b) ĐK : $-2\leq x\leq 1$
$\sqrt{1-x}-\sqrt{2+x}=1 \\ \Leftrightarrow 1-x + 2+x -2\sqrt{(1-x)(2+x)}=1 \\ \Leftrightarrow (1-x)(2+x)=1$
c)
$\sqrt{(5-2\sqrt{6})^x}+\sqrt{(5+2\sqrt{6})^x}=10 \\ \Leftrightarrow \sqrt{(\sqrt{3}-\sqrt{2})^{2x}}+\sqrt{(\sqrt{3}+\sqrt{2})^{2x}}=10 \\ \Leftrightarrow (\sqrt{3}-\sqrt{2})^x + (\sqrt{3}+\sqrt{2})^x=10$
Xét thấy $x=4$ là nghiệm của pt
d)
$\sqrt[3]{x-1}+\sqrt[3]{x-2}=\sqrt[3]{2x-3}\\\Leftrightarrow x-1+x-2+3\sqrt[3]{x-1}\sqrt[3]{x-2}(\sqrt[3]{x-1}+\sqrt[3]{x-2})=2x-3\\\Leftrightarrow \sqrt[3]{x-1}\sqrt[3]{x-2}(\sqrt[3]{x-1}+\sqrt[3]{x-2})=0$
f)
$\sqrt{x+3+4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=5 \\ \sqrt{(\sqrt{x-1}+2)^2}+\sqrt{(\sqrt{x-1}-3)^2}=5 $
g)
$x^2+4x+5=2\sqrt{2x+3}\\\Leftrightarrow x^2+4x+5+2x+4=2x+3+2\sqrt{2x+3}+1 \\ \Leftrightarrow (x+3)^2=(\sqrt{2x+3}+1)^2$
h)
$\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\dfrac{1}{2}(x+y+z)\\\Leftrightarrow x+y+z=2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}\\\Leftrightarrow x-2\sqrt{x}+1+y-1-2\sqrt{y-1}+1+z-2-2\sqrt{z-2}+1=0\\ \Leftrightarrow (\sqrt{x}-1)^2+(\sqrt{y-1}-1)^2+(\sqrt{z-2}-1)^2=0$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
