# [Toán 10] Giải phương trình

E

#### eye_smile

a,Ta có:
$\left( {x - 7} \right)\sqrt {{x^2} - 5x + 4} = 2x - 14\left( {x \ge 4/hoac/x \le 1} \right)$
$\leftrightarrow \left( {x - 7} \right)\left( {2 - \sqrt {{x^2} - 5x + 4} } \right) = 0$
$\leftrightarrow \left[ \begin{array}{l} x = 7 \\ {x^2} - 5x + 4 = 4 \\ \end{array} \right.$
$\leftrightarrow \left[ \begin{array}{l} x = 7 \\ x = 5 \\ x = 0 \\ \end{array} \right.\left( {thoa - man} \right)$
b,Đặt ${x^2} + x = t$
$\to pt \leftrightarrow \sqrt {t + 4} + \sqrt {t + 1} = \sqrt {2t + 9} \left( {t \ge - 1} \right)$
$\leftrightarrow 2t + 5 + 2\sqrt {\left( {t + 4} \right)\left( {t + 1} \right)} = 2t + 9$
$\leftrightarrow 2\sqrt {{t^2} + 5t + 4} = 4$
$\leftrightarrow {t^2} + 5t + 4 = 4$
$\leftrightarrow t\left( {t + 5} \right) = 0$
$\leftrightarrow \left[ \begin{array}{l} t = 0 \\ t = - 5 \\ \end{array} \right. \leftrightarrow \left[ \begin{array}{l} {x^2} + x = 0 \\ {x^2} + x + 5 = 0 \\ \end{array} \right. \leftrightarrow \left[ \begin{array}{l} x = 0 \\ x = - 1 \\ \end{array} \right.\left( {thoa - man} \right)$

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T

#### tranvanhung7997

$c. \sqrt[]{x + 3 - 4\sqrt[]{x - 1}} + \sqrt[]{x + 8 - \sqrt[]{x - 1}} = 1$
Điều kiện: $x \ge 1$
PT <=> $\sqrt[]{(\sqrt[]{x - 1} - 2)^2} + \sqrt[]{(\sqrt[]{x - 1} - 3)^2} = 1$
<=> $|\sqrt[]{x - 1} - 2| + |3 - \sqrt[]{x - 1}| = 1$
Ta có BĐT: $|x| + |y| \ge |x + y|$. Dấu = có $x.y \ge 0$
Áp dụng ta được: $|\sqrt[]{x - 1} - 2| + |3 - \sqrt[]{x - 1}| \ge 1$
Dấu = có: $(\sqrt[]{x - 1} - 2)(3 - \sqrt[]{x - 1}) \ge 0$
<=> $2 \le \sqrt[]{x - 1} \le 3$
<=> $5 \le x \le 10$ T/m
Vậy tập nghiệm PT là: $T = [5 ; 10]$

V

#### vansang02121998

$\sqrt{x^2+x+4}+\sqrt{x^2+x+1}=\sqrt{2x^2+2x+9}$

$\Leftrightarrow (\sqrt{x^2+x+4}+\sqrt{x^2+x+1})^2=\sqrt{2x^2+2x+9}^2$

$\Leftrightarrow 2x^2+2x+5+2\sqrt{(x^2+x+4)(x^2+x+1)}=2x^2+2x+9$

$\Leftrightarrow (x^2+x+4)(x^2+x+1)=4$

$\Leftrightarrow (x^2+x+1)^2+3(x^2+x+1)-4=0$

$\Leftrightarrow (x^2+x+1)^2-(x^2+x+1)+4(x^2+x+1)-4=0$

$\Leftrightarrow (x^2+x)(x^2+x+5)=0$

$\Leftrightarrow x(x+1)(x^2+x+5)=0$