Toán Lớp 8: Giải các bất phương trình .
Bài 1
a) (x – 1)(x + 2) > (x – 1)2 + 3 ; b) x (2x – 1) – 8 < 5 – 2x (1 – x );
c)(2x + 1)2 + (1 - x )3x <= (x+2)2 ; d) (x – 4)(x + 4) >= (x + 3)2 + 5
Bài 2
a) (x – 3)2 < x^2 – 5x + 4 b) (x – 3)(x + 3) < = (x + 2)2 + 3
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a)\left( {x – 1} \right)\left( {x + 2} \right) > {\left( {x – 1} \right)^2} + 3\\
\Leftrightarrow {x^2} + 2x – x – 2 > {x^2} – 2x + 1 + 3\\
\Leftrightarrow x + 2x > 4 + 2\\
\Leftrightarrow 3x > 6\\
\Leftrightarrow x > 2\\
Vậy\,x > 2\\
b)x\left( {2x – 1} \right) – 8 < 5 – 2x\left( {1 – x} \right)\\
\Leftrightarrow 2{x^2} – x – 8 < 5 – 2x + 2{x^2}\\
\Leftrightarrow – x + 2x < 5 + 8\\
\Leftrightarrow x < 13\\
Vậy\,x < 13\\
c){\left( {2x + 1} \right)^2} + \left( {1 – x} \right).3x \le {\left( {x + 2} \right)^2}\\
\Leftrightarrow 4{x^2} + 4x + 1 + 3x – 3{x^2} \le {x^2} + 4x + 4\\
\Leftrightarrow 7x – 4x \le 4 – 1\\
\Leftrightarrow 3x \le 3\\
\Leftrightarrow x \le 1\\
Vậy\,x \le 1\\
d)\left( {x – 4} \right)\left( {x + 4} \right) \ge {\left( {x + 3} \right)^2} + 5\\
\Leftrightarrow {x^2} – 16 \ge {x^2} + 6x + 9 + 5\\
\Leftrightarrow 6x \le – 16 – 14\\
\Leftrightarrow 6x \le – 30\\
\Leftrightarrow x \le – 5\\
Vậy\,x \le – 5\\
B2)a){\left( {x – 3} \right)^2} < {x^2} – 5x + 4\\
\Leftrightarrow {x^2} – 6x + 9 < {x^2} – 5x + 4\\
\Leftrightarrow – 5x + 6x > 9 – 4\\
\Leftrightarrow x > 5\\
Vậy\,x > 5\\
b)\left( {x – 3} \right)\left( {x + 3} \right) \le {\left( {x + 2} \right)^2} + 3\\
\Leftrightarrow {x^2} – 9 \le {x^2} + 4x + 4 + 3\\
\Leftrightarrow 4x \ge – 9 – 7\\
\Leftrightarrow x \ge \dfrac{{ – 16}}{4}\\
\Leftrightarrow x \ge – 4\\
Vậy\,x \ge – 4
\end{array}$