Giải phương trình:
a, √(2x – 1) + √(x – 2) = √(x + 1)
b, √(3x + 15) – √(4x + 17) = √(x + 2)
c, √(x – 2) + √(3 + x) = √(2x + 1)
d, √(2x – 1) = √(x – 1) + √(x – 4)
Giải phương trình: a, √(2x – 1) + √(x – 2) = √(x + 1) b, √(3x + 15) – √(4x + 17) = √(x + 2) c, √(x – 2) + √(3 + x) = √(2x + 1) d, √(
By Allison
Đáp án:
\(\eqalign{
& a)\,\,x = 2 \cr
& b)\,\,x = – 2 \cr
& c)\,\,x = 2 \cr
& d)\,\,x = 5 \cr} \)
Giải thích các bước giải:
$$\eqalign{
& a)\,\,\sqrt {2x – 1} + \sqrt {x – 2} = \sqrt {x + 1} \,\,\left( {x \ge 2} \right) \cr
& \Leftrightarrow 2x – 1 + x – 2 + 2\sqrt {\left( {2x – 1} \right)\left( {x – 2} \right)} = x + 1 \cr
& \Leftrightarrow 2x – 4 + 2\sqrt {\left( {2x – 1} \right)\left( {x – 2} \right)} = 0 \cr
& \Leftrightarrow \sqrt {\left( {2x – 1} \right)\left( {x – 2} \right)} = 2 – x\,\,\,\left( * \right) \cr
& Do\,\,x \ge 2 \Leftrightarrow 2 – x \le 0 \cr
& \left( * \right) \Rightarrow 2 – x = 0 \Leftrightarrow x = 2 \cr
& Thu\,\,lai: \cr
& \sqrt {2.2 – 1} + \sqrt {2 – 2} = \sqrt {2 + 1} \cr
& \Leftrightarrow \sqrt 3 = \sqrt 3 \,\,\left( {dung} \right) \cr
& Vay\,\,x = 2 \cr
& b)\,\,\sqrt {3x + 15} – \sqrt {4x + 17} = \sqrt {x + 2} \,\,\left( {x \ge – 2} \right) \cr
& \Leftrightarrow {{3x + 15 – 4x – 17} \over {\sqrt {3x + 15} + \sqrt {4x + 17} }} = \sqrt {x + 2} \cr
& \Leftrightarrow {{x + 2} \over {\sqrt {3x + 15} + \sqrt {4x + 17} }} + \sqrt {x + 2} = 0 \cr
& \Leftrightarrow \sqrt {x + 2} \left( {{{\sqrt {x + 2} } \over {\sqrt {3x + 15} + \sqrt {4x + 17} }} + 1} \right) = 0 \cr
& \Leftrightarrow x + 2 = 0 \Leftrightarrow x = – 2\,\,\left( {tm} \right) \cr
& Vay\,\,x = – 2 \cr
& c)\,\,\sqrt {x – 2} + \sqrt {3 + x} = \sqrt {2x + 1} \,\,\left( {x \ge 2} \right) \cr
& \Leftrightarrow x – 2 + 3 + x + 2\sqrt {\left( {x – 2} \right)\left( {3 + x} \right)} = 2x + 1 \cr
& \Leftrightarrow 2\sqrt {\left( {x – 2} \right)\left( {3 + x} \right)} = 0 \cr
& \Leftrightarrow \left[ \matrix{
x = 2\,\,\left( {tm} \right) \hfill \cr
x = – 3\,\,\left( {ktm} \right) \hfill \cr} \right. \cr
& Vay\,\,x = 2 \cr
& d)\,\,\sqrt {2x – 1} = \sqrt {x – 1} + \sqrt {x – 4} \,\,\left( {x \ge 4} \right) \cr
& \Leftrightarrow 2x – 1 = x – 1 + x – 4 + 2\sqrt {\left( {x – 1} \right)\left( {x – 4} \right)} \cr
& \Leftrightarrow \sqrt {\left( {x – 1} \right)\left( {x – 4} \right)} = 2 \cr
& \Leftrightarrow \left( {x – 1} \right)\left( {x – 4} \right) = 4 \cr
& \Leftrightarrow {x^2} – 5x + 4 = 4 \cr
& \Leftrightarrow {x^2} – 5x = 0 \Leftrightarrow \left[ \matrix{
x = 0\,\,\left( {ktm} \right) \hfill \cr
x = 5\,\,\left( {tm} \right) \hfill \cr} \right. \cr
& Vay\,\,x = 5 \cr} $$
Đáp án:
Giải thích các bước giải: c) Đặt $\sqrt[]{x-2}$ =a, $\sqrt[]{3+x}$ =b, $\sqrt[]{2x+1}$ =c.
=> $\left \{ {{a+b=c} \atop {a^2+b^2=c^2}} \right.$ .
=> ab+bc+ca=0.
=> a=b=c=0.
Giải tiếp là ra x.