ĐKXĐ: x>=\(-\frac{1}{3}\)
PT \(\Leftrightarrow x^2+2x-3=\sqrt{2x+2}-\sqrt{3x+1} \)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=\frac{2x+2-3x-1}{\sqrt{2x+2}+\sqrt{3x+1}}\Leftrightarrow\left(x-1\right)\left(x+3+\frac{1}{\sqrt{2x+2}+\sqrt{3x+1}}\right)=0\)
dễ nhận thấy x=1 là nghiệm duy nhất của pt
gpt:
\(\sqrt{2x^2-1}+\sqrt{x^2-3x-2}=\sqrt{2x^2+2x+3}+\sqrt{x^2-x+2}\)
GPT
\(x^2+6x+1=\left(2x+1\right)\sqrt{x^2+2x+3}\)3
\(\sqrt{3x-2}-\sqrt{x-1}=2x^2-x-3\)
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 bằng pp đặt ẩn phụ k hoàn toàn:
1, \(2\sqrt{2x+4}+4\sqrt{2-x}=\sqrt{9x^2+16}\)
2, \(\left(3x+2\right)\sqrt{2x-3}=2x^2+3x-6\)
GPT :
\(3\sqrt{x^3+8}=2x^2-3x+10\)
GPT : \(\sqrt{3x^2+3x}+\sqrt{x-x^2}=2x+1\)
GPT
A,\(\sqrt{X+1}+\sqrt{4-X}+\sqrt{4+3X-X^2=5}\)
B,\(\sqrt{X^2-2X+5}=2-\left(X^2-1\right)^4\)
C,\(\left(X-5\right)^3=5\sqrt[3]{2X-9}-3X+6\)
D,\(\left(2X+1\right)\sqrt{\frac{X+1}{X}}=X+2+\sqrt[3]{2X^2+X^3}\)
MN ƠI GIÚP E MAI E ĐI HOK RỒI
ETICH CHO AI TRẢ LỜ NHANH NHẤT
GPT: \(x^3+5x^2+2x=3\left(x+1\right)\sqrt{3x+2}.\)
