giải phương trình
a, 2 $\sqrt[]{x+3}$ + $\sqrt[]{9x+27}$ =5 -$\frac{1}{3}$$\sqrt[]{x+3}$
b, $\sqrt[]{x-2\sqrt[]{x-1}}$ =2
giải phương trình a, 2 $\sqrt[]{x+3}$ + $\sqrt[]{9x+27}$ =5 -$\frac{1}{3}$$\sqrt[]{x+3}$ b, $\sqrt[]{x-2\sqrt[]{x-1}}$ =2
By Valentina
$@phamnhuy6a1$
$@gaumatyuki$
$a)$ $2$ $\sqrt[]{x+3}$ + $\sqrt[]{9x+27}$ = $5-$ $1/3$ $\sqrt[]{x+3}$
⇔ $2$ $\sqrt[]{x+3}$ + $\sqrt[]{3².(x+3)}$+ $1/3$ $\sqrt[]{x+3}$ = $5$
⇔ $2$ $\sqrt[]{x+3}$ + $3$ $\sqrt[]{x+3}$+ $1/3$ $\sqrt[]{x+3}$ = $5$
⇔ $(2+3+1/3)$. $\sqrt[]{x+3}$ =$5$
⇔ $16/3$. $\sqrt[]{x+3}$ =$5$
⇔ $\sqrt[]{x+3}$ = $15/16$
⇔ $x+3=225/256$
⇔$x= -543/256$
$Vậy$ $pt$ $có$ $nghiệm$ $x=-543/256$
$b)$ $\sqrt[]{x-2\sqrt[]{x-1}}$ $=2$
⇔ $\sqrt[]{(x-1)-2\sqrt[]{x-1}+1}$ $=2$
⇔ $\sqrt[]{(\sqrt[]{x-1}-1)²}$ $=2$
⇔$\sqrt[]{x-1}$ $-1)²$ $=2$
⇔ $\sqrt[]{x-1}$ $-1=2$
⇔ $\sqrt[]{x-1}$ $=1$
⇔ $x-1=1$
⇔$x=2$
$Vậy$ $pt$ $có$ $nghiệm$ $x=2$
$Chúc$ $bạn$ $học$ $tốt!$
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