) = (x – 2)(-3x – 1)$
$(x – 2)(x – 2 + 3x + 1) = 0$

$(x – 2)(4x – 1) = 0$

$\Leftrightarrow$ $\left[ \begin{array}{l}x-2=0\\4x-1=0\end{array} \right.$

$\Leftrightarrow$ $\left[ \begin{array}{l}x=2\\x=\dfrac{1}{4}\end{array} \right.$

Vậy S = {2; 1/4}

ngocbichchau · Answer

Lời giải và giải thích chi tiết+Giải đáp: k) (x+3)(2x-1)=(x+3)(x-2) <=>(x+3)(2x-1)-(x+3)(x-2)=0 <=>(x+3)(2x-1-x+2)=0 <=>(x+3)(x+1)=0 <=> ( left[ begin{array}{l}x+3=0 x+1=0 end{array} right. ) <=> ( left[ begin{array}{l}x=-3 x=-1 end{array} right. ) Vậy: S={-3;-1} m) x^2-4x+4=(2-x)(3x+1) <=>(x-2)^2=(2-x)(3x+1) <=>(x-2)(x-2)-(2-x)(3x+1)=0 <=>(x-2)(x-2)+(x-2)(3x+1)=0 <=>(x-2)(x-2+3x+1)=0 <=>(x-2)(4x-1)=0 <=> ( left[ begin{array}{l}x-2=0 4x-1=0 end{array} right. ) <=> ( left[ begin{array}{l}x=2 x=1/4 end{array} right. ) Vậy: S={2;1/4}