Toán 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 20/07/2021 By Audrey 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/ $(x-2)(x+2)-x(x-1)=-4$ $⇔ x^2-4-x^2+x+4=0$ $⇔ x=0$ Trả lời
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 Trả lời