BT1: Tìm x:
a) (x-2) × (x+2) – (x+1) ² =0
b) (3x+1) × (x+1) – (3x+1) × (3x-1) =5
BT2: Chứng minh:
a) (a+b) × ( a² – ab+b² ) =a³+ b³
b) (a-b) × (a² + ab+b²) =a³-b³
BT1: Tìm x:
a) (x-2) × (x+2) – (x+1) ² =0
b) (3x+1) × (x+1) – (3x+1) × (3x-1) =5
BT2: Chứng minh:
a) (a+b) × ( a² – ab+b² ) =a³+ b³
b) (a-b) × (a² + ab+b²) =a³-b³
a,
(x-2) × (x+2) – (x+1) ² =0
⇔(x-2) ²-(x+1) ² =0
⇔(x-2-x-1).(x-2+x+1)=0
⇔(-3).(2x-1)=0
⇔2x-1=0
⇔2x=1
⇔x=1/2
b,
(3x+1) × (x+1) – (3x+1) × (3x-1) =5
⇔(3x+1).[ (x+1)-(3x-1)]=5
⇔(3x+1).(-2x+2)=5
⇔(3x+1).(-2x+2)∈Ư(5)
lập bảng ta có :
3x+ 1 1 (-1)
-2x +2 5 (-5)
x 0,3 -2/ 3
x 3/(-2) (-2)
⇔x∈{0,3 và 3/−2 ; −2/3-2}
2
a) (a+b) × ( a² – ab+b² ) =a³-a²b+b²a+a²b-b²a+b³=a³+b³
b) (a-b) × (a² + ab+b²) =a³-a²b+b²a+a²b-b²a-b³=a³-b³
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left( {x – 2} \right)\left( {x + 2} \right) – {\left( {x + 1} \right)^2} = 0\\
\Leftrightarrow \left( {{x^2} – {2^2}} \right) – \left( {{x^2} + 2x + 1} \right) = 0\\
\Leftrightarrow {x^2} – 4 – {x^2} – 2x – 1 = 0\\
\Leftrightarrow – 2x – 5 = 0\\
\Leftrightarrow – 2x = 5\\
\Leftrightarrow x = – \frac{5}{2}\\
b,\\
\left( {3x + 1} \right)\left( {x + 1} \right) – \left( {3x + 1} \right).\left( {3x – 1} \right) = 5\\
\Leftrightarrow \left( {3{x^2} + 4x + 1} \right) – \left( {9{x^2} – 1} \right) = 5\\
\Leftrightarrow – 4{x^2} + 4x + 2 = 5\\
\Leftrightarrow 4{x^2} – 4x + 3 = 0\\
\Leftrightarrow \left( {4{x^2} – 4x + 1} \right) + 2 = 0\\
\Leftrightarrow {\left( {2x – 1} \right)^2} + 2 = 0\,\,\,\,\,\,\,\,\,\,\,\left( {vn} \right)\\
2,\\
\left( {a + b} \right)\left( {{a^2} – ab + {b^2}} \right)\\
= {a^3} – {a^2}b + a{b^2} + {a^2}b – a{b^2} + {b^3}\\
= {a^3} + {b^3}\\
b,\\
\left( {a – b} \right)\left( {{a^2} + ab + {b^2}} \right)\\
= {a^3} + {a^2}b + a{b^2} – {a^2}b – a{b^2} + {b^3}\\
= {a^3} + {b^3}
\end{array}\)