1. ( x + 2 ) ² – x ( x + 1 ) = 6
2. ( 2 – 2x ) – 4 ( x ² + 1 ) = 5
3. ( x – 3 )( x + 3 ) – x( x + 2 ) = 8
4. ( 2x + 3 ) – 4 ( x ² + 2 ) = 10
5. ( 1- x ) ² – 4( x + 3 ) = -6
6. ( x – 2 )( x + 2 ) – x(x – 1 ) = -4
1. ( x + 2 ) ² – x ( x + 1 ) = 6
2. ( 2 – 2x ) – 4 ( x ² + 1 ) = 5
3. ( x – 3 )( x + 3 ) – x( x + 2 ) = 8
4. ( 2x + 3 ) – 4 ( x ² + 2 ) = 10
5. ( 1- x ) ² – 4( x + 3 ) = -6
6. ( x – 2 )( x + 2 ) – x(x – 1 ) = -4
Đáp án:
1/ $x=\dfrac{2}{3}$
2/ $x=-\dfrac{5}{8}$
3/ $x=-\dfrac{17}{2}$
4/ $x=\dfrac{3}{4}$
5/ $x=3+\sqrt{14}$ hoặc $x=3-\sqrt{14}$
6/ $x=0$
Giải thích các bước giải:
1/ $(x+2)^2-x(x+1)=6$
$⇔ x^2+4x+4-x^2-x-6=0$
$⇔ 3x-2=0$
$⇔ x=\dfrac{2}{3}$
2/ $(2-2x)^2-4(x^2+1)=5$
$⇔ 4-8x+4x^2-4x^2-4-5=0$
$⇔ -8x-5=0$
$⇔ x=-\dfrac{5}{8}$
3/ $(x-3)(x+3)-x(x+2)=8$
$⇔ x^2-9-x^2-2x-8=0$
$⇔ -2x-17=0$
$⇔ x=-\dfrac{17}{2}$
4/ $(2x+3)^2-4(x^2+2)=10$
$⇔ 4x^2+12x+9-4x^2-8-10=0$
$⇔ 12x-9=0$
$⇔ x=\dfrac{9}{12}=\dfrac{3}{4}$
5/ $(1-x)^2-4(x+3)=-6$
$⇔ 1-2x+x^2-4x-12+6=0$
$⇔ x^2-6x-5=0$
$⇔ x^2-6x+9-14=0$
$⇔ (x-3)^2-14=0$
$⇔ (x-3-\sqrt{14})(x-3+\sqrt{14})=0$
$⇔ \left[ \begin{array}{l}x=3+\sqrt{14}\\x=3-\sqrt{14}\end{array} \right.$
6/ $(x-2)(x+2)-x(x-1)=-4$
$⇔ x^2-4-x^2+x+4=0$
$⇔ x=0$
Đáp án:
Bạn xem lại đề nhé có 1 số câu thiếu đề
1. (x + 2)² – x(x + 1) = 6
⇔ x² + 4x + 4 – x² – x = 6
⇔ 3x + 4 = 6
⇔ 3x = 2
⇔ x = $\frac{2}{3}$
2. (2 – 2x)² – 4(x² + 1) = 5
⇔ 4 – 8x + 4x² – 4x² – 4 = 5
⇔ – 8x = 5
⇔ x = $\frac{-5}{8}$
3. (x – 3)(x + 3) – x(x + 2) = 8
⇔ x² – 9 – x² – 2x = 8
⇔ -9 – 2x = 8
⇔ – 2x = 8 + 9
⇔ – 2x = 17
⇔ x = $\frac{-17}{2}$
4. (2x + 3)² – 4(x² + 2) = 10
⇔ 4x² + 12x + 9 – 4x² – 8 = 10
⇔ 12x + 1 = 10
⇔ 12x = 9
⇔ x = $\frac{3}{4}$
5. (1 – x)² – 4(x + 3) = – 6
⇔ 1 – 2x + x² – 4x – 12 = -6
⇔ x² – 6x – 5 = 0
⇔ x² – 6x + 9 – 14 = 0
⇔ (x – 3)² – (√14)²= 0
⇔ (x – 3 – √14)(x – 3 + √14) = 0
⇔ \(\left[ \begin{array}{l}x=3+√14\\x=3-√14\end{array} \right.\)
6. (x – 2)(x + 2) – x(x – 1) = -4
⇔ x² – 4 – x² + x = -4
⇔ x – 4 = – 4
⇔ x = 0