Em kiểm tra lại đề bài, chỗ \(2x+x\) bên vế phải
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26 tháng 11 2021 lúc 21:08
GPT:
1, \(6x^2+10x-92+\sqrt{\left(x+70\right)\left(2x^2+4x+16\right)}=0\)
2,\(x+3+\sqrt{1-x^2}=3\sqrt{x+1}+\sqrt{1-x}\)
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
\(x^2+6x+1=\left(2x+1\right)\sqrt{x^2+2x+3}\)3
\(\sqrt{3x-2}-\sqrt{x-1}=2x^2-x-3\)
9 tháng 11 2021 lúc 14:53
Gpt: \(\sqrt{x+5}+\sqrt{3-x}-2\left(\sqrt{15-2x-x^2}+1\right)=0\)
\(\sqrt{\left(x-2\right)\left(x+3\right)}=5\)
\(\sqrt{\left(2x+3\right)^2}=x-5\)
\(\sqrt{x^2-6x+9}=x+7\)
\(\sqrt{2x-3}=x-1\)
GPT : \(\sqrt{\left|x+1\right|}+\sqrt[4]{x^2-2x+5}=2\sqrt[4]{x^2+3}\)
GPT: \(2x^2+2x+1=\left(2x+3\right)\left(\sqrt{x^2+x+2}-1\right)\)
\(\left\{{}\begin{matrix}6x^2\sqrt{x^3-6x+5}=\left(x^2+2x-6\right)\left(x^3+4\right)\\x+\dfrac{2}{x}=1+\dfrac{2}{y^2}\end{matrix}\right.\)
\(\left(5\right)\sqrt{x+3-4\sqrt{x-1}}\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\left(6\right)2x^2+3x+\sqrt{2x^2+3x+9}=33\)
\(\left(7\right)\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
\(\left(8\right)x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)