ĐK : .... . .
Đặt \(\sqrt{x+3}=u;\sqrt{3-2x}=v\)
Ta có hệ phương trình \(\int^{x+u^2+v^2=x+x+3+3-2x=6}_{x+4u+2v=11}\Rightarrow u^2-4u+v^2-2v=-5\)
<=> \(\left(u-2\right)^2+\left(v-1\right)^2=0\)
<=> u = 2 và v = 1
<=> x = 1
GPT
:\(x+4\sqrt{x+3}+2\sqrt{3-2x}=11\)
GPT :
\(\sqrt[4]{2x^3+x^2-2x+8}+\sqrt[4]{3x^3+x^2-2x+10}=2\sqrt[4]{x^2-2x+4}\)
gpt:\(\sqrt{3x^2+6x+4}+\sqrt{2x^2+4x+11}=\left(1-x\right)\left(x+3\right)\)
\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-x^2-2x\)
\(\sqrt{x^2-x+2}+\sqrt{x^2-3x+6}=2x\)
GPT: \(\sqrt{2x+3-\sqrt{x+2}}+\sqrt{2x+4+\sqrt{x+2}}=1+2\sqrt{x+2}\)
GPT :
\(x^2+\sqrt[3]{x^4-x^2}=2x+1\)
GPT :
\(x^2+\sqrt[3]{x^4-x^2}=2x+1\)
GPT : \(\sqrt{\left|x+1\right|}+\sqrt[4]{x^2-2x+5}=2\sqrt[4]{x^2+3}\)
GPT:\(\sqrt{1-x}+\sqrt{4+x}=3...\)
\(x^2+4x+5=2\sqrt{2x+3}\)
GPT
\(\sqrt[]{2x+3+\sqrt{x+2}}+\sqrt[]{2x+2-\sqrt{2+x}}=1+2\sqrt{x+2}\)
